( 2 x   –   3 ) 2   –   4 x 2   +   9   =   0     ⇔   ( 2 x   –   3 ) 2   –   ( 4 x 2   –   9 )   =   0     ⇔   ( 2 x   –   3 ) 2   –   ( ( 2 x ) 2   –   3 2 )   =   0     ⇔   ( 2 x   –   3 ) 2   –   ( 2 x   –   3 ) ( 2 x   +   3 )   =   0

ó (2x – 3)(2x – 3 – 2x – 3) = 0

ó (2x – 3)(-6) = 0

ó 2x – 3 = 0

ó x = 3 2

Đáp án cần chọn là: C

1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

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

2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)

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

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

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

3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)

\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)

\(\Leftrightarrow-24x=11+1+25=37\)

hay \(x=-\dfrac{37}{24}\)

5) Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow3x^2+3x-8x-8=0\)

\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)

8) Ta có: \(\left|x-5\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)

1)
\(4x^2-4x+1-4x^2-16x-16=9\)
\(-20x-15=9\)
-20x=24
x=-1,2

3)
(2x+1)²=5²
2x+1=5
2x=4
x=2
 

\(1,\Rightarrow4x^2-4x+1-4x^2-16x-16=9\\ \Rightarrow-20x=23\Rightarrow x=-\dfrac{23}{20}\\ 2,\Rightarrow9x^2-6x+1+2x+6+11-11x^2=15\\ \Rightarrow2x^2+4x-3=0\\ \Rightarrow2\left(x^2+2x+1\right)-5=0\\ \Rightarrow2\left(x+1\right)^2-5=0\\ \Rightarrow\left[\sqrt{2}\left(x+1\right)-\sqrt{5}\right]\left[\sqrt{2}\left(x+1\right)+\sqrt{5}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{2}\left(x+1\right)=\sqrt{5}\\\sqrt{2}\left(x+1\right)=-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\\x+1=-\sqrt{\dfrac{5}{2}}=\dfrac{-\sqrt{10}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{2}\\x=\dfrac{-\sqrt{10}-2}{2}\end{matrix}\right.\)

\(3,\Rightarrow\left(2x+1\right)^2-25=0\Rightarrow\left(2x+1-5\right)\left(2x+1+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

\(4,\Rightarrow x^3+3x^2+3x+1-x^3-2x^2-2x+1-x^2=15\\ \Rightarrow x+2=15\Rightarrow x=13\)

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

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

\(\Leftrightarrow3\left(x-1\right)=0\Leftrightarrow x=1\)

2, \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)

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

\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\Leftrightarrow x=-\dfrac{5}{2};x=2\)

3, \(\left(5-2x\right)\left(2x+7\right)=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7+2x+5\right)=0\Leftrightarrow\left(4x+12\right)\left(5-2x\right)=0\Leftrightarrow x=-3;x=\dfrac{5}{2}\)

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

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

\(\Leftrightarrow x-1=0\)

hay x=1

2) Ta có: \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-3x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\)

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

a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)

a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)

\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)

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

b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)

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

c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)

Bài 2:

a) \(=x^2-4-x^2-2x-1=-2x-5\)

b) \(=8x^3-1-8x^3-1=-2\)

Bài 3:

a) \(\Rightarrow x^3+8-x^3+2x=15\)

\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)

b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)

\(\Rightarrow7x=14\Rightarrow x=2\)