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a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
\(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)
bài 2 áp dụng hằng đẳng thức bạn nhé
bài 3\(A=\left(x^3+3x^2+3x+1\right)+5\)
\(=\left(x+1\right)^3+5\) thay x=19 vào ta được
\(A=20^3+5=8005\)
\(B=\left(x^3-3x^2+3x-1\right)+1\)
\(=\left(x-1\right)^3+1\)
thay x=11 vào ta được
\(B=\left(11-1\right)^3+1=10^3+1=1001\)
a) \(\dfrac{1}{8}x^3y^3-27=\left(\dfrac{1}{2}xy\right)^3-3^3=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}x^2y^2+\dfrac{1}{6}xy+9\right)\)
b)\(\dfrac{8}{125}x^3+27y^3=\left(\dfrac{2}{5}x\right)^3+\left(3y\right)^3=\left(\dfrac{2}{5}x+3y\right)\left(\dfrac{4}{25}x^2-\dfrac{6}{5}xy+9y^2\right)\)
c) \(0.008x^6-27y^3=\left(0.2x^2\right)^3-\left(3y\right)^3=\left(0.2x^2-3y\right)\left(0.04x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d)\(\left(2x+y\right)^3-\left(x-y\right)^3=\left(2x+y-x+y\right)[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2]\\ =\left(x+2y\right)\left(4x^2+4xy+y^2+2x^2-2xy+xy-y^2+x^2-2xy+y^2\right)\\ =\left(x+2y\right)\left(6x^2+xy+y^2\right)\)
Bài 1:
a) \(\dfrac{1}{8}x^3y^3-27\)
\(=\left(\dfrac{1}{2}xy\right)^3-3^3\)
\(=\left(\dfrac{1}{2}xy-3\right)\left[\left(\dfrac{1}{2}xy\right)^2+\dfrac{1}{2}xy.3+3^2\right]\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}xy+\dfrac{3}{2}xy+9\right)\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{7}{4}xy+9\right)\)
b) \(\dfrac{8}{125}x^3+\dfrac{1}{8}y^3\)
\(=\left(\dfrac{2}{5}x\right)^3+\left(\dfrac{1}{2}y\right)^3\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left[\left(\dfrac{2}{5}x\right)^2-\dfrac{2}{5}x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\right]\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left(\dfrac{4}{25}x-\dfrac{1}{5}xy+\dfrac{1}{4}y\right)\)
c) \(0.008x^6-27y^3\)
\(=\left(\dfrac{1}{5}x^2\right)^3-\left(3y\right)^3\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left[\left(\dfrac{1}{5}x^2\right)^2+\dfrac{1}{5}x^2.3y+\left(3y\right)^2\right]\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left(\dfrac{1}{25}x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d) \(\left(2x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(2x+y\right)-\left(x-y\right)\right]\left[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(2x+y-x+y\right)\left(4x^2+4xy+y^2+2x^3-2xy+xy-y^2+x^2-2xy+y^2\right)\)
\(=\left(x-2y\right)\left(4x^2+2x^3+xy\right)\)
1) \(-4x^5\left(x^3-4x^2+7x-3\right)\)
\(=-4x^8+16x^7-28x^6+12x^5\)
2) \(3x^4\left(-2x^3+5x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)\)
\(=-6x^7+15x^6-2x^5+x^4\)
3) \(-5x^2y^4\left(3x^2y^3-2x^3y^2-xy\right)\)
\(=-15x^4y^7+10x^5y^6+5x^3y^5\)
4) \(4x^3y^2\left(-2x^2y+4x^4-3y^2\right)\)
\(=-8x^5y^3+16x^7y^2-12x^3y^4\)
1,2,3,4 không tính được.
`5)(2x-1/2)^2`
`=(2x)^2-2+(1/2)^2`
`=4x^2-2+1/4`
`6)(x+1/4)^2`
`=x^2+1/2x+1/16`
tính theo kiểu hằng đẳng thức đáng nhớ ý bạn :'(