(5 - \(x\))(9\(x^2\) - 4) =0

\(\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\9x^2=4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x=-\dfrac{2}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x\) \(\in\) { - \(\dfrac{2}{3}\); \(\dfrac{2}{3}\); \(5\)}

7^2\(x\)  + 7^{2\(x\) + 3} = 344

7^2\(x\)  \(\times\) ( 1 + 7³) = 344

7^2\(x\)  \(\times\) (1 + 343) = 344

7^2\(x\)  \(\times\) 344        = 344

7^2\(x\)                    = 344 : 344

7^2\(x\)                  = 1

7^2\(x\)                 =  7⁰

\(2x\)                  = 0

\(x\)                   = 0

Kết luận: \(x\) = 0

|2 - 2\(x\)| - 3,75 = (-0,5)²

|2 - 2\(x\)| - 3,75 = 0,25

|2- 2\(x\)|           =0,25 + 3,75

|2 - 2\(x\)|          = 4

\(\left[{}\begin{matrix}2-2x=-4\left(x>1\right)\\2-2x=4\left(x< 1\right)\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=6\left(x\ge1\right)\\2x=-2\left(x\le1\right)\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Kết luận: \(x\) \(\in\) { -1; 3}

a)

TH1: \(x< \dfrac{-2}{3}\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=-x-\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(2-0,5x+x+\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(c\right)\)

TH2: \(\dfrac{-2}{3}\le x< 4\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(2-0,5x-x-\dfrac{2}{3}=0< =>x=\dfrac{8}{9}\left(c\right)\)

TH3: \(x\ge4\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=0,5x-2\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(0,5x-2-x-\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(l\right)\)

KL: x \(\left\{\dfrac{-16}{3};\dfrac{8}{9}\right\}\)

b) TH1: \(x\ge-1< =>\left|x+1\right|=x+1\)

PT <=> 2x - x -1 = \(\dfrac{-1}{2}\)

<=> x = \(\dfrac{1}{2}\) (c)

TH2: x < -1 <=> \(\left|x+1\right|=-x-1\)

PT <=> 2x + x + 1 = \(\dfrac{-1}{2}\)

<=> x = \(\dfrac{-1}{2}\) (l)

KL: x \(\in\left\{\dfrac{1}{2}\right\}\)

a: =>|x-1/4|=3/4

=>x-1/4=3/4 hoặc x-1/4=-3/4

=>x=1 hoặc x=-1/2

b: \(\left|x+\dfrac{1}{2}\right|=\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{2-9}{4}=-\dfrac{7}{4}\)(vô lý)

c: \(\Leftrightarrow\left[{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{4}{3};-6\right\}\)

e: =>|3/2-x|=0

=>3/2-x=0

hay x=3/2

a: \(M=\dfrac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x+1}\)

b: x thuộc {0;0,5}

=>x=0 hoặc x=0,5

Khi x=0 thì M=1/0+1=1

Khi x=0,5 thì M=1/0,5+1=1/1,5=2/3

=>M min=2/3 và M max=1

a) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)

\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)

\(-\dfrac{2}{3}x=\dfrac{1}{10}\)

x=\(\dfrac{1}{10}:-\dfrac{2}{3}\)

\(x=-\dfrac{3}{20}\)

Vậy \(x=-\dfrac{3}{20}\).

b) \(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)

\(\dfrac{2}{3}:x=-7-\dfrac{1}{3}\)

\(\dfrac{2}{3}:x=-\dfrac{22}{3}\)

\(x=\dfrac{2}{3}:-\dfrac{22}{3}\)

\(x=-\dfrac{1}{11}\)

Vậy \(x=-\dfrac{1}{11}\).

c) \(60\%x=\dfrac{1}{3}\cdot6\dfrac{1}{3}\)

\(60\%x=\dfrac{19}{9}\)

\(\dfrac{3}{5}x=\dfrac{19}{9}\)

\(x=\dfrac{19}{9}:\dfrac{3}{5}\)

\(x=\dfrac{95}{27}\)

Vậy \(x=\dfrac{95}{27}\).

d) \(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\)

\(\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\)

\(\dfrac{2}{3}-x=\dfrac{3}{20}\)

\(x=\dfrac{2}{3}-\dfrac{3}{20}\)

\(x=\dfrac{31}{60}\)

Vậy \(x=\dfrac{31}{60}\).

e) \(-2x-\dfrac{-3}{5}:\left(-0.5\right)^2=-1\dfrac{1}{4}\)

\(-2x-\dfrac{-12}{5}=-1\dfrac{1}{4}\)

\(-2x=-1\dfrac{1}{4}+\dfrac{-12}{5}\)

\(-2x=-\dfrac{73}{20}\)

\(x=-\dfrac{73}{20}:\left(-2\right)\)

\(x=\dfrac{73}{40}\)

Vậy \(x=\dfrac{73}{40}\).

Từ trái qua phải nhé.

\(\begin{matrix}x=0;2\\y=-\dfrac{3}{2};\dfrac{3}{4};3\end{matrix}\)

a.

3x - 2 = 2x - 3

<=> 3x -2x = -3+2

<=> x = -1

Vậy.............

b.

\(5-\left(x-6\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-x+6=12-8x\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

Vậy..........

c.

\(\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)

\(\Leftrightarrow5\left(7x-1\right)+2x.30=6\left(16-x\right)\)

\(\Leftrightarrow35x-5+60x=96-6x\)

\(\Leftrightarrow101x=101\)

\(\Leftrightarrow x=1\)

Vậy ......

a: =>x^2+4x-4x+1=0

=>x^2+1=0

=>Loại

b: =>2x-6+4=2x+2

=>-2=2(loại)

c: =>2(x+3)-2x-1=1

=>6-1=1

=>5=1(loại)

d =>x+3=0

=>x=-3(loại)

e: =>x^2-3x^2+3x-3x-2=0

=>-2x^2-2=0

=>x^2+1=0

=>Loại

a: =>1/3x-2/5x-2/5=0

=>-1/15x=2/5

hay x=-6

b: =>2(x+2)=0,5(2x+1)

=>2x+4=x+0,5

=>x=-3,5

d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)

\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)

\(\Leftrightarrow-x^2-6x=0\)

\(\Leftrightarrow-x\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)

a: Ta có: \(3\left(x-1\right)-3=2\left(x+3\right)\)

\(\Leftrightarrow3x-3-3=2x+6\)

\(\Leftrightarrow x=12\)

b: Ta có: \(\dfrac{x+4}{4}-\dfrac{x+3}{3}=\dfrac{x+6}{6}\)

\(\Leftrightarrow3x+12-4x-12=2x+12\)

\(\Leftrightarrow-3x=12\)

hay x=-4

c: Ta có: \(\left(2x-1\right)^2-x^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

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