a) \(\sqrt[]{x^2-4x+4}=x+3\)

\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)

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

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

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

pt quá vĩ đại =.= cx trên OLM lun 

câu a biến đổi to lắm

\(\Leftrightarrow-\left(12x\sqrt{6x-1}-2\sqrt{6x-1}-2x^3-9x^2+6x-8\right)=0\)rồi sao nx 

cái này ra nghiệm là 

\(2-\sqrt{2}\)và\(\sqrt{2}+2\)

a) \(x\left(2x-1\right)-6x+3=0\)

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

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

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

b) \(x^2\left(x+1\right)-9x-9=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm\sqrt{9}=\pm3\end{cases}}\)

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

=> x(2x - 1) - 3(2x - 1) = 0

=> (x - 3)(2x - 1) = 0

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

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

b) x²(x + 1)  - 9(x + 1) = 0

=> (x² - 9)(x + 1) = 0

=> \(\orbr{\begin{cases}x^2-9=0\\x+1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\pm3\\x=-1\end{cases}}\)

a. x.(x+3)-x²+15=0

=> x^2 + 3x - x^2 + 15 = 0

=> 3x + 15 = 0

=> 3x = -15

=> x = -5

vậy_

b. (2x-1)(x+3) - x(2x-6) =15

=> 2x^2 + 6x - x - 3 - 2x^2 + 6x = 15

=> x - 3 = 15

=> x = 18

vậy_

c. x³ -36x = 0

=> x(x^2 - 36) = 0

=> x = 0 hoặc x^2 - 36 = 0

=> x = 0 hoặc x^2 = 36

=> x = 0 hoặc x = 6 hoặc x = -6

vậy_

d. 6x² + 6x =x²+2x+1

=> 6x(x + 1) = (x + 1)^2

=> 6x(x + 1) - (x + 1)^2 = 0

=> (x + 1)(6x - x - 1) = 0

=> (x + 1)(5x - 1) = 0

=> x = -1 hoặc 5x = 1

=> x = -1 hoặc x = 1/5

vậy_

e. x(3x+1)=1-9x² 

=> x(3x + 1) = (1 - 3x)(1 + 3x)

=> x(3x + 1) - (1 - 3x)(1 + 3x) = 0

=> (3x + 1)(x - 1 + 3x) = 0

=> (3x + 1)(4x - 1) = 0

=> 3x + 1 = 0 hoặc 4x - 1 = 0

=> 3x = -1 hoặc 4x = 1

=> x = -1/3 hoặc x = 1/4

vậy_

(x-1)(2x+5)(x^2+2)=0

<=> x-1=0 hoặc 2x+5 =0 hoặc x^2 +2=0

<=> x=1 hoặc x=\(\frac{-5}{2}\)hoặc x\(=\varnothing\)

Vậy x=1; x=\(-\frac{5}{2}\)

a) (x-1)(2x+5)(x²+2)=0

    \(\Rightarrow\hept{\begin{cases}x-1=0\\2x+5=0\\x^2+2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0+1=1\\2x=-5\Rightarrow x=-\frac{5}{2}\\x^2=2\end{cases}}\)

Vô lí vì 2 ko chuyển đc mũ 2

Vậy x= 1 hoặc \(x=-\frac{5}{2}\)

Giúp mình hai câu cuối với 

(x+2)(x+3)-(x-2)(x+5)=0

=> x²+5x+6-x²-3x+10=0

=>2x+16=0 

 =>2x=-16

=>x=-8

a) = (3x +1)² =0 

3x+1 =0

x = -1/3

b) = (5x)² -2² =0

(5x+2)(5x-2) = 0

5x+2 =0

x = -2/5

5x -2  =0

x= 2/5

xem đi rui lam tip

a) 9x² + 6x + 1 = 0  => (3x)² + 2 x 3x + 1 = 0  => (3x + 1)²  = 0  => 3x + 1 = 0  => x = \(\frac{-1}{3}\)

b) 25x² = 4  => x² = 4 : 25  => x² = 0,16  => x = 0,4 hoặc x = -0,4

c) 8 - 125x³ = 0  => 125x³ = 8  => x³ = 8 : 125  => x³ = \(\frac{8}{125}\)=> x = \(\frac{2}{5}\)

c) = 2³ - (5x)³ =0

(2-5x)(4 +10x +25x²) =0

2-5x=0

x = 2/5

4 + 10x +25x²  = 0 (máy tính giải dc)

d) = (( 2x+1) - 5(x+2))² =0

2x+1 -5x -10=0

3x= -9

x = -3

(hoàn toàn ad hđt đáng nhớ thui,bn à)

a) Ta có: \(x^4-16x^2=0\)

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

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

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

Vậy: \(x\in\left\{0;4;-4\right\}\)

b) Ta có: \(9x^2+6x+1=0\)

\(\Leftrightarrow\left(3x\right)^2+2\cdot3x\cdot1+1^2=0\)

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

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\frac{1}{3}\)

Vậy: \(x=-\frac{1}{3}\)

c) Ta có: \(x^2-6x=16\)

\(\Leftrightarrow x^2-6x-16=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{8;-2\right\}\)

d) Ta có: \(9x^2+6x=80\)

\(\Leftrightarrow9x^2+6x-80=0\)

\(\Leftrightarrow9x^2+6x+1-81=0\)

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

\(\Leftrightarrow\left(3x+1-9\right)\left(3x+1+9\right)=0\)

\(\Leftrightarrow\left(3x-8\right)\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=8\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{10}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{3};-\frac{10}{3}\right\}\)

e) Ta có: \(25\left(2x-1\right)^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(10x-5\right)^2-\left(3x+3\right)^2=0\)

\(\Leftrightarrow\left(10x-5-3x-3\right)\left(10x-5+3x+3\right)=0\)

\(\Leftrightarrow\left(7x-8\right)\left(13x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-8=0\\13x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=8\\13x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{7}\\x=\frac{2}{13}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{7};\frac{2}{13}\right\}\)

a, \(\left|2x-1\right|-7=0\Leftrightarrow\left|2x-1\right|=7\)

Với \(x\ge\frac{1}{2}\)phương trình có dạng : 

\(2x-1=7\Leftrightarrow x=4\)( tm ) 

Với \(x< \frac{1}{2}\)phương trình có dạng : 

\(-2x+1=7\Leftrightarrow x=-3\)( tm )

Vậy tập nghiệm của phương trình là S = { -3 ; 4 } 

b, \(\frac{9x^2}{2\left(1-9x^2\right)}=\frac{3x}{6x-2}-\frac{1+9x}{3+9x}\)ĐK : \(x\ne\pm\frac{1}{3}\)

\(\Leftrightarrow-\frac{9x^2}{2\left(3x-1\right)\left(3x+1\right)}=\frac{3x}{2\left(3x-1\right)}-\frac{1+9x}{3\left(3x+1\right)}\)

\(\Leftrightarrow\frac{-27x^2}{6\left(3x-1\right)\left(3x+1\right)}=\frac{9x\left(3x+1\right)}{6\left(3x-1\right)\left(3x+1\right)}-\frac{2\left(1-9x\right)\left(3x+1\right)}{6\left(3x-1\right)\left(3x+1\right)}\)

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

\(\Leftrightarrow108x^2-15x=0\Leftrightarrow3x\left(36x-5\right)=0\Leftrightarrow x=0;x=\frac{5}{36}\)( tm )

Vậy tập nghiệm của phương trình là S = { 0 ; 5/36 }