6) ĐKXĐ: \(x\le-6\)

\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)

\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)

Vậy \(x\le-6\)

7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)

\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)

Vậy \(x\ge\dfrac{2}{3}\)

8) ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)

\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)

9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4

=>2*căn(x+5)=4

=>căn (x+5)=2

=>x+5=4

=>x=-1

b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

=>2*căn x-1=16

=>x-1=64

=>x=65

c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)

TH₁: \(x\ge3\)

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

TH₂: \(2\le x< 3\)

\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH₃: \(0\le x< 2\)

\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

TH₄: \(x< 0\)

\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)

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

a, \(x^2-6x+9=\left(x-3\right)^2\)

b, \(x^2-12x+36=\left(x-4\right)^2\)

c, \(9x^2-25=\left(3x-5\right)\left(3x+5\right)\)

d, \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)

e, \(x^4-8x^2+16=\left(x^2-4\right)^2=\left[\left(x-2\right)\left(x+2\right)\right]^2\)

f, \(x^4-81=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)

g, \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=9\left(1-x\right)\left(x+1\right)\)

h, \(\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2\)

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

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

a) x² - 6x + 9 = (x-3)²

b) x² - 12 + 36 = (x-6)²

c) 9x² - 25 = (3x - 25)(3x + 25)

d) x² - x + 1/4 = (x - 1/2)²

1: =(x+y-3x)(x+y+3x)

=(-2x+y)(4x+y)

2: =(3x-1-4)(3x-1+4)

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

=3(x+1)(3x-5)

3: =(2x)^2-(x^2+1)^2

=-[(x^2+1)^2-(2x)^2]

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

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

4: =(2x+1+x-1)(2x+1-x+1)

=3x(x+2)

5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

=(2x^2+2)*4x

=8x(x^2+1)

6: =(5x-5y)^2-(4x+4y)^2

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

=(x-9y)(9x-y)

7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)

=(x^2+2xy+y^2)(x^2-y^2)

=(x+y)^3*(x-y)

8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)

=[(x-2y)^2-4][(x+2y)^2-36]

=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)

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

\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)

hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)

2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)

hay \(x\in\left\{1;5\right\}\)

3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

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

hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

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

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

hay \(x\in\left\{-4;3;-3\right\}\)

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

6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)

\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)

\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)

hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)

1.

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

\(\Leftrightarrow x+3=5x-2\)

\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)

2.

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)

\(\Leftrightarrow x^2+x+1=x^2-2x+16\)

\(\Leftrightarrow3x=15\Leftrightarrow x=5\)

3.

\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)

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

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

7.

\(\Leftrightarrow x^2+2x-15=0\)

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

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

8.\(\Leftrightarrow x^4+x^3+4x^3+4x^2=0\)

\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0;x=-4\end{matrix}\right.\)

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

\(\Leftrightarrow x+2=3-2x\)

\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)

Bài 3:

b. $B=(x+y)(2x-y)+(xy^4-x^2y^2):(xy^2)$

$=(2x^2-xy+2xy-y^2)+(y^2-x)$

$=2x^2+xy-y^2+y^2-x=2x^2+xy-x$

Bài 4:
a. $25x^3-10x^2+x=x(25x^2-10x+1)=x(5x-1)^2$
b. $x^2-9x+9y-y^2=(x^2-y^2)-(9x-9y)=(x-y)(x+y)-9(x-y)=(x-y)(x+y-9)$

c. $16-x^2-4y^2-4xy=16-(x^2+4y^2+4xy)$

$=4^2-(x+2y)^2=(4-x-2y)(4+x+2y)$

Bài 5:

a. $36-x^3=100$

$x^3=36-100=-64=(-4)^3$

$\Rightarrow x=-4$

b.

$(x+2)(x-2)-(x+1)^2=7$

$\Leftrightarrow (x^2-4)-(x^2+2x+1)=7$

$\Leftrightarrow -2x-5=7$

$\Leftrightarrow -2x=12$

$\Leftrightarrow x=-6$

1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20