tìm x a, x+1/10 + x+1/11 + x+1/12 = x+1/13 + x+1/14 b, x+4/2000 + x+3/2001 = x+2/2002 + x+1/2003

G

gokubluessj1

5 tháng 6 2017 - olm

Tìm x, biết :

a) x+1/10 + x+1/11 + x+1/12 = x+1/13 + x+1/14

b) x+3/2000 + x+3/2001 = x+2/2002 + x+2/2003

#Toán lớp 7

3

LI

love is a beautiful pain

5 tháng 6 2017

Sorry mink mới lớp 5 nên ko thể giúp bn lm bài toán này thành thật xin lỗi

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TT

Thanh Tùng DZ

5 tháng 6 2017

a) $\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$

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

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

Dễ thấy $\frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}$nên biểu thức trong ngoặc thứ hai $\ne$0

Do đó $x+1=0$$\Rightarrow x=0-1=-1$

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

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

$\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}$

$\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+4}{2003}=0$

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

Vì $\frac{1}{2000}>\frac{1}{2001}>\frac{1}{2002}>\frac{1}{2003}$nên biểu thức trong ngoặc thứ hai phải $\ne$0

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

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D

DD

17 tháng 6 2021 - olm

Tìm x:

x+1/10 + x+1/11 + x+1/12 = x+1/13 + x+1/14

x+4/2000 + x+3/2001 = x+2/2002 + x+1/2003

#Toán lớp 7

2

TV

Trần Viết Phương

17 tháng 6 2021

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

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

$\Rightarrow (x + 1) (\frac{1}{10} + 111 + 112 - 113 - 114)$

Vì 10<11<12<13<14 $\Rightarrow 110 > 111 > 112 > 113 > 114$

$\Rightarrow 110 + 111 + 112 - 113 - 114 > 0$

$\Rightarrow x + 1 = 0$

$\Rightarrow x = - 1$

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응 우옌 민 후엔

17 tháng 6 2021

Câu 1:x+1/10 + x+1/11 = x+1/12 + x+1/13 + x+1/14.

<-> (x+1)(1/10+1/11-1/12-1/13-1/14)=0

<-> x+1=0

<-> x=-1

Câu 2:

x+4/2000+x+3/2001=x+2/2002+x

⇔x+4/2000+1+x+3/2001=x+2/2002+1+x+1/2003

⇔x+2004/2000+x+2004/2001=x+2004/2002+x+2004/2003

⇔(x+2004)/(1/2000+1/2001−1/2002−1/2003)=0

⇔x+2004=0

⇔x=-2004

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LT

lê thị trà giang

3 tháng 9 2016 - olm

TFM SỐ HỮU TỶ X:

a) X+1/10 + X+1/11 + X+1/12 =X+1/13 + X+1/14

b) X+4/2000 + X+3/2001 = X+2/2002+ X+1/2003

#Toán lớp 7

4

FF

fan FA

3 tháng 9 2016

a) <=>(x+1)(1/10 + 1/11+1/12) =(x+1)(1/13 + 1/14)
<=>(x+1)(1/10 + 1/11+1/12 -1/13 -1/14)=0
<=> x+1=0(vì biểu thức 1/10 + 1/11 +1/12-1/13-1/14#0)
<=>x= -1

b) (x+4)/2000 + (x+3)/2001 = (x+2)/2002 + (x+1)/2003
<=> (x+4)/2000 + 1 + (x+3)/2001 +1 = (x+2)/2002 + 1 + (x+1)/2003 + 1 (thêm 2 vào mỗi vế )
<=> (x+4+2000)/2000 + (x+3+2001)/2001 = (x+2+2002)/2002 + (x+1+2003)/2003
<=> (x+2004)/2000 + (x+2004)/2001 - (x+2004)/2002 - (x+2004)/2003 = 0 ( chuyển vế )
<=> (x+2004)(1/2000 + 1/2001 - 1/2002 - 1/2003) = 0 ( nhóm hạng tử x + 2004)
vậy biể thức trên bằng 0 tại x+2004 = 0 hoặc 1/2000 + 1/2001 - 1/2002 - 1/2003 = 0
mà ta dễ thấy 1/2000 + 1/2001 - 1/2002 - 1/2003 khác 0
nên biểu thức trên bằng 0 tại x+2004=0
=> x = -2004
vậy S = { -2004}

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LM

Lê Minh Anh

3 tháng 9 2016

a/ $\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$

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

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

Mà: $\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1$

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NP

Nguyễn Phúc Hậu

14 tháng 6 2016 - olm

/3x-5/=4
x+1/10 + x+1/11 + x+1/12 = x+1/13 + x+1/14
x+4/2000 + x+3/2001 = x+2/2002 + x+1/2003

#Toán lớp 7

0

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

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

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

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

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

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

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

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

$\Leftrightarrow x=-2004$

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F

Fʊʑʑʏツ👻

13 tháng 10 2019 - olm

thực hiện phép tính:

a)$\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$

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

#Toán lớp 7

1

TN

Thảo Nguyễn『緑』

13 tháng 10 2019

a) $\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$

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

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

Mà $\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0$

nên x + 1 = 0 => x = -1

Vậy x = -1

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

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

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

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

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

Mà $\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0$

nên 2004 + x = 0 => x = -2004

Vậy x = -2004

=))

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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

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

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

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

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

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

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

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

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

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

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

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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...

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TT

Trần Thị Anh Thư

19 tháng 6 2016 - olm

Tìm số hữu tỉ x, biết:

a) $\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}+\frac{x+1}{13}+\frac{x+1}{14}$

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

#Toán lớp 7

1

LM

Lee Min Ho club

19 tháng 6 2016

sai đề rồi

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KT

Kim Taehyungie

23 tháng 9 2019

$Tìm$ $x,$ $biết$ $:$

$a)\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}+=\frac{x+1}{13}+\frac{x+1}{14}\\ b)\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}$

#Toán lớp 7

3