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\(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)
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
\(B=\left(11-1\right)^3+1=10^3+1=1001\)
\(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=x^3-4x^2+16x+4x^2-16x+64\)
\(=x^3+64\)
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^2+3x^2y+9xy^2-3x^2y-9xy^2-27y^3\)
\(=\)\(x^2-27y^3\)
\(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3xy}+4y^2\right)\)
\(=\)\(\frac{x^3}{27}-\frac{2}{9xy}+\frac{4xy^2}{3}+\frac{2x^2y}{9}-\frac{4y}{3xy}+8y^3\)
làm nốt nha
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 ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
Bài 1:
a) \(25\left(x+2y\right)^2-16\left(2x-y\right)^2\)
\(=\left[5\left(x+2y\right)\right]^2-\left[4\left(2x-y\right)\right]^2\)
\(=\left[5\left(x+2y\right)-4\left(2x-y\right)\right]\left[5\left(x+2y\right)+4\left(2x-y\right)\right]\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(14y-3x\right)\left(13x+6y\right)\)
b) \(0,25\left(x-2y\right)^2-4\left(x+y\right)^2\)
\(=\left[\dfrac{1}{2}\left(x-2y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)
\(=\left[\dfrac{1}{2}\left(x-2y\right)-2\left(x+y\right)\right]\left[\dfrac{1}{2}\left(x-2y\right)+2\left(x+y\right)\right]\)
\(=\left(\dfrac{1}{2}x-y-2x-2y\right)\left(\dfrac{1}{2}x-y+2x+2y\right)\)
\(=\left(-\dfrac{3}{2}x-3y\right)\left(\dfrac{5}{2}x+y\right)\)
\(=-3\left(\dfrac{1}{2}x+y\right)\left(\dfrac{5}{2}x+y\right)\)
c) \(\dfrac{4}{9}\left(x-3y\right)^2-0,04\left(x+y\right)^2\)
\(=\left[\dfrac{2}{3}\left(x-3y\right)\right]^2-\left[\dfrac{1}{5}\left(x+y\right)\right]^2\)
\(=\left[\dfrac{2}{3}\left(x-3y\right)-\dfrac{1}{5}\left(x+y\right)\right]\left[\dfrac{2}{3}\left(x-3y\right)+\dfrac{1}{5}\left(x+y\right)\right]\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x-2y+\dfrac{1}{5}x+\dfrac{1}{5}y\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x-\dfrac{9}{5}y\right)\)
\(=\dfrac{1}{5}\left(\dfrac{7}{3}x-11y\right).\dfrac{1}{5}\left(\dfrac{13}{3}x-9y\right)\)
\(=\dfrac{1}{25}\left(\dfrac{7}{3}x-11y\right)\left(\dfrac{13}{3}x-9y\right)\)
d) \(-25x^2+30x-9\)
\(=-\left(25x^2-30x+9\right)\)
\(=-\left[\left(5x\right)^2-2.5x.3+3^2\right]\)
\(=-\left(5x-3\right)^2\)
Bài 2:
a) \(x^3y^2-x^2y^3-2x+2y\)
\(=x^2y^2\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2y^2-2\right)\)
Thay x = -1 và y = -2 vào ta được
\(=\left[-1-\left(-2\right)\right]\left[\left(-1\right)^2\left(-2\right)^2-2\right]\)
\(=1\left(4-2\right)\)
\(=2\)
b) \(5x^2-3x+3y-5y^2\)
\(=5\left(x^2-y^2\right)-3\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
Thay x = 3 và y = 1 vào ta được
\(=5\left(3-1\right)\left(3+1\right)-3\left(3-1\right)\)
\(=5.2.4-3.2\)
\(=34\)
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 :'(