\(\left(x+2\right)^5-\left(x-2\right)^5=64\)

\(\Rightarrow x^5+10x^4+40x^3+80x^2+80x+32-\left(x^5-10x^4+40x^3-80x^2+80x-32\right)=64\)

\(\Rightarrow20x^4+160x^2+64=64\)

\(\Rightarrow20x^4+160x^2=0\)

\(\Rightarrow20x^2\left(x^2+8\right)=0\)

mà \(x^2+8>0\)

\(\Rightarrow x^2=0\Rightarrow x=0\)

\(\Leftrightarrow x^5+10x^4+40x^3+80x^2+80x+32-x^5+10x^4-40x^3+80x^2-80x+32=64\)

\(\Rightarrow20x^4+160x^2+54-64=0\)

\(\Rightarrow20x^4+160x^2=0\)

\(\Leftrightarrow20x^2\left(x^2+8\right)=0\)

\(\Leftrightarrow x=0\)

Do \(x^2+8=\ge0\)(luôn đúng)

Vây: \(x^2\ge-8\)

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

2) \(y\left(x-2y\right)^2+xy^2\left(2y-x\right)=\left(2y-x\right)\left(2\left(2y-x\right)+1\right)=\left(2y-x\right)\left(4y-2x+1\right)\)

3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)

4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)

Xét ΔMNQ và ΔNMP có

MN chung

NQ=MP

MQ=NP

=>ΔMNQ=ΔNMP

=>góc OMN=góc ONM

=>OM=ON 

OM+OP=MP

ON+OQ=NQ

mà MP=NQ và OM=ON

nên OP=OQ

a) \(y^2+49-14y\)

\(=y^2-14y+49\)

\(=y^2-2\cdot7\cdot y+7^2\)

\(=\left(y-7\right)^2\)

b) \(a^2+4b^2-4ab\)

\(=a^2-4ab+4b^2\)

\(=a^2-2\cdot2b\cdot a+\left(2b\right)^2\)

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

c) \(y^2+y+\dfrac{1}{4}\)

\(=y^2+2\cdot\dfrac{1}{2}\cdot y+\left(\dfrac{1}{2}\right)^2\)

\(=\left(y+\dfrac{1}{2}\right)^2\)

a: Sửa đề: AD=AB

AD=AB

AD=BC

=>AB=BC

b: AB=AD

=>góc ADB=góc ABD

=>góc ADB=góc BDC

=>DB là phân giác của góc ADC

`@` `\text {Ans}`

`\downarrow`

`a,`

`(2x - 1)^2 - 25 = 0`

`<=> (2x - 1)^2 = 25`

`<=> (2x - 1)^2 = (+-5)^2`

`<=>`\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

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

Vậy, `S = {-2; 3}`

`b,`

`8x^3 - 50x = 0`

`<=> x(8x^2 - 50) = 0`

`<=>`\(\left[{}\begin{matrix}x=0\\8x^2-50=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\8x^2=50\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x^2=\dfrac{25}{4}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=0\\x=\pm\dfrac{5}{2}\end{matrix}\right.\)

Vậy, `S = {-5/2; 0; 5/2}.`

a) (2x - 1)² - 25 = 0

(2x - 1)² - 5² = 0

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

(2x - 6)(2x + 4) = 0

2x - 6 = 0 hoặc 2x + 4 = 0

*) 2x - 6 = 0

2x = 6

x = 3

*) 2x + 4 = 0

2x = -4

x = -2

Vậy x = -2; x = 3

b) 8x³ - 50x = 0

2x(4x² - 25) = 0

2x[(2x)² - 5²] = 0

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

2x = 0 hoặc 2x - 5 = 0 hoặc 2x + 5 = 0

*) 2x = 0

x = 0

*) 2x - 5 = 0

2x = 5

x = 5/2

*) 2x + 5 = 0

2x = -5

x = -5/2

Vậy x = -5/2; x = 0; x = 5/2

\(x^2-2xy+y^2+3x-3y-4\)

\(=\left(x-y\right)^2-1+3x-3y-3\)

\(=\left[\left(x-y\right)^2-1^2\right]+\left(3x-3y-3\right)\)

\(=\left[\left(x-y\right)-1\right]\left[\left(x-y\right)+1\right]+3\left(x-y-1\right)\)

\(=\left(x-y-1\right)\left(x-y+1\right)+3\left(x-y-1\right)\)

\(=\left(x-y-1\right)\left[\left(x-y+1\right)+3\right]\)

\(=\left(x-y-1\right)\left(x-y+4\right)\)

a) 6x² + 7xy + 2y²

= 6x² + 4xy + 3xy + 2y²

= (6x² + 4xy) + (3xy + 2y²)

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

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

b) x² - y² + 10x - 6y + 16

= x² + 10x + 25 - y² - 6y - 9

= (x² + 10x + 25) - (y² + 6y + 9)

= (x + 5)² - (y + 3)²

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

= (x - y + 2)(x + y + 8)

c) 4x⁴ + y⁴

= 4x⁴ + 4x²y² + y⁴ - 4x²y²

= (2x² + y²)² - (2xy)²

= (2x² + y² - 2xy)(2x² + y² + 2xy)

a: =(x^2+x)^2+4(x^2+x)-12

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

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

b: =(x^2+8x)^2+22(x^2+8x)+120

=(x^2+8x+12)(x^2+8x+10)

=(x+2)(x+6)(x^2+8x+10)

c: =8x^2+12x-2x-3

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

a: =>x^2-6x+9+y^2+8y+16=0

=>(x-3)^2+(y+4)^2=0

=>x=3 và y=-4