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\(10^2=\left(3x-5y\right)^2=\left(\sqrt{3}.\sqrt{3}x-\sqrt{5}.\sqrt{5}y\right)^2\)
\(\Rightarrow100\le\left(3+5\right)\left(3x^2+5y^2\right)\)
\(\Rightarrow3x^2+5y^2\ge\frac{100}{8}=\frac{25}{2}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}y=-\frac{5}{4}\\x=\frac{5}{4}\end{matrix}\right.\)
1A,B,D
2 M=2
3 \(=\dfrac{3}{4x}\)
4 \(=\dfrac{4\left(x+y\right)}{x-y}=\dfrac{4x+4y}{x-y}\)
5 K rút gọn đc
6 \(=\dfrac{4\left(x-1\right)+2\left(x-1\right)}{6\left(x-1\right)}=\dfrac{6\left(x-1\right)}{6\left(x-1\right)}=1\)
a,3x3y3-15x2y2=3x2y2(xy-5)
b,2x(x-5y)+8y(5y-x)=2x(x-5y)-8y(x-5y)=(x-5y).(2x-8y)
c,(3x-1)2-16=(3x-1)2-42=(3x-1+4)(3x-1-4)=(3x+3)(3x-5)
d,x3-3x2+3x-1=x3-1-(3x2+3x)=x3-1-3x(x+1)=(x3-1-3x)(x+1)
e,125x3+1=(5x)3+13=(5x+1)(25x2-5x.1+12)
f,x3+6x2y+12xy2+8y3=x3+3.x2.2y+3.x.(2y)2+(2y)3=(x+2y)3
\(a,\left(6x+5y\right)\left(6x-5y\right)\)
\(=\left(6x\right)^2-\left(5y\right)^2\)
\(=36x^2-25x^2\)
\(b,\left(-4xy-5\right)\left(5-4xy\right)\)
\(=-\left(5+4xy\right)\left(5-4xy\right)\)
\(=-[5^2-\left(4xy\right)^2]\)
\(=-\left(25-16xy^2\right)\)
\(c,\left(3x-4\right)^2+2.\left(3x-4\right).\left(4-x\right)+\left(4-x\right)^2\)
\(=\left(3x-4\right)\left(3x-4+2\right)\left(4-x\right)\left(1+4-x\right)\)
\(=\left(3x-4\right)\left(3x-2\right)\left(4-x\right)\left(5-x\right)\)
(3x+5y) (3x - 5y )
=9x2-15xy+15xy-25y2
=9x2-25y2(-15xy+15xy)
=9x2-25y2
\(\left(3x+5y\right)\left(3x-5y\right)=9x^2-25y^2\)