Bài 2: 

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

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

Nhìn mãi mới hiểu cái đề bài @-@

`a)đk:` $\begin{cases}\sqrt{x^2-2x} \ge 0\\x+\sqrt{x^2-2x} \ne 0\\x-\sqrt{x^2-2x} ne 0\\\end{cases}$
`<=>` $\begin{cases}x \ge 2\,or\,x<0\\x \ne 0\end{cases}$
`b)A=(x+sqrt{x^2-2x})/(x-sqrt{x^2-2x})-(x-sqrt{x^2-2x})/(x+sqrt{x^2+2x})`
`=((x+sqrt{x^2-2x})^2-(x-sqrt{x^2-2x})^2)/((x+sqrt{x^2-2x})(x-sqrt{x^2-2x}))`
`=(x^2+x^2-2x+2sqrt{x^2-2x}-x^2-x^2+2x+2sqrt{x^2-2x})/(x^2-x^2+2x)`
`=(4sqrt{x^2-2x})/(2x)`
`=(2sqrt{x^2-2x})/x`
`c)A<2`
`<=>2sqrt{x^2-2x}<2x`
`<=>sqrt{x^2-2x}<x(x>=2)`(BP 2 vế thì x>=2)
`<=>x^2-2x<x^2`
`<=>2x>0`
`<=>x>0`
`<=>x>=2`
Vậy `x>=2` thì `A<2`.

Mấy bài dài dài kia tí mình làm cho :) 

( x - 1 )³ - x( x - 2 )² + 1 

= x³ - 3x² + 3x - 1 - x( x² - 4x + 4 ) + 1

= x³ - 3x² + 3x - x³ + 4x² - 4x

= x² - x = x( x - 1 )

2x( 3x + 2 ) - 3x( 2x + 3 )

= 6x² + 4x - 6x² - 9x

= -5x

( x + 2 )³ + ( x - 3 )² - x²( x + 5 )

= x³ + 6x² + 12x + 8 + x² - 6x + 9 - x³ - 5x²

= 2x² + 6x + 17

( 2x + 3 )( x - 5 ) + 2x( 3 - x ) + x - 10

= 2x² - 7x - 15 + 6x - 2x² + x - 10

= -25

( x + 5 )( x² - 5x + 25 ) - x( x - 4 )² + 16x

= x³ + 5³ - x( x² - 8x + 16 ) + 16x

= x³ + 125 - x³ + 8x² - 16x + 16

= 8x² + 125

( -x - 2 )³ + ( 2x - 4 )( x² + 2x + 4 ) - x²( x - 6 )

= -x³ - 6x² - 12x - 8 + 2x³ - 16 - x³ + 6x²

= -12x - 24 = -12( x + 2 )

Tương tự ... 

a, \(\left(x-1\right)^3-x\left(x-2\right)^2+1=x^3-3x^2+3x-1-x^3+4x^2-4x+1=x^2-x\)

b, \(2x\left(3x+2\right)-3x\left(2x+3\right)=6x^2+4x-6x^2-9x=-5x\)

c, \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)=x^3+6x^2+12x+8+x^2+6x+9-x^3-5x^2=2x^2+18x+17\)

Bạn cố gắng giúp mình nha bạnミ★NQ_NP:3★彡cảm ơn bạn nhiều

a) (2x+3)²-2(2x+3)(2x+5)+(2x+5)²

=4x²+12x+9-(4x+6)(2x+5)+4x²+20x+25

=4x²+12x+9-(8x²+12x+20x+30)+4x²+20x+25

=4x²+12x+9-8x²-12x-20x-30+4x²+20x+25

=4

b) (x²+x+1)(x²-x+1)(x²-1)

=((x²+1)²-x²)(x²-1)

=(x⁴+x²+1)(x²-1)

=x⁶+x⁴+x²-x⁴-x²-1

=x⁶-1

c)(a+b-c)²+(a-b+c)²-2(b-c)²

=a²+b²+c²+2ab-2ac-2bc+a²+b²+c²-2ab+2ac-2bc-2(b²-2bc+c²)

=2a²+2b²+2c²-4bc-2b²+4bc-2c²

=2a²

d) (a+b+c)²+(a-b-c)²+(b-c-a)²+(c-a-b)²

= a²+b²+c²+2ab+2ac+2bc+a²+b²+c²-2ab-2ac+2bc+a²+b²+c²+2bc-2ab+2ac+a²+b²+c²-2ac-2bc+2ab

=4a²+4b²+4c²+4ab+4bc

a) Ta có: \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)

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

=4

b) Ta có: \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)

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

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

\(=x^6-1\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a) \(4x\left(a-b\right)+6xy\left(b-a\right)\)

\(=4x\left(a-b\right)-6xy\left(a-b\right)\)

\(=\left(4x-6xy\right)\left(a-b\right)\)

\(=2x\left(2-3y\right)\left(a-b\right)\)

b) \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

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

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

\(=\left(8-2x\right)\left(2x+1\right)\)

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

g: \(\left(3x-1\right)^2-\left(x+3\right)^2\)

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

\(=\left(2x-4\right)\left(4x+2\right)\)

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