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

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

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

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

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

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$

het thoirui pan oi

b: Ta có: \(\left(x-3\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+9\left(x+2\right)^2\)

\(=x^3-9x^2+27x-27-x^3-8+9x^2+36x+36\)

\(=53x+1\)

\(2x^3-50x=0\)

<=>  \(2x\left(x^2-25\right)=0\)

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

đến đây

bạn tự giải nhé

hk tốt   

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

1,\(5x^2=13x\Leftrightarrow5x^2-13x=0\Leftrightarrow x\left(5x-13\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{13}{5}\end{cases}}\)

2,\(\left(5x^2+3x-2\right)^2=\left(4x^2-3x-2\right)^2\Leftrightarrow\orbr{\begin{cases}5x^2+3x-2=4x^2-3x-2\\5x^2+3x-2=-4x+3x+2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+6x=0\\9x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\left(x+6\right)=0\\\left(3x\right)^2=2^2\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=0or-6\\x=-\frac{2}{3}or\frac{2}{3}\end{cases}}\)

3,\(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9+x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2+4x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2+4x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x\left(x+4\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=0or-4\end{cases}}\)

4,\(5x\left(x-2000\right)-x+2000=0\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}\)

5,\(5x\left(x-2\right)-x+2=0\Leftrightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\5x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)

6,\(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-8\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\4x-8=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

7,\(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(2x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x-4=0\\2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

tí làm nửa kia 

8,\(x^2-6x+8=0\Leftrightarrow x^2-6x+9-1=0\Leftrightarrow\left(x-3\right)^2-1^2=0\)

\(\Leftrightarrow\left(x-3-1\right)\left(x-3+1\right)=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

9,\(9x^2+6x-8=0\Leftrightarrow9x^2+6x+1-9=0\Leftrightarrow\left(3x+1\right)^2-3^2=0\)

\(\Leftrightarrow\left(3x+1-3\right)\left(3x+1+3\right)=0\Leftrightarrow\left(3x-2\right)\left(3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\3x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{4}{3}\end{cases}}\)

10,\(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}\Leftrightarrow}x=-1\)

11,\(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

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

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

\(\Leftrightarrow\left(5-2x\right)\left(2x+7-5-2x\right)=0\Leftrightarrow\left(5-2x\right).2=0\Leftrightarrow5-2x=0\Leftrightarrow x=\frac{5}{2}\)

13,\(x\left(2x-1\right)+\frac{1}{3}.\frac{2}{3}x=0\Leftrightarrow x\left(2x-1\right)+\frac{2}{9}x=0\)

\(\Leftrightarrow x\left(2x-1+\frac{2}{9}\right)=0\Leftrightarrow x\left(2x-\frac{7}{9}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\2x=\frac{7}{9}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{18}\end{cases}}\)

14,\(4\left(2x+7\right)-9\left(x+3\right)^2=0\Leftrightarrow8x+28-9x^2-54x-81=0\)

\(\Leftrightarrow-9x^2+\left(8x-54x\right)+\left(28-81\right)=0\Leftrightarrow-9x^2-46x-53=0\)

\(\Leftrightarrow9x^2+46x+53=0\)Ta có : \(\Delta'=\frac{2116}{4}-477=529-477=52\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-23+\sqrt{52}}{9}\\x=\frac{-23-\sqrt{52}}{9}\end{cases}}\)

\(A.\left(2,3x-6,5\right)\left(0,1x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2,3x-6,5=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2,3x=6,5\\0,1x=-2\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6,5}{2,3}\\x=-20\end{cases}}\)

1, x= 2

2, x = 4

**** bạn mình trước nhé

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