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

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

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

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

=>2x-1=0

=>\(x=\dfrac{1}{2}\)

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

=>\(x^2+4x+4-x^2+3x=2\)

=>7x+4=2

=>7x=-2

=>\(x=-\dfrac{2}{7}\)

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

=>\(x^2-10x+25-x^2-2x=5\)

=>-12x+25=5

=>-12x=5-25=-20

=>\(x=\dfrac{20}{12}=\dfrac{5}{3}\)

4: \(\left(x-1\right)^2+x\left(4-x\right)=11\)

=>\(x^2-2x+1+4x-x^2=11\)

=>2x+1=11

=>2x=10

=>x=5

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

=>\(x^2-9=x^2-10x+25\)

=>-10x+25=-9

=>-10x=-25-9=-34

=>\(x=\dfrac{34}{10}=\dfrac{17}{5}\)

6: \(\left(2x+1\right)^2-4x\left(x-1\right)=17\)

=>\(4x^2+4x+1-4x^2+4x=17\)

=>8x+1=17

=>8x=16

=>x=2

7: \(\left(3x+1\right)^2-9x\left(x-2\right)=25\)

=>\(9x^2+6x+1-9x^2+18x=25\)

=>24x+1=25

=>24x=24

=>x=1

8: \(\left(3x-2\right)\left(3x+2\right)-9x\left(x-1\right)=0\)

=>\(9x^2-4-9x^2+9x=0\)

=>9x-4=0

=>9x=4

=>\(x=\dfrac{4}{9}\)

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

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

=>4(x+2)=0

=>x+2=0

=>x=-2

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

=>\(x^2+4x+4-\left(x^2-9\right)+3=0\)

=>\(x^2+4x+7-x^2+9=0\)

=>4x+16=0

=>4x=-16

=>x=-4

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

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

=>7(3x+2)=0

=>3x+2=0

=>3x=-2

=>\(x=-\dfrac{2}{3}\)

12: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

=>\(x^2+6x+9-x^2+4=4x+17\)

=>6x+13=4x+17

=>2x=4

=>x=2

13: \(3\left(x-1\right)^2+\left(x+5\right)\left(-3x+2\right)=-25\)

=>\(3\left(x^2-2x+1\right)+2x-3x^2+10-15x=-25\)

=>\(3x^2-6x+3-3x^2-13x+10=-25\)

=>-19x+13=-25

=>-19x=-38

=>x=2

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

=>\(x^2+6x+9+x^2-4x+4=2x^2\)

=>2x=-13

=>\(x=-\dfrac{13}{2}\)

a) 

\(P=4x^4+y^4\\ =4x^4+4x^2y^2+y^4-4x^2y^2\\ =\left(4x^4+4x^2y^2+y^2\right)-4x^2y^2\\ =\left(2x^2+y^2\right)^2-\left(2xy\right)^2\\ =\left(2x^2-2xy+y^2\right)\left(2x^2+2xy+y^2\right)\) 

b) 

\(Q=x^4+64\\ =x^4+16x^2+64-16x^2\\ =\left(x^4+16x^2+64\right)-16x^2\\ =\left(x^2+8\right)^2-\left(4x\right)^2\\ =\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

$(-25).(-17).4+(-20)$

$=25.17.4-20$

$=(25.4).17-20$

$=100.17-20$

$=1700-20=1680$

(-25).(-17).4+(-20)

=25.17.4+(-20)

=25.4.17+(-20)

=100.17+(-20)

=1700+(-20)

=1700-20

= 1680

Sửa đề: \(a^2+2ab+b^2-2a-2b+1\)

\(=\left(a^2+2ab+b^2\right)-2\left(a+b\right)+1\)

\(=\left(a+b\right)^2-2\left(a+b\right)\cdot1+1^2\)

\(=\left(a+b-1\right)^2\)

a) $(x-3)(x^2+3x+9)-x^3$

$=(x-3)(x^2+x.3+3^2)-x^3$

$=x^3-3^3-x^3$

$=-3^3=-27$

b) $(2x-y)(4x^2+2xy+y^2)+y^3-7x^3$

$=(2x-y)[(2x)^2+2x.y+y^2]+y^3-7x^3$

$=(2x)^3-y^3+y^3-7x^3$

$=8x^3-7x^3=x^3$

#$\mathtt{Toru}$

\(a.x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left[x^2+x\cdot2y+\left(2y\right)^2\right]\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ b.x^3-64\\ =x^3-4^3\\ =\left(x-4\right)\left(x^2+4\cdot x+4^2\right)\\ =\left(x-4\right)\left(x^2+4x+16\right)\\ c.8x^3-y^3\\ =\left(2x\right)^3-y^3\\ =\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\\ =\left(2x-y\right)\left(4x^2+2xy+y^2\right)\\ d.\dfrac{1}{27}x^6-125\\ =\left(\dfrac{1}{3}x^2\right)^3-5^3\\=\left(\dfrac{1}{3}x^2-5\right)\left[\left(\dfrac{1}{3}x^2\right)^2+\dfrac{1}{3}x^2\cdot5+5^2\right]\\ =\left(\dfrac{1}{3}x^2-5\right)\left(\dfrac{1}{9}x^4+\dfrac{5}{3}x^2+25\right)\)

a) $x^3-(2y)^3$

$=(x-2y)[x^2+x.2y+(2y)^2]$

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

b) $x^3-64$

$=x^3-4^3$

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

$=(x-4)(x^2+4x+16)$

c) $8x^3-y^3$

$=(2x)^3-y^3$

$=(2x-y)[(2x)^2+2x.y+y^2]$

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

d) $\frac{1}{27}x^6-125$

$=\left(\frac13 x^2\right)^3-5^3$

$=\left(\frac13 x^2-5\right)\left[\left(\frac13 x^2\right)^2+\frac13  x^2.5+5^2\right]$

$=\left(\frac13 x^2-5\right)\left(\frac 19x^4+\frac53 x^2+25\right)$

#$\mathtt{Toru}$

a) 

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

b) 

\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-y^3-26x^3\\ =\left(3x+y\right)\left[\left(3x\right)^2-3x\cdot y+y^2\right]-y^3-26x^3\\ =\left(3x\right)^3+y^3-y^3-26x^3\\ =27x^3-26x^3\\ =x^3\)

a) $x^3+2^3$

$=(x+2)(x^2-x.2+2^2)$

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

b) $8x^3+1$

$=(2x)^3+1^3$

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

$=(2x+1)(4x^2-2x+1)$

c) $\frac{1}{27}+x^3$

$=\left(\frac13\right)^3+x^3$

$=\left(\frac13+x\right)\left[\left(\frac13\right)^2-\frac13.x+x^2\right]$

$=\left(\frac13+x\right)\left(\frac19-\frac13 x+x^2\right)$

d) $x^6+y^6$

$=(x^2)^3+(y^2)^3$

$=(x^2+y^2)[(x^2)^2-x^2.y^2+(y^2)^2]$

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

#$\mathtt{Toru}$

\(a.x^3+2^3\\ =\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\\ =\left(x+2\right)\left(x^2-2x+4\right)\\ b.8x^3+1\\ =\left(2x\right)^3+1^3\\= \left(2x+1\right)\left[\left(2x\right)^2-2x\cdot1+1^2\right]\\ =\left(2x+1\right)\left(4x^2-2x+1\right)\\ c.\dfrac{1}{27}+x^3\\ =\left(\dfrac{1}{3}\right)^3+x^3\\ =\left(\dfrac{1}{3}+x\right)\left[\left(\dfrac{1}{3}\right)^2-\dfrac{1}{3}x+x^2\right]\\ =\left(\dfrac{1}{3}+x\right)\left(\dfrac{1}{9}-\dfrac{1}{3}x+x^2\right)\\ d.x^6+y^6\\ =\left(x^2\right)^3+\left(y^2\right)^3\\ =\left(x^2+y^2\right)\left[\left(x^2\right)^2-x^2\cdot y^2+\left(y^2\right)^2\right]\\ =\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\)

Xét tứ giác ABCD có \(\widehat{ABC}+\widehat{ADC}=180^0\)

nên ABCD là tứ giác nội tiếp

=>\(\widehat{DAC}=\widehat{DBC};\widehat{BAC}=\widehat{BDC}\)

mà \(\widehat{CDB}=\widehat{CBD}\)(CB=CD)

nên \(\widehat{DAC}=\widehat{BAC}\)

=>AC là phân giác của góc BAD

Sửa đề: $x=y+z$

Với \(x,y,z\ne0\):

\(\dfrac{1}{x}-\dfrac{1}{y}-\dfrac{1}{z}=1\)

\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{y}-\dfrac{1}{z}\right)^2=1^2\)

\(\Leftrightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}-\dfrac{2}{xy}+\dfrac{2}{yz}-\dfrac{2}{zx}=1\)

\(\Leftrightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}-2.\left(\dfrac{z-x+y}{xyz}\right)=1\)

\(\Leftrightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}-2.\dfrac{z-\left(y+z\right)+y}{xyz}=1\)

\(\Leftrightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=1\) (đpcm)