`a)|2x-15|=13`

`**2x-15=13`

`<=>2x=28`

`<=>x=14.`

`**2x-15=-13`

`<=>2x=-2`

`<=>x=-1.`

`b)|7x+3|=66`

`**7x+3=66`

`<=>7x=63`

`<=>x9`

`**7x+3=-66`

`<=>7x=-69`

`<=>x=-69/7`

`c)|5x-2|=0`

`<=>5x-2=0`

`<=>5x=2`

`<=>x=2/5`

\(a,\Leftrightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

Vậy ...

\(b,\Leftrightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)

Vậy ...

\(c,\Leftrightarrow5x-2=0\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy ...

a \(\Rightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x=18\\2x=-8\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

b \(\Rightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}7x=63\\7x=-69\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)

c \(\Rightarrow5x-2=0\Rightarrow x=\dfrac{2}{5}\)

\(a,\Rightarrow\left(35x+3\right)\cdot19=152\\ \Rightarrow35x+3=8\\ \Rightarrow x=\dfrac{1}{7}\\ b,\Rightarrow3\left(x+7\right)=42\\ \Rightarrow x+7=14\Rightarrow x=7\\ c,\Rightarrow3\left(x+1\right)=48\\ \Rightarrow x+1=16\Rightarrow x=15\\ d,\Rightarrow120-5x+100\cdot2:5=4\cdot15\\ \Rightarrow120-5x+40=60\\ \Rightarrow5x=100\Rightarrow x=20\\ e,\Rightarrow4x-10=30\\ \Rightarrow4x=40\\ \Rightarrow x=10\\ g,\Rightarrow10x+10=70\\ \Rightarrow10x=60\\ \Rightarrow x=6\)

a) Ta có : x=0 không là nghiệm của phương trình.          Chia cả hai vế của phương trình cho \(^{x^2}\) ta có:

    \(x^2-2x-1-\frac{2}{x}+\frac{1}{x^2}=0\) \(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)-1=0\)  (1)

Đặt \(x+\frac{1}{x}=t\)  \(\left(t>2\right)\) hoăc \(\left(t

\(a,4\left(x-3\right)=7^2-1^3\)

\(4\left(x-3\right)=49-1\)

\(4\left(x-3\right)=48\)

\(\Rightarrow x-3=48:4\)

\(x-3=24\)

\(\Rightarrow x=27\)

\(5x+x=\frac{150}{2}-3\)

\(6x=75-3\)

\(6x=72\)

\(\Rightarrow x=12\)

a, 4( x - 3) = 7² - 1³

<=> 4x - 12 = 49 - 1

<=> 4x = 49 - 1 +12

<=> x = 15

Vậy....

b, 5x + x = \(\frac{150}{2}\)+ 3 

<=> 5x + x = 75 + 3

<=> 6x = 78

<=> x = 13

Vậy......

`a)sqrt{5x-2}=3(x>=2/5)`

`<=>5x-2=9`

`<=>5x=11`

`<=>x=11/5(tm)`

`b)sqrt{x^2-4x+4}-5=0`

`<=>\sqrt{(x-2)^2}=5`

`<=>|x-2|=5`

`<=>` \(\left[ \begin{array}{l}x-2=5\\x-2=-5\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=7\\x=-3\end{array} \right.\) 

`c)3sqrt{4x+8}-sqrt{9x+18}+9sqrt{(x+2)/9}=sqrt{72}(x>=-2)`

`<=>6sqrt{x+2}-3sqrt{x+2}+3sqrt{x+2}=sqrt{72}`

`<=>6sqrt{x+2}=6sqrt2`

`<=>sqrt{x+2}=sqrt2`

`<=>x+2=2`

`<=>x=0(tm)`

\(a,ĐK:x\ge\dfrac{2}{5}\)

\(\Leftrightarrow5x-2=9\)

\(\Leftrightarrow5x=11\)

\(\Leftrightarrow x=\dfrac{11}{5}\)

\(b,\)

\(\Leftrightarrow x^2-5x+4=25\)

\(\Leftrightarrow x^2-5x-21=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{109}}{2}\\x=\dfrac{5-\sqrt{109}}{2}\end{matrix}\right.\)

\(c,\)

\(\Leftrightarrow6\sqrt{x+2}-3\sqrt{x+2}+9\cdot\sqrt{\dfrac{x+2}{9}}=6\sqrt{2}\)

\(\Leftrightarrow2\sqrt{x+2}-\sqrt{x+2}+3\cdot\sqrt{\dfrac{x+2}{9}}=2\sqrt{2}\)

Đặt \(\sqrt{x+2}=a\) ta có (1)

\(2a-a+3\cdot\dfrac{a}{\sqrt{9}}=2\sqrt{2}\)

\(\Leftrightarrow a+3\cdot\dfrac{a}{3}=2\sqrt{2}\)

\(\Leftrightarrow2a=2\sqrt{2}\)

\(\Leftrightarrow a=\sqrt{2}\)

Thay \(a=\sqrt{2}\) vào (1) ta có

\(\sqrt{x+2}=\sqrt{2}\)

\(\Leftrightarrow x+2=2\)

\(\Leftrightarrow x=0\)