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

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

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

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

\(\Leftrightarrow2x-5=0\)

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

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

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

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

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

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

\(\Rightarrow2x-5=0\)

\(\Rightarrow2x=5\)

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

\(b,2x^3+3x^2+2x+3=0\)

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

\(\Rightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\)

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

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

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

\(\Rightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)^3+\left(x+3\right)\left(x-9\right)=0\)

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

\(\Rightarrow\left(x+3\right).x^3=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x^3=0\end{matrix}\right.\)

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

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

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

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

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

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

a, (3x+2)(2x+9) - (x+2)(6x+1) = (x+1)-(x-6)                                                           b,  3(2x-1)(3x-1) - (2x-3)(9x-1) = 0

 =>   6x²+4x+27x+18-6x²-12x-x-2 = x+1-x+6                                                             =>   18x² -9x-6x+3-18x²+27x+2x-3 = 0

=>                                      18x+16 = -5                                                                     =>                        14x                        = 0

=>                                     18x        = -5-16                                                                =>                             x                        = 0

=>                                      18x       = -18

=>                                           x      = -1

X+1-X+5=7 MÀ

Lần sau đăng thì chia thành nhiều câu hỏi nhé

\(16^2-9.\left(x+1\right)^2=0\)

\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)

\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)

\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)

\(\left[13-3x\right].\left[19+3x\right]=0\)

\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)

KL:..............................

Nhiều câu hỏi mà bn ??

a, Ta có 

\(x^2+2\left(2x+3\right)-\left|x^2-2x+2\right|=77\)

\(\Rightarrow x^2+4x+4+2-\left|x^2-2x+1+1\right|=77\)

\(\Rightarrow\left(x+2\right)^2+2-\left|\left(x-1\right)^2+1\right|=77\)

Vì \(\left(x+1\right)^2\ge0\left(\forall x\right)\) \(\Rightarrow\left(x+1\right)^2+1>0\Rightarrow\left|\left(x+1\right)^2+1\right|=\left(x+1\right)^2+1\)

\(\Rightarrow\left(x+2\right)^2-\left(x-1\right)^2+1=75\) \(\Rightarrow\left(x+2\right)^2-\left(x-1\right)^2=74\Rightarrow\left(x+2-x+1\right).\left(x+2+x+1\right)=74\)

\(\Rightarrow3.\left(x+3\right)=74\Rightarrow x+3=\frac{74}{3}\Rightarrow x=\frac{65}{3}\)

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

Vì \(\left(x-1\right)^2\ge0;x^2\ge0\Rightarrow2x^2\ge0\)

\(\Rightarrow\left(x-1\right)^2+2x^2\ge0\)

\(\Rightarrow\hept{\begin{cases}2x^2=0\\\left(x-1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=1\end{cases}}}\) ( vô lý )

Vậy ko có gt x thỏa mãn

a) \(x^2\left(x-3\right)+12-4x=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=3\\x\in\left\{\pm2\right\}\end{cases}}\)

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

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

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

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

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

c) \(\left(2x-1\right)^2-25=0\)

\(\left(2x-1\right)^2-5^2=0\)

\(\left(2x-1-5\right)\left(2x-1+5\right)=0\)

\(\left(2x-6\right)\left(2x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\2x+4=0\end{cases}}\)

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

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

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

\(\left(x-2\right)\left(5x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\5x-8=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{5}\end{cases}}\)

a. \(9\left(x+2\right)-3\left(x+2\right)=0\)

\(\Leftrightarrow9x+18-3x-6=0\)

\(\Leftrightarrow6x+12=0\)

\(\Leftrightarrow x=-2\)

e. \(\left(2x-1\right)^2-45=0\)

\(\Leftrightarrow4x^2-2x+1-45=0\)

\(\Leftrightarrow4x^2-2x-44=0\)

Đến đó tự giải tiếp nha!

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

\(\Leftrightarrow4x-10-3x=0\)

\(\Leftrightarrow x-10=0\)

\(\Leftrightarrow x=10\)

g. \(2x^2-6x=0\)

\(\Leftrightarrow2x\left(x-3\right)=0\)

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

sao làm nhung cau de the

a, (x-5).(x-1) >0
<=> x-5>0 và x-1>0
<=> x-5>0
<=> x>5
x-1>0
<=> x>1
Vậy x>5
b, (2x-3).(x+1) <0
<=> 2x-3<0 và x+1<0
2x-3<0 <=> 2x<3 <=> x<2/3
x+1<0 <=> x<-1
Vậy x<2/3
c, 2x² - 3x +1>0
<=> 2x² - 2x- x +1>0
<=>(x-1). (2x-1) >0
<=> x-1>0 và 2x-1>0
x-1>0 <=> x>1
2x-1>0 <=> 2x>1 <=> x>1/2
Vậy x>1/2