Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
4:
a: 2003*2005=(2004-1)(2004+1)=2004^2-1<2004^2
b: 8(7^2+1)(7^4+1)(7^8+1)
=1/6*(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^2-1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^16-1)<7^16-1
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
mik chỉ biết bài 5 thôi !
\(a,=3x\left(y-z\right)-y\left(y-z\right)=\left(3x-y\right)\left(y-z\right)\\ b,=x^3\left(x-1\right)+x\left(x-1\right)=x\left(x^2+1\right)\left(x-1\right)\\ c,=x\left(y+z\right)+y\left(y+z\right)=\left(x+y\right)\left(y+z\right)\\ d,=\left(x-3\right)^2\\ e,=\left(x+2\right)^3\\ f,=\left(2x-x+y\right)\left(2x+x-y\right)=\left(x+y\right)\left(3x-y\right)\\ g,=\left(y+1\right)\left(5x-2\right)\\ h,=\left(x+2\right)^2\\ i,=x^2\left(x^2-2\right)\\ k,=3x\left(x-4y\right)\)
\(a,=\left(5x+1\right)^2-y^2=\left(5x-y+1\right)\left(5x+y+1\right)\\ b,Sửa:25\left(2x+1\right)^2-\left(3x+2\right)^2\\ =\left[5\left(2x+1\right)-3x-2\right]\left[5\left(2x+1\right)+3x+2\right]\\ =\left(10x+5-3x-2\right)\left(10x+5+3x+2\right)\\ \left(7x+3\right)\left(13x+7\right)\\ c,=2x^2-8x+3x-12=\left(x-4\right)\left(2x+3\right)\)
a: \(25x^2-\dfrac{10}{3}xy+\dfrac{1}{9}y^2=\left(5x-\dfrac{1}{3}y\right)^2\)
b: \(25x^2-15x+\dfrac{9}{4}=\left(5x-\dfrac{3}{2}\right)^2\)
c: \(\left(2x+\dfrac{1}{2}y\right)\left(4x^2-xy+\dfrac{1}{4}y^2\right)=8x^3+\dfrac{1}{8}y^3\)
d: \(\left(x^2-\dfrac{2}{3}\right)\left(x^4+\dfrac{2}{3}x^2+\dfrac{4}{9}\right)=x^6-\dfrac{8}{27}\)
a) \(-y^2+\dfrac{1}{9}\)
= \(-\left(y^2-\dfrac{1}{9}\right)\)
= \(-\left(y-\dfrac{1}{3}\right)\left(y+\dfrac{1}{3}\right)\)
b) \(4\left(x-3\right)^2-9\left(x+1\right)^2\)
= \(\left(2x-3\right)^2-\left(3x+3\right)^2\)
= \(\left(2x-3+3x+3\right)\left(2x-3-3x-3\right)\)
= \(5x\left(-x-6\right)\)
c) \(25x^2-20xy+4y^2\)
= \(\left(5x-2y\right)^2\)
d) \(-9x^2+12xy-4y^2\)
= \(-\left(9x^2-12xy+4y^2\right)\)
= \(-\left(3x-2y\right)^2\)
e) \(25x^2-\dfrac{1}{8}x^2y^2\)
= \(\left(5x-\dfrac{\sqrt{2}}{4}xy\right)\left(5x+\dfrac{\sqrt{2}}{4}xy\right)\)
f) \(9x^2+6x+1\)
= \(\left(3x+1\right)^2\)
okey! Vì you t sẽ chăm thêm 1 lần nữa!!!^^
\(a.-y^2+\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2-y^2=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\)
\(b,4\left(x-3\right)^2-9\left(x+1\right)^2=\left[2\left(x-3\right)\right]^2-\left[3\left(x+1\right)\right]^2=\left(2x-6\right)^2-\left(3x+3\right)^2=\left(2x-6-3x-3\right)\left(2x-6+3x+3\right)=\left(-x-9\right)\left(5x-3\right)\)\(c,25x^2-20xy+4y^2=\left(5x-2y\right)^2\)
\(d,-9x^2+12xy-4y^2=-\left(3x-2y\right)^2\)
\(e,25x^2-\dfrac{1}{8}x^2y^2=\left(5x-\dfrac{\sqrt{2}}{4}xy\right)\left(5x+\dfrac{\sqrt{2}}{4}xy\right)\)\(f,9x^2+6x+1=\left(3x+1\right)^2\)
