\(\left|\left(x+\dfrac{1}{2}\right)\left|2x-\dfrac{3}{4}\right|\right|=2x-\dfrac{3}{4}\) Ai nhan...

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NN

Ngố's Ngố's

10 tháng 11 2018

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

Ai nhanh mị tik

Thím nào là ARMY thì kb với mị

2

DN

Dương Ngọc Nguyễn

10 tháng 11 2018

Violympic toán 7

LQ

Lương Quang Trung

10 tháng 11 2018

Violympic toÃ¡n 7

Các câu hỏi dưới đây có thể giống với câu hỏi trên

PT

Phạm Thị Phương Linh

10 tháng 11 2018 - olm

\(\left|\left(x+\frac{1}{2}\right)\left|2x-\frac{3}{4}\right|\right|=2x-\frac{3}{4}\)

Giúp mị !

Thím nào là ARMY thì kb vs mị

0

TC

Thảo Cỏ

30 tháng 10 2017

Tìm x:

\(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9=5\right|\)

\(\left|x-2\right|+\left|x-3\right|+\left|2x-8\right|=9\\ \left|x+2\right|+\left|x+3\right|+\left|x+1\right|=4\\ \left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.13}\right|+...+\left|x+\dfrac{1}{397.401}\right|=101x\)

1

PT

Phương Trâm

30 tháng 10 2017

\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)

\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)

\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)

\(\Rightarrow x=\dfrac{100}{401}\)

M

mmmm

20 tháng 6 2017

Tìm các số hữu tỷ x,biết rằng:

a,\(\left(x-\dfrac{5}{3}\right):-1\dfrac{3}{4}=0\)

b,\(\left(x-\dfrac{1}{5}\right)\left(1\dfrac{3}{5}+2x\right)=0\)

c,\(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)

d,(2x-3):\(\left(x+1\dfrac{3}{4}\right)< 0\)

1

AT

Aki Tsuki

20 tháng 6 2017

a/dễ --> tự lm

b/ \(\left(x-\dfrac{4}{7}\right)\left(1\dfrac{3}{5}+2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)

Vậy...............

c/ \(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)

TH1: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{4}{7}\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{4}{7}\)

TH2: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)

Vậy \(x>\dfrac{4}{7}\) hoặc \(x< -\dfrac{1}{2}\) thì thỏa mãn đề

d/ \(\left(2x-3\right):\left(x+1\dfrac{3}{4}\right)< 0\)

TH1: \(\left\{{}\begin{matrix}2x-3>0\\x+1\dfrac{3}{4}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1,5\\x< -\dfrac{7}{4}\end{matrix}\right.\)(vô lý)

TH2: \(\left\{{}\begin{matrix}2x-3< 0\\x+1\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< 1,5\\x>-\dfrac{7}{4}\end{matrix}\right.\)\(\Rightarrow-\dfrac{7}{4}< x< 1,5\)

Vậy...................

MT

Mitsuha Taki

11 tháng 9 2017

x< -7/4(vô lí ) vì sao bạn

PL

PhươngAnh Lê

24 tháng 7 2017

Bài 4 : Tìm x,y biết...

2

NH

Nguyễn Huế

24 tháng 7 2017

mình làm lại câu b) nha

b) |x-3|=-4

th1: x-3=-4

x=3+(-4)

x=-1

th2: x-3=4

x=3+4

x=7

NH

Nguyễn Huế

24 tháng 7 2017

b) \(\left|x-3\right|=-4\)

t/h1:\(x-3=-4\)

\(x=3-\left(-4\right)\)

\(x=7\)

t/h2:\(x-3=4\)

\(x=3-4\)

\(x=-1\)

NM

Nguyễn Minh Bảo Anh

1 tháng 7 2017

Tìm x a)\(\left(x+5\right)^3\)=-64 b)\(\left(2x-3\right)^2\)=9 c)\(\dfrac{x}{\dfrac{4}{2}}\)=\(\dfrac{4}{\dfrac{x}{2}}\) d)\(\dfrac{-32}{\left(-2\right)^n}\)=4 e)\(\dfrac{8}{2x}=4\) g)\(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\) h)\(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\) i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\) ...

2

LV

Lê Vương Kim Anh

1 tháng 7 2017

a) ( x + 5 )³= -64

x + 5 = - 4

x = - 4 - 5

x = -9

b) (2x - 3)²=9

2x - 3 = 3

2x = 3+3

2x = 6

x = 6 : 2

x = 3

e) \(\dfrac{8}{2x}=4\)

=> 4 . 2x = 8

8x =8

x = 8 : 8

x = 1

g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}:\left(\dfrac{1}{2}\right)^1=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}:\dfrac{1}{2}=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{8}.\dfrac{1}{2}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{16}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\left(\dfrac{1}{2}\right)^{2.2}\)

=> x = 2

h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)

\(\dfrac{1}{4}.x=\dfrac{1}{32}\)

x = \(\dfrac{1}{32}:\dfrac{1}{4}\)

x = \(\dfrac{1}{8}\)

i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)

\(x=\dfrac{-1}{27}\)

HN

Hải Ngân

2 tháng 7 2017

a) (x + 5)³ = -64

=> (x + 5)³ = (-4)³

x + 5 = -4

x = -4 - 5

x = -9

b) (2x - 3)² = 9

=> (2x - 3)² = (\(\pm\)3)²

=> 2x - 3 = 3 hoặc 2x - 3 = -3

*2x - 3 = 3

2x = 3 + 3

2x = 9

x = \(\dfrac{9}{2}\)

*2x - 3 = -3

2x = -3 + 3

2x = 0

x = 0 : 2

x = 0

Vậy x \(\in\left\{\dfrac{9}{2};0\right\}\)

c) \(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)

=> \(x.\dfrac{x}{2}=4.\dfrac{4}{2}\)

\(\dfrac{x}{2}=8\)

x = 8 : 2

x = 4

d) \(\dfrac{-32}{\left(-2\right)^n}=4\)

\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)

=> (-2)ⁿ . (-2)²= (-2)⁵

(-2)ⁿ = (-2)⁵: (-2)²

(-2)ⁿ = (-2)³

Vậy n = 3

e) \(\dfrac{8}{2x}=4\)

=> 2x . 4 = 8

2x = 8 : 4

2x = 2

x = 1

g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^3\)

2x - 1 = 3

2x = 3 + 1

2x = 4

x = 4 : 2

x = 2

h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)

\(x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\)

\(x=\left(\dfrac{1}{2}\right)^3\)

\(x=\dfrac{1}{8}\)

i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)

\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)\)

\(x=\left(\dfrac{-1}{3}\right)^3\)

\(x=\dfrac{-1}{27}\).

NH

Nguyễn Hoàng Linh

28 tháng 2 2017

Thu gọn các đơn thức :

a) \(\left(\dfrac{a}{2}\right)^33xy\left(4a^2x^3\right)\left(\dfrac{13}{3}ay^2\right)\)

b)\(\left(2x^2y^3z^4\right)^k\left(-\dfrac{1}{2}xy^2\right)^2\)

c) \(\left(\dfrac{7}{3}x^2y^3\right)^{10}\left(\dfrac{3}{7}x^5y^4\right)^{10}\)

3

NH

Nguyễn Hoàng Linh

28 tháng 2 2017

giúp mình với mọi người ơi

N

ngonhuminh

28 tháng 2 2017

a) \(A=\left(\dfrac{1}{2^3}.3.\dfrac{13}{3}\right)\left(a^{3+2+1}\right)\left(x^{1+3}\right)\left(y^{1+2}\right)=\dfrac{13}{8}.a^6.x^4.y^3\)

\(B=\left[2^k.\left(-\dfrac{1}{2}\right)^2\right]\left(x^{2k+2}\right)\left(y^{3k+2.2}\right)\left(z^{4k+}\right)=2^{k-2}.x^{2\left(k+1\right)}.y^{3k+4}.z^{4k}\)

HD

Hòa Đình

25 tháng 1 2018

\(tìmx\)

a,\(4\left|3x-1\right|+\left|x\right|-2\left|x-5\right|+7\left|x-3\right|=12\)

\(b,\left|x+1\right|+\left|x-2\right|+\left|x+3\right|=6\)

\(\left|\left(x+\dfrac{1}{2}\right)\backslash2x-\dfrac{3}{4}\ \right|=2x-\dfrac{3}{4}\)

0

NP

Nguyễn Phúc Nguyên

24 tháng 8 2017

Tìm x biết:

a) \(\left|x-\dfrac{1}{2}\right|+\left|x+2\right|=\dfrac{3}{4}\)

b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\)

c) \(\left|x-1\right|+2\left(x+4\right)=10\)

d) \(\dfrac{11}{12}+\dfrac{11}{12.23}+...+\dfrac{11}{89.100}+x=1\dfrac{2}{3}\)

e) \(\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{19.21}\right)-x+4\dfrac{221}{231}=2\dfrac{1}{3}\)

3

TT

Tiểu Thư họ Nguyễn

24 tháng 8 2017

a) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)

Lập bảng xét dấu:

x		-2		\(\dfrac{1}{2}\)
x + 2	-	0	+		+
x - \(\dfrac{1}{2}\)	-		-	0	+

TH : Xét x < -2

Ta có : - ( x+ 2) - (x - \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)

-x - 2 -x + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)

- 2x - 2 + \(\dfrac{1}{2}\)= \(\dfrac{3}{4}\)

-2x = 2\(\dfrac{1}{4}\)

=> x = \(-1\dfrac{1}{8}\) ( loại )

TH 2: \(-2\le x< \dfrac{1}{2}\)

Ta có : x + 2 + ( -x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)

=> \(2,5=\dfrac{3}{4}\) ( loại )

TH3 : \(x\ge\dfrac{1}{2}\)

x+ 2 + x - \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)

2x + 1,5 = \(\dfrac{3}{4}\)

x = -0,375( loại )

vậy ....

TT

Tiểu Thư họ Nguyễn

24 tháng 8 2017

b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)

c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)

TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)

\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)

TH2 : \(x-1< 0\Rightarrow x< 1\)

=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)

Vậy x = 1

DD

Đức Đạt Đỗ (Đạt 301 Channel)

25 tháng 2 2019 - olm

Bài 1: Thu gọn đơn thức ( a là hằng só )

a) \(1\dfrac{1}{4}x^2y\left(\dfrac{-5}{6}xy\right)^0.\left(-2\dfrac{1}{3}xy\right)\)

b) \(\dfrac{1}{2}x.\dfrac{1}{4}x^2.\dfrac{x^3}{8}.2y.4y^2.x^3\)

c) \(\left(\dfrac{-9}{2}\right)^3.3xy\left(4a^2x^3\right)\left(4\dfrac{1}{3}ay^2\right)\)

d) \(\left(\dfrac{1}{3}xy^2\right)^3\left(\dfrac{-3}{7}x^4y\right)^2\)

1

K

KAl(SO4)2·12H2O

25 tháng 2 2019

T giải thử thôi nhé :w

a) \(1\frac{1}{4}x^2y\left(\frac{-5}{6}xy\right)^0.\left(-2\frac{1}{3}xy\right)\)

\(=\frac{5}{4}x^2y\left(\frac{-5}{6}xy\right)^0.\left(-\frac{5}{2}xy\right)\)

\(=1.\frac{5}{4}x^2y\left(-\frac{5}{2}xy\right)\)

\(=-\frac{5}{4}x^2y.1.\frac{5}{2}xy\)

\(=-1.\frac{5}{4}.\frac{5}{2}x^3y^2\)

\(=-1.\frac{25x^3y^2}{8}\)

\(=-\frac{25x^3y^2}{8}\)