dài thế ai trả lời đc hả ?

tu lam di luoi vua thoi

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

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

(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0

(3x+2)(x+1)(1-2x)=0

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

Bài 5: 

a. 1 - 2y + y²

= (1 - y)²

b. (x + 1)² - 25

= (x + 1)² - 5²

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

= (x - 4)(x + 6)

c. 1 - 4x²

= 1² - (2x)²

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

d. 8 - 27x³

= 2³ - (3x)³

= (2 - 3x)(4 + 6x + 9x²)

e. (đề hơi khó hiểu ''x3'' !?)

g. x³ + 8y³

= (x + 2y)(x² - 2xy + y²)

Yêu cầu đề là gì vậy bạn?

A) -2x(3x+2)(3x-2)+5(x+2)² - (x-1)(2x+1)(2x+1)

= -2x(9x²-4)+5(x²+4x+4) - (x-1)(4x²-1)

= -18x³+8x+5x²+20x+20-(4x³-x-4x²+1)

= -18x³+5x²+28x+20-4x³+x+4x²+1

= -22x³+9x²+29x+21

B) (7x-8)(7x+8)-10(2x+3)²+5x(3x-2)²-4x(x-5)²

= 49x² - 64 -10(4x²+ 12x + 3) + 5x(9x² - 12x +4) - 4x(x² - 10x +25)

= 49x² - 64 -40x² - 120x - 30 + 45x³ - 60x² - 20x - 4x³ + 40x² -100x

= 41x³ -11x² -240x -94

C) \(\left(x^2-3\right)\left(x^2+3\right)-5x^2\left(x+1\right)^2-\left(x^2-3x\right)\left(x^2-2x\right)+4x\left(x+2\right)^2\)

\(\left(x^4-9\right)-5x^2\left(x^2+2x+1\right)-\left(x^4-2x^3-3x^3+6x^2\right)+4x\left(x^2+4x+4\right)\)

\(x^4-9-5x^4-10x^3-5x^2-x^4+5x^3-6x^2+4x^3+16x^2+16x\)

\(-5x^4-x^3+5x^2+20x-9\)

D) \(-6x^2\left(x+5\right)^2-\left(x-3\right)^2+\left(x^2-2\right)\left(2x^2+1\right)-4x^2\left(3x-4\right)^2\)

\(-6x^2\left(x^2+10x+25\right)-\left(x^2-6x+9\right)+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)

\(-6x^4-60x^3+150x^2-x^2+6x-9+2x^4-3x^2-2-36x^4+96x^3-64x^2\)

\(-40x^4+36x^3+82x^2+6x-11\)

Bài 1. Tính:

a) \(x^2\left(x-2x^3\right)\)

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

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

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

c. \(\left(x-2\right)\left(x^2+3x-4\right)\)

\(=x^3+3x^2-4x-2x^2-6x+8\)

\(=x^3+x^2-10x+8\)

d) \(\left(x-2\right)\left(x-x^2+4\right)\)

\(=x^2-x^3+4x-2x+2x^2-8\)

\(=3x^2-x^3+2x-8\)

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

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

f) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)

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

\(=\left(6x^2+x-2\right)\left(3-x\right)\)

\(=18x^2+3x-6-6x^3-x^2+2x\)

\(=17x^2+5x-6-6x^3\)

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

\(=x^3+3x^2-5x+3x^2+9x-15\)

\(=x^3+6x^2+4x-15\)

h) \(\left(xy-2\right)\left(x^3-2x-6\right)\)

\(=x^4y-2x^2y-6xy-2x^3+4x+12\)

i) \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)

\(=20x^3-5x^4+10x^3-4x^4+x^3-2x^2+8x^3-2x^2+4x-12x^2+3x-6\)

\(=39x^3-9x^4-16x^2+7x-6\)

Bài 5: Tìm x, biết

1) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2-9\right)-6=0\)

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow-4x+7=0\)

\(\Leftrightarrow-4x=-7\)

\(\Leftrightarrow x=\dfrac{-7}{-4}=\dfrac{7}{4}\)

Vậy \(x=\dfrac{7}{4}\)

2) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)-10=0\)

\(\Leftrightarrow4x^2-24x+36-4x^2+1-10=0\)

\(\Leftrightarrow-24x+27=0\)

\(\Leftrightarrow-24x=-27\)

\(\Leftrightarrow x=\dfrac{-27}{-24}=\dfrac{9}{8}\)

Vậy \(x=\dfrac{9}{8}\)

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

\(\Leftrightarrow\left(x^2-8x+16\right)-\left(x^2-4\right)-6=0\)

\(\Leftrightarrow x^2-8x+16-x^2+4-6=0\)

\(\Leftrightarrow-8x+14=0\)

\(\Leftrightarrow-8x=-14\)

\(\Leftrightarrow x=\dfrac{-14}{-8}=\dfrac{7}{4}\)

Vậy \(x=\dfrac{7}{4}\)

5) \(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)

\(\Leftrightarrow9\left(x^2+2x+1\right)-\left(9x^2-4\right)-10=0\)

\(\Leftrightarrow9x^2+18x+9-9x^2+4-10=0\)

\(\Leftrightarrow18x+3=0\)

\(\Leftrightarrow18x=-3\)

\(\Leftrightarrow x=\dfrac{-3}{18}=\dfrac{-1}{6}\)

Vậy \(x=\dfrac{-1}{6}\)

Bài 8: Phân tích đa thức thành nhân tử

1) \(5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=5\left(x-y-2x\right)\left(x-y+2x\right)\)

2 \(16x-5x^2-3\)

\(=-5x^2+16x-3\)

\(=-5x^2+15x+x-3\)

\(=-\left(5x^2-15x\right)+\left(x-3\right)\)

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

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

3) \(x^2-5x+5y-y^2\)

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

\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-5\right)\)

4) \(3x^2-6xy+3y^2-12z^2\)

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

\(=3\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)

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

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

5) \(x^2+4x+3\)

\(=x^2+x+3x+3\)

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

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

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

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

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

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

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

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

7) \(x^2-4x-5\)

\(=x^2+x-5x-5\)

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

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

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

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)