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Bài 1:
\(a,\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(x^2-xy+1\right)=\left(x^3y^2+\dfrac{1}{2}x^2y^3\right)\left(x^2-xy+1\right)=x^5y^2-x^4y^3+x^3y^2+\dfrac{1}{2}x^3y^3-\dfrac{1}{2}x^3y^4+\dfrac{1}{2}x^2y^3\)
\(b,\left(\dfrac{1}{2}x-1\right)\left(2x-3\right)=x^2-\dfrac{3}{2}x-2x+3=x^2-\dfrac{7}{2}x+3\)\(c,\left(x-7\right)\left(x-5\right)=x^2-5x-7x+35=x^2-12x+35\)\(f,\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x-1\right)=\left(x^2-\dfrac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\dfrac{1}{4}\)Bài 2 ,
\(\left(x-1\right)\left(x^2+x+1\right)=x^3+x^2+x-x^2-x-1=x^3-1\Rightarrowđpcm\)\(b,\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4+x^3y+x^2y^2+y^3x+x^3y-x^2y^2-xy^3-y^4=x^4-y^4\)
2: \(=a^2\left(a+3\right)+4\left(a+3\right)=\left(a+3\right)\left(a^2+4\right)\)
3: \(=\left(2a-1\right)^2-4b^2\)
\(=\left(2a-1-2b\right)\left(2a-1+2b\right)\)
4: \(=-\left(x^2+x-2\right)=-\left(x+2\right)\left(x-1\right)\)
5: \(=7\left(x^2-2xy^2+y^4\right)=7\left(x-y^2\right)^2\)
6: \(=\left(x+2\right)^2-y^2=\left(x+2+y\right)\left(x+2-y\right)\)
Bài 1:
\(\left(2x+3\right)^2+\left(2x-3\right)^2+2\left(2x+3\right)\left(2x-3\right)\)
\(=\left(2x+3+2x-3\right)^2=\left(4x\right)^2=16x^2\)
Bài 2:
a, \(\left(x^2+xy+y^2\right)\left(x-y\right)+\left(x^2-xy+y^2\right)\left(x+y\right)\)
\(=x^3-y^3+x^3+y^3=2x^3\)
b, \(\left(2a-b\right)\left(4a^2+2ab+b^2\right)\)
\(=\left(2a\right)^3-b^3=8a^3-b^3\)
c, \(13x\left(3-x\right)-12\left(x+1\right)\)
\(=39x-13x^2-12x-12=-13x^2-27x-12\)
d, \(\left(2x-1\right)\left(x+12\right)\left(x^2+14\right)\)
\(=\left(2x^2+24x-x-12\right)\left(x^2+14\right)\)
\(=2x^4+23x^3-12x^2+28x^2+322x-168\)
\(=2x^4+23x^3+16x^2+322x-168\)
e, Giống câu b
Chúc bạn học tốt!!!
\(A=a^4+a^3+a^3b+a^2b\)
\(A=a\left(a^3+a^2\right)+b\left(a^3+a^2\right)\)
\(A=\left(a+b\right)\left(a^3+a^2\right)\)
\(A=a^2\left(a+1\right)\left(a+b\right)\)
Bài 1:
\(A=x^2+6x+5=x^2+5x+x+5=x\left(x+5\right)+\left(x+5\right)=\left(x+1\right)\left(x+5\right)\)
Đặt \(a=x^2-x+2\) ta có:
\(B=\left(a-1\right).a-12=a^2-a-12=a^2+3a-4a-12=a\left(a+3\right)-4\left(a+3\right)=\left(a+3\right)\left(a-4\right)\)
Thay a = x2 - x + 2 vào ta được:
\(\left(x^2-x+2-4\right)\left(x^2-x+2+3\right)=\left(x^2-x-2\right)\left(x^2-x+5\right)=\left(x+1\right)\left(x-2\right)\left(x^2-x+5\right)\)
a: \(=\left(x^2-4\right)\left(x^2+4\right)-x^2+3\)
\(=x^4-16-x^2+3\)
\(=x^4-x^2-13\)
b: \(=x^3-6x^2+12x-8-x^3-1+6x^2-12x+6\)
\(=-3\)
c: \(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2-b^3-6a^2b\)
\(=2b^2\)