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5 tháng 8 2016

2: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

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

Đặt \(x^2+x+1=a\)ta có

\(a\left(a+1\right)-12=a^2+a-12=a^2+4a-3a-12=a\left(a+4\right)-3\left(a+4\right)=\left(a+4\right)\left(a-3\right)\)

Thay \(a=x^2+x+1\)ta được

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

\(=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)Kl...

3. \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+7+8\right)+15\)

Đặt \(x^2+8x+7=a\) Ta có

\(a\left(a+8\right)+15=a^2+8a+15=a^2+5a+3a+15=a\left(a+5\right)+3\left(a+5\right)=\left(a+5\right)\left(a+3\right)\)

Thay \(a=x^2+8x+15\)ta được

\(\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x^2+6x+2x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x+6\right)\left(x+2\right)\left(x^2+8x+10\right)\)

29 tháng 11 2023

bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)

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

\(=\dfrac{x^2-2x-x^2-2x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{4x}{x^2-4}\)

Bài 2:

1: \(x^2y^2-8-1\)

\(=x^2y^2-9\)

\(=\left(xy-3\right)\left(xy+3\right)\)

2: \(x^3y-2x^2y+xy-xy^3\)

\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)

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

\(=xy\left[\left(x-1\right)^2-y^2\right]\)

\(=xy\left(x-1-y\right)\left(x-1+y\right)\)

3: \(x^3-2x^2y+xy^2\)

\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)

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

4: \(x^2+2x-y^2+1\)

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

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

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

5: \(x^2+2x-4y^2+1\)

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

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

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

6: \(x^2-6x-y^2+9\)

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

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

14 tháng 8 2018

mới lớp 1 thôi

14 tháng 8 2018

Làm tử tế giúp mình đi,,,,,,,

17 tháng 8 2018

\(\left(x^2+x\right)^2-2x^2-2x-15\)

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

\(=\left(x^2+x\right)^2-\left[\left(2x^2+2x\right)+15\right]\)

\(=\left(x^2+x\right)^2-\left[2.\left(x^2+x\right)+15\right]\)

\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\) \(\left(1\right)\)

đặt \(x^2+x=t\)

\(\left(1\right)\)\(=\)  \(t^2-2t-15\)

            \(=\left(t-1\right)^2-16\)

            \(=\left(t-1-4\right)\left(t-1+4\right)\)

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

thay \(t=x^2+x\) ta có

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

các câu còn lại tương tự nha

học tốt 

31 tháng 7 2017

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

\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)

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

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

2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)

Đặt  \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)

\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)

Thay a , ta có :

\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)