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\(a.\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)\\ =x^4-\frac{4}{25}y^2\)
\(b.\left(2x+y^2\right)^3\\ =8x^3+12x^2y^2+6xy^4+y^6\)
\(c.\left(3x^2-2y\right)^3\\ =27x^6-54x^4y+36x^2y^2-8y^3\)
\(\left(x+4\right)\left(x^2-4x+16\right)\\ =x^3+64\)
\(e.\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)\\ =x^6-\frac{1}{27}\)
6) Ta có
\(A=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\)
\(=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{xy+2xz+yz+2xy+zx+2yz}\)
\(\Leftrightarrow A\ge\frac{1}{3\left(xy+yz+zx\right)}\ge\frac{1}{3\left(x^2+y^2+z^2\right)}=\frac{1}{3}\)
\(1,\left(\frac{a}{3}+4y\right)^2=\frac{a^2}{9}+\frac{8ay}{3}+16y^2\)
\(2,\)Bạn xem lại đề bài giùm mk nhé
\(\left(x^2+\frac{2}{5}y\right).\left(x^2-\frac{2}{5}y\right)=\left(x^2\right)^2-\left(\frac{2}{5}y\right)^2=x^4-\frac{4}{25}y^2\)
b) Ta có: \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(2x^2+2y^2-x^2-2xy-y^2\ge0\)
\(x^2-2xy+y^2\ge0\)
\(\left(x-y\right)^2\ge0\) luôn đúng \(\forall x;y\)
Vậy \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\left(đpcm\right)\)
a)=(x+1/4)^2=100^2=10000
b) =x^2-(y+1)^2=(x-y-1)(x+y+1)=86.100=8600
bằng 20
a. x2 + \(\frac{1}{2}x\)+ \(\frac{1}{16}\)
= x2 + 2.x\(\frac{1}{4}\)+ \(\left(\frac{1}{4}\right)^2\)
= (x + \(\frac{1}{4}\))2
= (99,75 + 0,25)2
= 1002 = 10000
b. x2 - y2 - 2y - 1
= x2 - (y2 + 2y + 1)
= x2 - (y + 1)2
= (x + y + 1)(x - y - 1)
= (93 + 6 + 1)(93 - 6 - 1)
= 100 . 86 = 8600
a)\(x^2+\frac{1}{2}x+\frac{1}{16}=x^2+2.\frac{1}{4}x+\frac{1}{4^2}\)\(=\left(x+\frac{1}{4}\right)^2=\left(99,75+\frac{1}{4}\right)^2\)\(=100^2=1000\)
b)\(x^2-y^2-2y-1=x^2-\left(y+1\right)^2\)\(=\left(x-y-1\right)\left(x+y+1\right)\)\(=\left(93-6-1\right).\left(93+6+1\right)\)
=86.100=8600