a: \(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a, \(\sqrt{\left(2x-3\right)^2}=7\\ \Rightarrow\left|2x-3\right|=7\\ \Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\sqrt{x+3}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

\(a,\Rightarrow\left(4x-1\right)^2=25=5^2=\left(-5\right)^2\\ \Rightarrow\left[{}\begin{matrix}4x-1=5\\4x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\\ b,\Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:9=16=2^4\Rightarrow x=4\\ c,\Rightarrow3^{2x+3}=3^{2\left(x+3\right)}\\ \Rightarrow2x+3=2x+6\Rightarrow0x=3\left(vô.lí\right)\\ \Rightarrow x\in\varnothing\)

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)

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

f: Ta có: \(2x^2+7x+3=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)

\(\Leftrightarrow-11x=-22\)

hay x=2

b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)

\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)

\(\Leftrightarrow x=-5\)

\(a,\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\\ \Rightarrow26x=26\Rightarrow x=1\\ b,\Rightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\\ \Rightarrow39x=-39\Rightarrow x=-1\)

b: =>3x+9=0 và y^2-9=0 và x+y=0

=>x=-3; y=3

a: (2x-5)(y+3)=-22

mà x,y là số nguyên

nên \(\left(2x-5;y+3\right)\in\left\{\left(1;-22\right);\left(11;-2\right);\left(-1;22\right);\left(-11;2\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;-25\right);\left(8;-5\right);\left(2;19\right);\left(-3;-1\right)\right\}\)

a) Ta có: \(x\left(x-1\right)-x^2+2x=5\)

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

hay x=5

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

\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-x+1\right)=0\)

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

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

c) Ta có: \(\left(x+3\right)\cdot\left(x^2-3x+9\right)-x\left(x-2\right)^2=19\)

\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-19=0\)

\(\Leftrightarrow x^3+8-x^3+4x^2-4x=0\)

\(\Leftrightarrow4x^2-4x+8=0\)(Vô lý)

`a)(x+5)^3=-64`

`(x+5)^3=(-4)^3`

`x+5=-4`

`x=-4-5=-9`

Vậy `x=-9`

`2)(2x-3)^3=8`(9 không được)

`(2x-3)^3=2^3`

`2x-3=2`

`2x=5`

`x=5/2`

Vậy `x=5/2`

a)\(\left(x+5\right)^3=64\)

\(\left(x+5\right)^3=4^3\)

\(x+5=4\)

\(x=4-5\)

\(x=-1\)

b) \(\left(2x-3\right)^3=9\)

\(\left(2x-3\right)^3=3^3\)

\(2x-3=3\)

\(2x=3+3\)

\(2x=6\)

\(x=\dfrac{6}{2}\)

\(x=3\)

Giải:

a) \(\left(x+5\right)^3=-64\) 

\(\Rightarrow\left(x+5\right)^3=\left(-4\right)^3\) 

\(\Rightarrow x+5=-4\) 

\(\Rightarrow x=-4-5\) 

\(\Rightarrow x=-9\) 

b) \(\left(2x-3\right)^3=8\) 

\(\Rightarrow\left(2x-3\right)^3=2^3\) 

\(\Rightarrow2x-3=2\) 

\(\Rightarrow2x=2+3\) 

\(\Rightarrow2x=5\) 

\(\Rightarrow x=5:2\) 

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