a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

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

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

a)  (x - 3)² - 5.(x - 2) + 5 = 0.

<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0

<=> x^2 - 11x + 24 = 0

<=> (x-3)(x-8)=0

<=> x = 3 hoặc x = 8

b) (2x - 1)² - 3.(x - 2).(x + 2) - 25 = 0.

<=> 4x^2 - 4x + 1 - 3x^2 + 12 - 25 = 0

<=> x² - 4x - 12 = 0

<=> (x+2)(x-6) = 0

<=> x = -2 hoặc x = 6

d)  x² - 4x + 5 = 0.

<=> (x - 2)² = -1 (vô lý)

Vậy phương trình vô nghiệm

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

a) x = 5.                        b) x = 1 2 .

c) x = 3 5  hoặc x = 1.     d) x = 3.

\(a,x^2-10x=-25\)

\(< =>x^2-10x+25=0\)

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

b, \(4x^2-4x=-1\)

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

\(< =>\left(2x-1\right)^2=0< =>x=\frac{1}{2}\)

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

\(< =>\left(1-2x\right)^2-\left(3x-2\right)^2=0\)

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

\(< =>\left(-5x+3\right)\left(x-1\right)=0\)

\(< =>\orbr{\begin{cases}x=\frac{3}{5}\\x=1\end{cases}}\)

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

\(< =>\left(x-2+5-2x\right)\left(x^2-4x+4+5x-2x^2-10+4x+25-20x+4x^2\right)=0\)

\(< =>\left(3-x\right)\left(-5x^2-15x+19\right)=0\)

\(< =>\left(x-3\right)\left(5x^2+15x-19=0\right)\)

\(< =>\orbr{\begin{cases}x=3\\x^2+3x-\frac{19}{5}=0\end{cases}}\)

Xét phương trình \(x^2+3x-\frac{19}{5}=0< =>\left(x^2+2.x.\frac{3}{2}+\frac{9}{4}\right)-\left(\frac{19}{5}+\frac{9}{4}\right)=0\)

\(< =>\left(x+\frac{3}{2}\right)^2=\frac{29}{5}+\frac{1}{4}\)

\(< =>\orbr{\begin{cases}x=\sqrt{\frac{29}{5}+\frac{1}{4}}-\frac{3}{2}\\x=-\sqrt{\frac{29}{5}+\frac{1}{4}}-\frac{3}{2}\end{cases}}\)

Vậy .........

4, x^2-10x+8x-80=0

x(x-8)+10(x-8)=0

x+10=0         =)x=-10

hoặc

x-8=0        =)x=8

1, =(x+2)(x-2)=0

x+2=0     =)x=-2

hoặc

x-2=0      =)x=2

2,3(x^2-5^2)=0

=x+5=0    =)x=-5

hoặc 

 x-5=0   =)x=5

3,(3+2)^2=25

5^2=25

5, x^2-x-11x+11=0

x(x-1)-11(x-1)=0

x-11=0    =)x=11

hoặc 

x-1=0    =)x=1

xl nheee mk làm nhầm câu 4 trc

là mũ hay nhân vậy ?

bn có thể viết đề rõ hơn k nhìn có vẻ rối

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

câu b thiếu bn ơi

ok

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

a: Ta có: \(\left(2x-1\right)^2-25=0\)

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

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

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

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

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

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