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...................

x< -7/4(vô lí ) vì sao bạn

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

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

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

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

\(x=7\)

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

\(x=3-4\)

\(x=-1\)

a)

\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)

b)

\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)

tiếp đi bạn

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

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

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

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

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

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

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

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

T giải thử thôi nhé :w

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

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

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

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

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

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

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

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

Lập bảng xét dấu:

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 ....

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

c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)

=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)

=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)

=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)

d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)

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

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

=> x= \(\dfrac{1}{6}\)

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây