MỌI NGƯỜI GIÚP EM VỚIBài 1: tìm xa)$\left|3x-5\right|=4$b)$\dfrac{x 1}{10} \dfrac{x 1}{11} \dfrac{x 1}{12}=\dfrac{x 1}{13} \dfrac{x 1}{14}$c)\(\dfrac{x 4}{2000} \dfrac{x 3}{2001}=\dfrac{x 2}{2002}...

AH

ANH HOÀNG

18 tháng 9 2021

MỌI NGƯỜI GIÚP EM VỚIBài 1: tìm xa)$\left|3x-5\right|=4$b)$\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$c)$\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$Bài 2:...

Đọc tiếp

#Toán lớp 7

2

LL

Lấp La Lấp Lánh

18 tháng 9 2021

Bài 1:

a) $\left|3x-5\right|=4$

$\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.$

c) $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$

$\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0$

$\Leftrightarrow x=-2004$( do $\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0$)

Bài 2:

a) $=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}$

$=\dfrac{1}{4}+\dfrac{3}{4}=1$

b) $=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)$

$=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)$

$=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}$

Đúng(1)

EC

Edogawa Conan

18 tháng 9 2021

Bài 1:

a) $\left|3x-5\right|=4$ (1)

$\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.$

b) $\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

$\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0$

$\Leftrightarrow x+1=0$ $\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)$

$\Leftrightarrow x=-1$

c) $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)$

$\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$

$\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0$

$\Leftrightarrow x+2004=0$ $\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)$

$\Leftrightarrow x=-2004$

Đúng(1)

LQ

Lê Quang Dũng

13 tháng 7 2017

13 Tìm x, biết :

a) $\dfrac{2}{3}x+4=-12$; b) $\dfrac{3}{4}+\dfrac{1}{4}:x=-3$; c) $\left|3x-5\right|=4$

d) $\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$ ;

e) $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

#Toán lớp 7

1

MS

Mashiro Shiina

13 tháng 7 2017

Các câu dễ tự làm :v

$\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

$\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$

$\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0$

$\Rightarrow x+1=0\Rightarrow x=-1$

$\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1$

$\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$

$\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$

$\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0$

$\Rightarrow x+2004=0\Rightarrow x=-2004$

Đúng(0)

NB

Nguyễn Bá Hải

17 tháng 8 2017

Tìm x:

a) $\dfrac{\left(x+1\right)}{10}+\dfrac{\left(x+1\right)}{11}+\dfrac{\left(x+1\right)}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

b) $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

#Toán lớp 7

2

TT

Trần Thị Hương

17 tháng 8 2017

a. $\dfrac{\left(x+1\right)}{10}+\dfrac{\left(x+1\right)}{11}+\dfrac{\left(x+1\right)}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

$\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$

$\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0$

Vì $\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0$

$\Rightarrow x+1=0$

$x=-1$

b, $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\\ \left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\\ \dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\\ x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)$

vì $\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\\ \Rightarrow x+2004=0\\ x=-2004$

Đúng(0)

DH

Đức Hiếu

17 tháng 8 2017

a, $\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

$\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$

$\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0$

Vì $10< 11< 12< 13< 14$ nên $\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}>\dfrac{1}{14}$

$\Rightarrow\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0$

$\Rightarrow x+1=0\Rightarrow x=-1$

Vậy.................

b, $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)$

$\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$

$\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$

$\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0$

Vì $2000< 2001< 2002< 2003$ nên $\dfrac{1}{2000}>\dfrac{1}{2001}>\dfrac{1}{2002}>\dfrac{1}{2003}$

$\Rightarrow\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}>0$

$\Rightarrow x+2004=0\Rightarrow x=-2004$

Vậy.................

Chúc bạn học tốt!!!

Đúng(0)

AN

Ánh Nắng Ban Mai

21 tháng 8 2017

Tìm số hữu tỉ x, biết rằng:

a. $\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

b. $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

#Toán lớp 7

3

MS

Mashiro Shiina

21 tháng 8 2017

$\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

$\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$

$\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0$

$\Rightarrow x+1=0\Rightarrow x=-1$

$\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1$

$\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$

$\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$

$\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0$

$\Rightarrow x+2004=0\Rightarrow x=-2004$

Đúng(0)

NH

Nguyễn Huy Tú

21 tháng 8 2017

a, $\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

$\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$

$\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0$

Do $\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0$

$\Rightarrow x+1=0\Rightarrow x=-1$

Vậy x = -1

b, $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$

$\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0$

Vì $\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0$

$\Rightarrow x+2004=0\Rightarrow x=-2004$

Vậy...

Đúng(0)

AD

Ánh Dương

27 tháng 2 2019

Giải các phương trình sau:

a) $\dfrac{15-x}{2000}+\dfrac{14-x}{2001}=\dfrac{13-x}{2002}+\dfrac{12-x}{2003}$

b) $\dfrac{x-5}{2010}+\dfrac{x-4}{2011}=\dfrac{x-2010}{5}+\dfrac{x-2011}{4}$

c) $\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0$

giúp mk vs

#Toán lớp 8

2

QD

Quốc Đạt

27 tháng 2 2019

a) $\dfrac{15-x}{2000}+\dfrac{14-x}{2001}=\dfrac{13-x}{2002}+\dfrac{12-x}{2003}$

$\Leftrightarrow\dfrac{15-x}{2000}+1+\dfrac{14-x}{2001}+1=\dfrac{13-x}{2002}+1+\dfrac{12-x}{2003}+1$

$\Leftrightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}=\dfrac{2015-x}{2002}+\dfrac{2015-x}{2003}$

$\Rightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}-\dfrac{2015-x}{2002}-\dfrac{2015-x}{2003}=0$

$\Leftrightarrow\left(2015-x\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0$

$\Leftrightarrow2015-x=0$

<=> x=2015

Vậy phương trình có nghiệm là x=2015

Đúng(0)

QD

Quốc Đạt

27 tháng 2 2019

b) $\dfrac{x-5}{2010}+\dfrac{x-4}{2011}=\dfrac{x-2010}{5}+\dfrac{x-2011}{4}$

$\Leftrightarrow\dfrac{x-5}{2010}-1+\dfrac{x-4}{2011}-1=\dfrac{x-2010}{5}-1+\dfrac{x-2011}{4}-1$

$\Leftrightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}=\dfrac{x-2015}{5}+\dfrac{x-2015}{4}$

$\Rightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}-\dfrac{x-2015}{5}-\dfrac{x-2015}{4}=0$

$\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0$

$\Leftrightarrow x-2015=0$

=> x=2015

Vậy phương trình có nghiệm x=2015

Đúng(0)

TN

Trần Ngọc

26 tháng 6 2017

1.tìm số hữu tỉ x : a/|3x-5|=4 b/ $\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$ c/$\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$ 2/tìm số nguyên n để ps sau đây có giá trị nguyên và tính: A=$\dfrac{3n+9}{n-4}$ 3/tìm các số nguyên...

Đọc tiếp

#Toán lớp 7

1

Q

qwerty

26 tháng 6 2017

Bài 1:

a) $\left|3x-5\right|=4$

$\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.$

Vậy $x_1=\dfrac{1}{3};x_2=3$

b) $\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

cho đáp án tự làm (vì cách lm của mik bị ném đá khá nhiều lần òi :D)

$x=-1$

c) như câu b nhé :D

$x=-2004$

Đúng(0)

TN

Trần Ngọc

26 tháng 6 2017

thank bạn!!

Đúng(0)

QN

quan nguyen hoang

13 tháng 2 2022

$a,\left(\dfrac{37}{9}+\dfrac{13}{4}\right)x\dfrac{9}{4}+\dfrac{11}{4}$ b,$1+\left(\dfrac{9}{10}-\dfrac{-4}{5}\right):\dfrac{19}{6}$

c,$\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}x\dfrac{12}{5}$

Giúp mik nha:>

#Toán lớp 6

1

NL

Nguyễn Lê Phước Thịnh

13 tháng 2 2022

a: $=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}$

b: $=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}$

c: $=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}$

Đúng(1)

TT

Trần Thị Tú Anh 8B

1 tháng 2 2019

bài 1 : Giải cac phuong trinh sau a, $\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}$ b, $\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}$ c, $2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)$ d, $\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5$ e, $\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}$ g, $\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}$ Các bạn ơi ! giúp mik với !!! chiều nay học...

Đọc tiếp

#Toán lớp 8

2

SS

Songoku Sky Fc11

1 tháng 2 2019

a)MTC 15

$\dfrac{\left(x-3\right)\times3}{15}=\dfrac{6.15-\left(1-2x\right)\times5}{15}=\dfrac{3x-9}{15}=\dfrac{90-5-10x}{15}=3x-9=90-5-10x\Leftrightarrow3x+10x=90-5+9$

Đúng(0)

KH

Kiêm Hùng

1 tháng 2 2019

Chưa nghỉ tết à :))

$a,\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}$

$\Rightarrow3\left(x-3\right)=6.15-5\left(1-2x\right)$

$\Leftrightarrow3x-9=90-5+10x$

$\Leftrightarrow3x-10x=90-5+9$

$\Leftrightarrow-7x=94$

$\Leftrightarrow x=-\dfrac{94}{7}$

Vậy.....

$b,\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}$

$\Rightarrow2\left(3x-2\right)-5.12=3\left[3-2\left(x+7\right)\right]$

$\Leftrightarrow6x-4-60=-6x-33$

$\Leftrightarrow6x+6x=-33+60+4$

$\Leftrightarrow12x=31$

$\Leftrightarrow x=\dfrac{31}{12}$

Vậy.....

$c,2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)$

$\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x$

$\Leftrightarrow2x+x=5-\dfrac{13}{5}-\dfrac{6}{5}$

$\Leftrightarrow3x=\dfrac{6}{5}$

$\Leftrightarrow x=\dfrac{2}{5}$

Vậy.....

$d,\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5$

$\Rightarrow28\left[5\left(x-1\right)+2\right]-42\left(7x-1\right)=24\left[2\left(2x+1\right)\right]-5.168$

$\Leftrightarrow140x-84-294x+42=96x+48-840$

$\Leftrightarrow140x-294x-96x=48-840-42+84$

$\Leftrightarrow-250x=-750$

$\Leftrightarrow x=3$

Vậy.....

$e,\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}$

$\Rightarrow6\left(x-1\right)+3\left(x-1\right)=12-4\left[2\left(x-1\right)\right]$

$\Leftrightarrow6x-6+3x-3=12-8x+8$

$\Leftrightarrow6x+3x+8x=12+8+3+6$

$\Leftrightarrow17x=29$

$\Leftrightarrow x=\dfrac{29}{17}$

Vậy.....

$g,\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}$

$\Leftrightarrow\dfrac{2}{2001}-\dfrac{x}{2001}-1=\dfrac{1}{2002}-\dfrac{x}{2002}-\dfrac{x}{2003}$

$\Leftrightarrow-\dfrac{x}{2001}+\dfrac{x}{2002}+\dfrac{x}{2003}=\dfrac{1}{2002}+1-\dfrac{2}{2001}$

$\Leftrightarrow x\left(-\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\right)=1+\dfrac{1}{2002}-\dfrac{2}{2001}$

$\Leftrightarrow x=\dfrac{\left(1+\dfrac{1}{2002}-\dfrac{2}{2001}\right)}{\left(-\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\right)}=2003$

Vậy.....

Đúng(0)

NQ

Nguyễn Quý Trung

25 tháng 7 2021

Cần giúp nhanh vsBài 1. Tìm xa) $\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}$b) $\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}$c) $3x.\left(x+\dfrac{2}{3}\right)=0$d) $x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)$Bài 2. Tính...

Đọc tiếp

#Toán lớp 7

5

AH

Akai Haruma

Giáo viên

25 tháng 7 2021

Bài 1:

a.

$|x+\frac{7}{4}|=\frac{1}{2}$

$\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.$

b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$

$|2x+1|=\frac{11}{15}$

$\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.$

c.

$3x(x+\frac{2}{3})=0$

$\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.$

d.

$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$

$\Leftrightarrow x=\frac{2}{5}$

Đúng(2)

AH

Akai Haruma

Giáo viên

25 tháng 7 2021

Nguyễn Quý Trung:

$x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}$

Bạn bớt 2 vế đi 1/3 thì $x=\dfrac{2}{5}$

Đúng(2)

NN

Nguyễn Nhi

12 tháng 8 2017

Giải PT

a) x⁴ = 4x + 1

b) x² = $\dfrac{4x^2}{(x+2\left(\right)^{ }2}$ = 12

Bài 2: Giải PT

$\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}$

#Toán lớp 8

2

LP

Linh Phương

12 tháng 8 2017

Mở đầu về phương trình

Đúng(1)

MV

Mới vô

12 tháng 8 2017

2.

$\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0$

Vì $\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0$

$\Rightarrow x>0$

Vậy $x>0$

Đúng(1)

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP