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\(a,=\left(3+x\right)\left(9-3x+x^2\right)\\ b,=\left(4x+0,1\right)\left(16x^2-0,4x+0,01\right)\\ c,=\left(2-3x\right)\left(4+6x+9x^2\right)\\ d,=\left(\dfrac{x}{5}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{25}+\dfrac{xy}{15}+\dfrac{y^2}{9}\right)\)
2x y 2 + x 2 y 4 + 1 = x y 2 2 + 2.x y 2 .1 + 1 2 = x y 2 + 1 2
a,16x2-9
=42x2-32
=(4x-3)(4x+3) HĐT thứ 3
b,9a2-25b2
=32a2-52b2
=(3a-5b)(3a+5b) HĐT thứ 3
c,81-y4
=32.32-y2.y2
=(32-y2)
=(3-y)(3+y) HĐT thứ 3
d,(2x+y)2-1
=(2x+y-1)(2x+y-1) HĐT thứ 3
e,(x+y+z)2-(x-y-z)2
cái này là HĐT thứ 8 mở rộng bạn lên mạng tìm nha
a) \(16x^2-9=\left(4x\right)^2-3^2=\left(4x-3\right).\left(4x+3\right)\)
b) \(9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right).\left(3a+5b^2\right)\)
c) \(81-y^4=9^2-\left(y^2\right)^2=\left(9-y^2\right).\left(9+y^2\right)\)
d)\(\left(2x+y\right)^2-1=\left(2x+y\right)^2-1^2=\left(2x+y-1\right).\left(2x+y+1\right)\)
Ta có: \(81+\left(9-x^2\right)^2\)
Tương tự: \(81+-x^4+18x^2+-81\)
\(=\left(-x^4\right)+\left(18x^2\right)+\left(81+-81\right)\)
\(=-x^4+18x^2\)
\(81-\left(9-x^2\right)^2\)
\(=\)\(9^2-\left(3^2-x^2\right)^2\)
\(=\)\(9^2-\left(9-x^2\right)^2\)
\(=\)\(\left(9-9+x\right)\left(9+9-x\right)\)
\(=\)\(x\left(18-x\right)\)
Chúc bạn học tốt ~
\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+5x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ c,=\left(x+\dfrac{1}{5}y\right)\left(x^2-\dfrac{1}{5}xy+\dfrac{1}{25}y^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ e,=\left(x-1+3\right)\left[\left(x-1\right)^2-3\left(x-1\right)+9\right]\\ =\left(x+2\right)\left(x^2-2x+1-3x+3+9\right)\\ =\left(x+2\right)\left(x^2-5x+13\right)\\ f,=\left(\dfrac{x^2}{2}-y^2\right)\left(\dfrac{x^4}{4}+\dfrac{x^2y^2}{2}+y^4\right)\)
\(\left(2x+3\right)^2-y^2=\left(2x+y+3\right)\left(2x-y+3\right)\)
\(a,27-x^3\)
\(=3^3-x^3\)
\(=\left(3-x\right)\left(9+3x+x^2\right)\)
Các câu còn lại lm tương tự nhé.
hok tốt!
a) \(27-x^3=\left(3-x\right)\left(9+3x+x^2\right)\)
b) \(8x^3+0,001=\left(2x+0,1\right)\left(4x^2-0,2x+0,01\right)\)
c) \(\frac{x^3}{125}-\frac{y^3}{27}=\left(\frac{x}{5}-\frac{y}{3}\right)\left(\frac{x^2}{25}+\frac{xy}{15}+\frac{y^2}{9}\right)\)
p/s: chúc bạn học tốt
\(a,=16^2-3^2=\left(16+3\right)\left(16-3\right)\)
\(c,=9^2-\left(y^2\right)^2=\left(9+y^2\right)\left(9-y^2\right)\)
\(b,\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right)\left(3a+5b^2\right)\)
a/ 16^2-9=16^2-3^2=(16-9)(16+9)
b/ 9a^2-25b^4=(3a)^2-(5b)^2=(3a-5b)(3a+5b)
c/ 81-y^4=9^2-(y^2)^2=(9-y^2)(9+y^2)=(3-y)(3+y)(9+y^2)