a)        \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)

\(\Leftrightarrow\)\(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)

Vậy...

\(2x^3+3x^2-32x-48=0\)

\(\Leftrightarrow2x^3-32x+3x^2-48=0\)

\(\Leftrightarrow2x\left(x^2-16\right)+3\left(x^2-16\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x^2-16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm4\\x=-\frac{3}{2}\end{matrix}\right.\)

b/ \(\Leftrightarrow10x^2-15x+4x-6=0\)

\(\Leftrightarrow5x\left(2x^2-3\right)+2\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{2}{5}\end{matrix}\right.\)

Lời giải:

a)

$10x^2-11x-6=0$

$\Leftrightarrow 10x^2-15x+4x-6=0$

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

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

$\Rightarrow 2x-3=0$ hoặc $5x+2=0$

$\Rightarrow x=\frac{3}{2}$ hoặc $x=-\frac{2}{5}$

b)

$2x^3+3x^2-32x=48$

$\Leftrightarrow 2x^3+3x^2-32x-48=0$

$\Leftrightarrow 2x^3-8x^2+11x-44x+12x-48=0$

$\Leftrightarrow 2x^2(x-4)+11x(x-4)+12(x-4)=0$

$\Leftrightarrow (x-4)(2x^2+11x+12)=0$

$\Leftrightarrow (x-4)[2x(x+4)+3(x+4)]=0$

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

$\Rightarrow x-4=0; x+4=0$ hoặc $2x+3=0$

$\Rightarrow x=\pm 4$ hoặc $x=\frac{-3}{2}$

xin lỗi mọi người mk nhấn nhầm toán lớp 8 nha

a) \(2x^3-32x=0\)

\(2x\left(x^2-16\right)=0\)

\(2x\left(x-4\right)\left(x+4\right)=0\)

\(\Rightarrow2x=0\)hoặc \(\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

  vậy \(x=0\) hoặc \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

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

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

\(\left(2x-7\right)\left(4x+3\right)=0\)

\(\orbr{\begin{cases}2x-7=0\\4x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)

  vậy \(\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)

c) \(2\left(x+3\right)-x^2-3x=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

    vậy \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

d) \(4x^2-25-\left(2x+5\right)\left(x+7\right)=0\)

\(\left(4x^2-25\right)-\left(2x+5\right)\left(x+7\right)=0\)

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

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

\(\left(2x+5\right)\left(x-12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+5=0\\x-12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)

 vậy \(\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)

Chọn A.

a: \(\left(2x-3\right)\left(3x^2+1\right)-6x\left(x^2-x+1\right)+3x^2-2x=10\)

\(\Leftrightarrow6x^3+2x-9x^2-3-6x^3+6x^2-6x+3x^2-2x=10\)

\(\Leftrightarrow-6x-3=10\)

=>-6x=13

hay x=-13/6

b: \(\Leftrightarrow3x^2-3x+x-2-3x^2+5x=-8-5x\)

=>3x-2=-5x-8

=>8x=-6

hay x=-3/4

c: \(\Leftrightarrow64x^3-27-64x^3+32x^2-32x^2+x=20\)

=>x-27=20

hay x=47

\(4\left(8x^3-12x^2+6x-1\right)+3\left(2x-1\right)-3\sqrt{1-y}-4\left(1-y\right)\sqrt{1-y}=0\)

\(\Leftrightarrow4\left(2x-1\right)^3+3\left(2x-1\right)-3\sqrt{1-y}-4\left(1-y\right)\sqrt{1-y}=0\)

Đặt \(\left\{{}\begin{matrix}2x-1=a\\\sqrt{1-y}=b\end{matrix}\right.\)

\(4\left(a^3-b^3\right)+3\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left[4\left(a^2+ab+b^2\right)+3\right]=0\)

\(\Rightarrow a=b\Rightarrow2x-1=\sqrt{1-y}\Rightarrow x\ge\frac{1}{2}\)

\(y=1-\left(2x-1\right)^2\) thay xuống pt dưới