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24 tháng 7 2018

c,4x2-20x+25-4x2-20x-25+40x-1

=-1

Bài 1:

a) Ta có: \(\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)

\(=4x^2-4x+1+4\left(x^2+2x-3\right)-2\left(25-30x+9x^2\right)\)

\(=4x^2-4x+1+4x^2+8x-12-50+60x-18x^2\)

\(=-10x^2+64x-61\)

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

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

\(=-4a^2\)

c) Ta có: \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)

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

\(=\left(4x\right)^2=16x^2\)

d)

Sửa đề: \(\left(x^2+5x-1\right)^2+2\left(5x-1\right)\left(x^2+5x-1\right)+\left(5x-1\right)^2\)

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

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

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

\(=x^4+100x^2+4+20x^3-40x-4x^2\)

\(=x^4+20x^3+96x^2-40x+4\)

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

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

\(=x^3-x-x^3-1\)

=-x-1

f) Ta có: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x\left(x^2-16\right)-\left(x^4-1\right)\)

\(=x^3-16x-x^4+1\)

2 tháng 1 2022

tách nhỏ câu hỏi ra bạn

2 tháng 1 2022

cảm ơn yeu

21 tháng 6 2017

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

\(=5\left(4x^2-4x+1\right)+\left(4x-4\right)\cdot\left(x+3\right)+2\left(25-30x+9x^2\right)\)

\(=20x^2-20x+5+4x^2+12x-4x-12+50-60+18x^2\)

\(=42x^2-72x+43\)

2) \(C=\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a+1\right)^2\)

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

\(=4a^4+2a^2-4a^2+2a^2+1-4a^2-4a-1\)

\(=4a^4-4a^2-4a\)

3) Sky Sơn Tùng làm đúng rồi nhé.

4) \(E=\left(x^2-5x+1\right)^2+2\left(5x-1\right)\left(x^2-5x+1\right)\left(5x-1\right)^2\)

\(=x^4+27x^2+1-10x^3+250x^5-1400x^4+1030x^3-302x^2+40x-2\)

\(=-1399x^4-275x^2-1+1020x^3+250x^5+40x\)

5) \(F=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2\)

\(=\left[a^2+b^2-c^2-\left(a^2-b^2+c^2\right)\right]\cdot\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\)

\(=\left(a^2+b^2-c^2-a^2+b^2-c^2\right)\cdot2a^2\)

\(=\left(2b^2-2c^2\right)\cdot2a^2\)

\(=2\left(b^2-c^2\right)\cdot2a^2\)

\(=2\left(b-c\right)\left(b+c\right)\cdot2a^2\)

\(=2\cdot2a^2\cdot\left(b-c\right)\left(b+c\right)\)

\(=4a^2\cdot\left(b-c\right)\left(b+c\right)\)

6) \(G=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+\left(-c\right)^2+2ab-2ac-2bc-2\left(a^2+2ab+b^2\right)\)

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

\(=0+0+c^2+0+c^2\)

\(=2c^2\)

7) \(H=\left(a+c\right)\left(a-c\right)-\left(a-b-c\right)\left(a-b+c\right)+b\left(b-2x\right)\)

\(=a^2-c^2-\left[\left(a-b\right)^2-c^2\right]+b^2-2bx\)

\(=a^2-c^2-\left(a^2-2ab+b^2-c^2\right)+b^2-2bx\)

\(=a^2-b^2-a^2+2ab-b^2+c^2+b^2-2bx\)

\(=2ab-2bx\)

21 tháng 6 2017

\(D=\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)=\left(9x-1+1-5x\right)^2=\left(4x\right)^2=16x^2\)

26 tháng 5 2023

\(a,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=\left(3x^2+5x^2-8x^2\right)+\left(-6x-5x\right)+24\)

\(=0-11x+24\)

\(=-11x+24\)

\(b,\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x\)

\(=14x^2+7x-6x-3-5x^2-20x+2x+8-9x^2+17x\)

\(=\left(14x^2-5x^2-9x^2\right)+\left(7x-6x-20x+2x+17x\right)+\left(-3+8\right)\)

\(=0+0+5\)

\(=5\)

3 tháng 10 2018

a) Ta có x 6 + 2 x 3 + 3 x 3 − 1 . 3 x x + 1 . x 2 + x + 1 x 6 + 2 x 3 + 3 = 3 x x 2 − 1  

b) Gợi ý: a 3   +   2 a 2  - a - 2 = (a - 1)(a + 1) (a + 2)

Thực hiện phép tính từ trái qua phải thu được:  = 1 3

26 tháng 7 2021

a) A=(4-5x)2-(3+5x)2=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1

26 tháng 7 2021

B=(3x-1)(1+3x)-(3x+1)2=9x2-1-(3x+1)2=9x2-1-(9x2+6x+1)=9x2-1-9x2-6x-1=-6x-2=-2(3x+1)