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

Tính
a) ( 2x + 5 )² + ( 2x + 5 )² - 2(2x + 3 ) (2x + 5 )

=> Sai đề

b) ( x - 3 ) ( x + 5 ) - ( x - 3 ) ²

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

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

=(x-3).8

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) 3x² - x(3x - 5) = 9

3x² - 3x² + 5x = 9

5x = 9

x = 9/5

--------------------

B) 5x² + 9x - 2 = 0

5x² + 10x - x - 2 = 0

(5x² + 10x) - (x + 2) = 0

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

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

x + 2 = 0 hoặc 5x - 1 = 0

*) x + 2 = 0

x = -2

*) 5x - 1 = 0

5x = 1

x = 1/5

Vậy x = -2; x = 1/5

---------------------

D) 4(5 - 3x) = 5x - 5

20 - 12x = 5x - 5

-12x - 5x = -5 - 20

-17x = -25

x = 25/17

--------------------

E) 2x² - 11x + 14 = 0

2x² - 4x - 7x + 14 = 0

(2x² - 4x) - (7x - 14) = 0

2x(x - 2) - 7(x - 2) = 0

(x - 2)(2x - 7) = 0

x - 2 = 0 hoặc 2x - 7 = 0

*) x - 2 = 0

x = 2

*) 2x - 7 = 0

2x = 7

x = 7/2

Vậy x = 2; x = 7/2

Câu C và F ghi đề bằng công thức đúng lại em

c: =>x^2-(x-2)(x+5)=2x-1

=>x^2-x^2-5x+2x+10=2x-1

=>3x+10=2x-1

=>x=-11

f: =>3x-5=4(2x+3)

=>8x+12=3x-5

=>5x=-17

=>x=-17/5

\(a,\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\left(x\ne2;x\ne3\right)\\ \Leftrightarrow\dfrac{5}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}=0\\\Leftrightarrow\dfrac{5-\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=0 \\ \Leftrightarrow5-x^2+9=0\\ \Leftrightarrow14-x^2=0\\ \Leftrightarrow x^2=14\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{14}\\x=-\sqrt{14}\end{matrix}\right.\)

\(b,\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\left(x\ne-1;x\ne3\right)\\ \Leftrightarrow\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{x}{2\left(3-x\right)}\\ \Leftrightarrow\dfrac{x\left(x-3\right)-2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\\ \Leftrightarrow x^2-3x-4x=-x^2-x\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(c,\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\left(x\ne1\right)\\ \Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ \Leftrightarrow-2x^2+x+1=2x^2-2x\\ \Leftrightarrow4x^2-3x-1=0\\ \Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)

\(d,\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}=\dfrac{5-x}{2x^2+10x}\left(x\ne5;x\ne-5\right)\\ \Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\\ \Leftrightarrow\dfrac{x^2+25x-2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x+5\right)\left(x-5\right)}\\ \Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=-\left(x^2-10x+25\right)\\ \Leftrightarrow x^2+25x-2x^2-20x-50=-x^2+10x-25\\ \Leftrightarrow-5x=25\\ \Leftrightarrow x=-5\)

Tick nha

f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)

\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)

\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)

\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)

\(-x^3=27\)

\(x=-3\)

Bài 1:

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(6x-9+4-2x=-3\)

\(4x=-2\)

\(x=-\frac{1}{2}\)

b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)

\(2x^3-4x+x^2-2x^3-x^2=-12\)

\(-4x=-12\)

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

Bài 2:

a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(6x^2+9x-6x^2-15x+4x+10=8\)

\(-2x=8\)

\(x=-4\)

b/ \(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)

\(4x^2-4x-3x^2+15-x^2=-7\)

\(-4x=-22\)

\(x=\frac{11}{2}\)

c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)

\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)

\(6x-4x-10-12x+6x+12=-6\)

\(-4x=-8\)

\(x=2\)

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)