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a) \(\left(2x^3-y^2\right)^3\)
\(=\left(2x^3\right)^3-3\cdot\left(2x^3\right)^2\cdot y^2+3\cdot2x^3\cdot\left(y^2\right)^{^2}-\left(y^2\right)^3\)
\(=8x^9-3\cdot4x^6y^2+3\cdot2x^3y^4-y^6\)
\(=8x^9-12x^6y^2+6x^3y^4-y^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x+2y+z\right)\left(x+2y-z\right)\)
\(=\left(x+2y\right)^2-z^2\)
\(=x^2+4xy+4y^2-z^2\)
d) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y-\dfrac{1}{2}x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\dfrac{3}{2}x^7y-\dfrac{1}{8}x^6\)
e) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=\left(x^2-3\right)\left(4x^2+9\right)\)
\(=4x^4+9x^2-12x^2-27\)
\(=4x^4-3x^2-27\)
f) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(=\left(2x\right)^3-1^3\)
\(=8x^3-1\)
\(a,\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)\(b,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)\(d,\left(2x^3y-0,5x^2\right)^3=8x^9y^3-6x^4y^2x^2+3x^3yx^4-0,125x^6=8x^9y^3-6x^6y^2+3x^7y-0,125x^6\)
Bài 1:
a) \(\left(x+y\right)^2-y^2=x^2+2xy+y^2-y^2=x^2+2xy=x\left(x+2y\right)\)
b) Sửa đề: \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)^2\left(x+y\right)^2\)
c) \(x\left(x-3y\right)^2+y\left(y-3x\right)^2=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\)
\(=x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y\)
\(=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3\)
Bài 2:
a) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)
\(=2a\left(a^2+3b^2\right)\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(b^2+3a^2\right)\)
\(a,=x^2+x+\dfrac{1}{4}\\ b,=4x^2+2x+\dfrac{1}{4}\\ c,=x^2-2+\dfrac{1}{x^2}\\ d,=4x^2+\dfrac{8}{3}x+\dfrac{4}{9}x^2\\ e,=a^2-1\\ f,=25x^4-4\)
\(a,\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)
\(b,\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)
\(c,\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)
\(d,\left(\dfrac{2x+2}{3x}\right)^2=\dfrac{\left(2x+2\right)^2}{9x^2}=\dfrac{4x^2+8x+4}{9x^2}\)
\(e,\left(a-1\right).\left(a+1\right)=a^2-1\)
\(f,\left(5x^2-2\right).\left(5x^2+2\right)=25x^4-4\)
1.a (3x-2y)2= (3x)2 - 2. 3x . 2y - (2y)2 = 9x2 - 12xy - 4y2
2.b (2x - 1/2)2 = (2x)2 - 2.2x.1/2 - (1/2)2= 4x2 - 2 - 1/4
3.c (x/2 - y) (x/2+y)= (x/2)2 - (y)2 = x/4 - y2
Bài 1 :
\(\left(3x-2y\right)^2=9x^2-12xy+4y^2\)
\(\left(2x-\frac{1}{2}\right)^2=4x^2-4x+\frac{1}{4}\)
\(\left(\frac{x}{2}-y\right)\left(\frac{x}{2}+y\right)=\frac{x^2}{4}-y^2\)
\(\left(x+\frac{1}{3}\right)^3=x^3+x^2+\frac{1}{3}x+\frac{1}{27}\)
\(\left(x-2\right)\left(x^2+2x+2^2\right)=x^3-8\)
a) Ta có: \(A=\left(x^3-x^2y+xy^2-y^3\right)\left(x+y\right)\)
\(=x^4+x^3y-x^3y-x^2y^2+x^2y^2+xy^3-xy^3-y^4\)
\(=x^4-y^4\)
Thay x=2 và \(y=-\frac{1}{2}\) vào biểu thức \(A=x^4-y^4\), ta được:
\(A=2^4-\left(-\frac{1}{2}\right)^4\)
\(=16-\frac{1}{16}\)
\(=\frac{255}{16}\)
Vậy: \(\frac{255}{16}\) là giá trị của biểu thức \(A=\left(x^3-x^2y+xy^2-y^3\right)\left(x+y\right)\) tại x=2 và \(y=-\frac{1}{2}\)
b) Ta có: \(B=\left(a-b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\)
\(=a^5+a^4b+a^3b^2+a^2b^3+ab^4-a^4b-a^3b^2-a^2b^3-ab^4-b^5\)
\(=a^5-b^5\)
Thay a=3 và b=-2 vào biểu thức \(B=a^5-b^5\), ta được:
\(B=3^5-\left(-2\right)^5\)
\(=243-\left(-32\right)\)
\(=243+32=275\)
Vậy: 275 là giá trị của biểu thức \(B=\left(a-b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\) tại a=3 và b=-2
c) Ta có: \(C=\left(x^2-2xy+2y^2\right)\left(x^2+y^2\right)+2x^3-3x^2y^2+2xy^3\)
\(=x^4+x^2y^2-2x^3y-2xy^3+2x^2y^2+2y^4+2x^3-3x^2y^2+2xy^3\)
\(=x^4-2x^3y+2y^4+2x^3\)
Thay \(x=y=\frac{-1}{2}\) vào biểu thức \(C=x^4-2x^3y+2y^4+2x^3\), ta được:
\(C=\left(-\frac{1}{2}\right)^4-2\cdot\left(-\frac{1}{2}\right)^3\cdot\frac{-1}{2}+2\cdot\left(-\frac{1}{2}\right)^4+2\cdot\left(-\frac{1}{2}\right)^3\)
\(=\frac{1}{16}-2\cdot\frac{-1}{8}\cdot\frac{-1}{2}+2\cdot\frac{1}{16}+2\cdot\frac{-1}{8}\)
\(=\frac{1}{16}-\frac{1}{8}+\frac{1}{8}-\frac{1}{4}\)
\(=\frac{1}{16}-\frac{1}{4}=\frac{1}{16}-\frac{4}{16}=\frac{-3}{16}\)
Vậy: \(-\frac{3}{16}\) là giá trị của biểu thức \(C=\left(x^2-2xy+2y^2\right)\left(x^2+y^2\right)+2x^3-3x^2y^2+2xy^3\) tại \(x=y=\frac{-1}{2}\)
\(A=4x^2-5xy+3y^2\\\Rightarrow 2A=2\cdot(4x^2-5xy+3y^2)\\\Rightarrow2A=8x^2-10xy+6y^2\\B=3x^2+2xy+y^2\\\Rightarrow3B=3\cdot(3x^2+2xy+y^2)\\\Rightarrow3B=9x^2+6xy+3y^2\\C=-x^2+3xy+2y^2\)
Khi đó: $2A-3B-C$
$=(8x^2-10xy+6y^2)-(9x^2+6xy+3y^2)-(-x^2+3xy+2y^2)$
$=8x^2-10xy+6y^2-9x^2-6xy-3y^2+x^2-3xy-2y^2$
$=(8x^2-9x^2+x^2)+(-10xy-6xy-3xy)+(6y^2-3y^2-2y^2)$
$=-19xy+y^2$
2A-3B-C
\(=2\left(4x^2-5xy+3y^2\right)-3\left(3x^2+2xy+y^2\right)+x^2-3xy-2y^2\)
\(=8x^2-10xy+6y^2-9x^2-6xy-3y^2+x^2-3xy-2y^2\)
\(=-19xy+y^2\)
