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16 tháng 9 2016

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

\(\left(2x+3\right)^2+\left(2x+3\right)\left(2x-6\right)+\left(x-3\right)^2=\left(2x+3\right)^2+2\left(2x+3\right)\left(x-3\right)+\left(x-3\right)^2=\left(2x+3+x-3\right)^2=\left(3x\right)^2=9x^2\)

máy pa lag ko vào đc tn

a: \(=x\sqrt{2}-\sqrt{\left(x\sqrt{2}+1\right)^2}=x\sqrt{2}-\left|x\sqrt{2}+1\right|\)

b: Khi A=-3 thì \(\left|x\sqrt{2}+1\right|=x\sqrt{2}+3\)

\(\Leftrightarrow x\sqrt{2}+1=-x\sqrt{2}-3\)

\(\Leftrightarrow2x\sqrt{2}=-4\)

hay \(x=-\sqrt{2}\)

a: \(=\dfrac{\left(x+1\right)\left[\left(3x-2\right)-\left(2x+5\right)\left(x-1\right)\right]}{x+1}\)

=3x-2-2x^2+2x-5x+5

=-2x^2+3

b: \(=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\)

c: =x^3-3x^2+3x-1-x^3-1+9x^2-1

=6x^2+3x-3

24 tháng 6 2023

\(a,\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]:\left(x+1\right)\)

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

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

\(=\left[\left(x+1\right)\left(3x-2-2x^2+2x-5x+5\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(-2x^2+3\right)\right].\dfrac{1}{x+1}\)

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

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

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

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

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

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

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

\(=-3x^2+3x-2-3x+9x^2-1+3x\)

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

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

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

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

\(=4x^2\)

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

\(=3x^2+12x+12-3x^2-16x-5+x^2+10x+25\)

\(=x^2+6x+32\)

a: =x^3+6x^2+12x+8-(x^3+3x^2+3x+1)

=x^3+6x^2+12x+8-x^3-3x^2-3x-1

=3x^2+9x+7

b: =x^3-9x^2+27x-27-x(x^2-6x+9)

=x^3-9x^2+27x-27-x^3+6x^2-9x

=-3x^2+18x-27

c: =x^3+3x^2+3x+1-x^3-x^2-2x^2-4x

=-x+1

HQ
Hà Quang Minh
Giáo viên
18 tháng 8 2023

\(a,\left(x+2\right)^3-\left(x+1\right)^3\\ =\left(x+2-x-1\right)\left(x^2+4x+4+x^2+3x+2+x^2+2x+1\right)\\ =3x^2+9x+7\\ b,\left(x-3\right)^3-x\left(x-3\right)^2\\ =x^3-6x^2+9x-27-x^3+6x^2-9x\\ =-27\)

 

23 tháng 12 2020

a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)

\(=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}-\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)

\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)

\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)

b) Ta có: \(x=3-2\sqrt{2}\)

\(=2-2\cdot\sqrt{2}\cdot1+1\)

\(=\left(\sqrt{2}-1\right)^2\)

Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được: 

\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)

\(=\dfrac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)

\(=\dfrac{\sqrt{2}-\sqrt{2}+1}{\sqrt{2}-1}\)

\(=\dfrac{1}{\sqrt{2}-1}\)

\(=\sqrt{2}+1\)

Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)

23 tháng 12 2020

cái x-3 ở tử phân tích kiểu j ra đc cái kia v bạn

 

22 tháng 10 2023

1:

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

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

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

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)

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

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

\(=9+10x-15=10x-6\)

2: 

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

\(=9x^2-12x+4-5x^2+20x+4x-4\)

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

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

29 tháng 10 2023

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

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

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

\(=6x^2+12x-16\)

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

\(=x^2-2x+1-9x^2-12x-4\)

\(=-8x^2-14x-3\)

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

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

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