\(x^2-36=0\Rightarrow x^2=36\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

\(3x^2-75=0\)

\(\Rightarrow3\left(x^2-25\right)=0\)

\(\Rightarrow x^2-25=0\Rightarrow x^2=25\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

\(4x^2-4x+1=0\)

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

\(\Rightarrow2x-1=0\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)

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

\(\Rightarrow\left(x+3\right)^2=4\)

\(\Rightarrow\left[{}\begin{matrix}x+3=2\\x+3=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)

a) \(x^2-36=0\Leftrightarrow x^2=36\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{36}\\x=-\sqrt{36}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

vậy \(x=6;x=-6\)

b) \(3x^2-75=0\Leftrightarrow3\left(x^2-25\right)=0\Leftrightarrow x^2-25=0\Leftrightarrow x^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{25}\\x=-\sqrt{25}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\) vậy \(x=5;x=-5\)

c) \(4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)

d) \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x+3=\sqrt{4}\\x+3=-\sqrt{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=2\\x+3=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\) vậy \(x=-1;x=-5\)

\(X=\)\(-2\)

\(X=3\)

\(X=-4\)

\(X=1,5\)

a/ \(\left(x-4\right)^2-36=0\)

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

<=> \(\left(x-10\right)\left(x+2\right)=0\)

<=> \(\orbr{\begin{cases}x-10=0\\x+2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)

b/ \(\left(x+8\right)^2=121\)

<=> \(\left(x+8\right)^2-121=0\)

<=> \(\left(x+8-11\right)\left(x+8+11\right)=0\)

<=> \(\left(x-3\right)\left(x+19\right)=0\)

<=> \(\orbr{\begin{cases}x-3=0\\x+19=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=3\\x=-19\end{cases}}\)

d/ \(4x^2-12x+9=0\)

<=> \(\left(2x\right)^2-2.2x.3+3^2=0\)

<=> \(\left(2x-3\right)^2=0\)

<=> \(2x-3=0\)

<=> \(x=\frac{3}{2}\)

4x(x-2005)-(x+2005)=0

4x(x-2005)+(x-2005)=0

(x-2005)(4x+1)=0

<=>x-2005=>x=2005

4x+1=0=>x=-1/4

b, (x+1)²-x-1=0

(x+1)²-(x+1)=0

(x+1)(x+1-1)=0

(x+1)x=0

<=>x+1=0=>x=-1

x =0

a)\(x^2-2x-24=0\Leftrightarrow x^2-2x+1-25=0\)

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

\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=6\\x=-4\end{cases}}\)

b)\(x^2+8x+12=0\Leftrightarrow x^2+8x+16-4=0\)

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

\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=0\Leftrightarrow\hept{\begin{cases}x=-2\\x=-6\end{cases}}\)

c)\(4x^2+4x-63=0\Leftrightarrow4x^2+4x+1-64=0\)

\(\Leftrightarrow\left(2x+1\right)^2-8^2=0\Leftrightarrow\left(2x+1-8\right)\left(2x+1+8=0\right)\)

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

Bài 1 :

a, \(\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)

=> \(\left(x-3\right)\left(4x-1-5x-2\right)=0\)

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

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

Vậy phương trình có nghiệm là \(x=\pm3\) .

b, \(\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\)

=> \(\left(x+3\right)\left(x-5+3x-4\right)=0\)

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

=> \(\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)

Vậy phương trình có nghiệm là \(x=-3,x=\frac{9}{4}\) .

c, \(\left(x+6\right)\left(3x-1\right)+x^2-36=0\)

=> \(\left(x+6\right)\left(3x-1\right)+\left(x-6\right)\left(x+6\right)=0\)

=> \(\left(x+6\right)\left(3x-1+x-6\right)=0\)

=> \(\left(x+6\right)\left(4x-7\right)=0\)

=> \(\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)

Vậy phương trình có nghiệm là \(x=-6,x=\frac{7}{4}\) .

a) ( 4x - 1 ) ( x - 3 ) - ( x - 3 ) ( 5x + 2 ) = 0

⇔ ( x - 3 ) ( 4x - 1 - 5x - 2 ) = 0

⇔ ( x - 3 ) ( -x - 3 ) = 0

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

Ý b) tương tự ý a) thôi.

c) ( x + 6 ) ( 3x - 1 ) + x² - 36 = 0

⇔ ( x + 6 ) ( 3x - 1 ) + ( x + 6 ) ( x - 6 ) = 0

⇔ (x+6)(3x-1+x-6)=0

⇔ (x+6)(4x-7)=0

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)

BÀI 1:

 a)   \(ĐKXĐ:\) \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}}\) \(\Leftrightarrow\)\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)

b)  \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)

\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)^2}{8}\)

\(=\frac{2x+4-2x+4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}\)

\(=\frac{x+2}{x-2}\)

c)  \(A=0\)  \(\Rightarrow\)\(\frac{x+2}{x-2}=0\)

                      \(\Leftrightarrow\) \(x+2=0\)

                      \(\Leftrightarrow\)\(x=-2\) (loại vì ko thỏa mãn ĐKXĐ)

Vậy ko tìm đc  x   để  A = 0

p/s:  bn đăng từng bài ra đc ko, mk lm cho

giải nhanh giúp mik nha mn:)

a) \(\left(x+17\right).\left(25-x\right)=0\)

\(\Leftrightarrow x+17=0\)hoặc \(25-x=0\)

Từ \(x+17=0\Rightarrow x=0-17=-17\)

Từ \(25-x=0\Rightarrow x=25-0=25\)

Vậy \(x=-17\) hoặc \(25\)

a) ( 4x - 1 ) ( x - 2 ) = 0

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=2\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{4};2\right\}\)

b) 4x² - 12x = 0

<=> 4x ( x - 3 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}4x=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy \(x\in\left\{0;3\right\}\)

c) ( x - 5 )4 + 25 - x² = 0

( x - 5 ) 4 + ( 5 - x ) ( 5 + x ) = 0

( x - 5 ) ( 4 + 5 + x ) = 0

( x - 5 ) ( 9 + x ) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\9+x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-9\end{cases}}\)

Vậy \(x\in\left\{-9;5\right\}\)

a)x=0,25,x=2

b)x=3,x=0

1 , 

\(b,x^2-2x=0\)

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

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

KL :..

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

\(b,4x^2-4x+1=\left(2x\right)^2-2.2x+1\)

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