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

Bài 1

A= (x-2)(2x-1)-2x(x+3)=2x²-x-4x+2-2x²-6x=-11x+2

Bài 1:

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

\(A=2x^2-x-4x+2-2x^2-6x\)

\(A=-11x+2\)

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

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

\(B=-12x\)

c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)

\(C=12x^2+18x-12x^2+8x+3x-2\)

\(C=29x-2\)

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

\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)

\(D=36x-10\)

Bài 2: 

a: Ta có: \(2x\left(3x-5\right)\left(x+11\right)-3x\left(2x+3\right)\left(x+7\right)\)

\(=2x\left(3x^2+33x-5x-55\right)-3x\left(2x^2+14x+3x+21\right)\)

\(=6x^3+56x^2-110x-6x^2-51x^2-63x\)

\(=-117x\)

b: Ta có: \(\left(x^2+5x-6\right)\left(x-1\right)-\left(x+2\right)\left(x^2-x+1\right)-x\left(3x-10\right)\)

\(=x^3+4x^2-11x+6-\left(x^3-x^2+x+2x^2-2x+2\right)-3x^2+10x\)

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

=4

c: Ta có: \(\left(x^2+x+1\right)\left(x-1\right)-x^2\left(x+1\right)+x^2-5\)

\(=x^3-1-x^3-x^2+x^2-5\)

=-6

MK KO BT MK MỚI HO C LỚP 6

AI HỌC LỚP 6 CHO MK XIN

Bài 2 

a. (x-2y)2 =2x-4y

b. (2x^2 +3)2 =4x^2+6

c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)

d. (2x-1)3 = 6x-3

 Xin lỗi mik chỉ lm ổn bài 2 thôi!

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

2:

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

b: \(2\left(x-1\right)+x^2-x\)

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

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

c: \(3x^2+14x-5\)

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

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

3: 

a: \(2x\left(x-1\right)-2x^2=4\)

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

=>-2x=4

=>x=-2

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

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

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

=>-4x=3

=>x=-3/4

c: \(4x^2-25+\left(2x+5\right)^2=0\)

=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)

=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)

=>4x(2x+5)=0

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

bai dai dong qua

a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0

\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)

\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)

\(-5x-8=0\)

\(x=-\frac{8}{5}\)

Mai mik làm mấy bài kia sau

2/

b) ( cái bài này chịu)

c) (x+1)^3-(x-1)^3-6(x-1)^2=-10

(x+1-x+1)\(\left[\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)\(-6\left(x^2-2x+1\right)=-10\)

\(2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-6x^2+12x-6=-10\)

\(2\left(3x^2+1\right)-6x^2+12x-6=0\)

\(6x^2+2-6x^2+12x-6=-10\)

\(12x=-10+4\)

\(12x=-6=>x=-\frac{1}{2}\)

d) (5x-1)^2-(5x-4)(5x+4)=7

\(25x^2-10x+1-25x^2+16=7\)

-10x = 7 - 17

-10x = -10

x= 1

Câu còn lại bn làm tương tự

3/

a) 

Ta có: 

(a+b+c)^2=3(ab+bc+ca)

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 3ab + 3bc + 3ac

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 3ab - 3bc - 3ac = 0

a^2 + b^2 + c^2  - ac - bc - ab = 0

2a^2 + 2b^2 + 2c^2  - 2ac - 2bc - 2ab = 0

(a²-2ab+b²⁾+(a²-2ac+c²) + (b²-2bc +c²) = 0

(a-b)^2 + (a-c)^2 + (b-c)^2 =0

=> a=b=c