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\(\frac{\left(x-y\right)^3-3xy\left(x+y\right)+y^3}{x-6y}\)
\(=\frac{x^3-3x^2y+3xy^2-y^3-3x^2y-3xy^2+y^3}{x-6y}\)
\(=\frac{x^3-6x^2y}{x-6y}\)
\(=\frac{x^2\left(x-6y\right)}{x-6y}\)
\(=x^2\)
chúc bạn học giỏi
(x-y)3 - 3xy(x+y) + y3
x - 6y
x3 - 3x2y +3xy2 - y3 - 3x2y - 3xy2 + y3
x - 6y
x3 - 6x2y
x - 6y
x2(x-6y)
x - 6y
= x2
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a)ĐK: a>0 b>0 nhé bạn đề thiếu
(a-b)2\(\ge\)0
<=>a2+b2\(\ge\)2ab
<=>a2+2ab+b2\(\ge\)4ab
<=>(a+b)2\(\ge\)4ab
<=>\(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)
<=>\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
Dấu "=" xảy ra <=> (a-b)2=0<=>a=b
=>A\(\ge\)\(\left(a+b\right)\dfrac{4}{a+b}=4\)(đpcm)
b)\(B=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{a+c}{b}=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
Áp dụng bất đẳng thức cosi x+y\(\ge\)2\(\sqrt{xy}\)cho 2 số dương x;y ta có:
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{ac}{ca}}=2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\)
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\)
Dấu "=" xảy ra khi và chỉ khi:\(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{c}{a}\\\dfrac{b}{c}=\dfrac{c}{b}\\\dfrac{a}{b}=\dfrac{b}{a}\end{matrix}\right.\)\(\Leftrightarrow\)a=b=c
=>B\(\ge2+2+2=6\)(đpcm)
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a) \(VT=\left(x+y\right)^2-y^2=x^2+2xy+y^2-y^2\)
\(=x^2+2xy=x\left(x+2y\right)=VP\)
b) \(VT=\left(x^2+y^2\right)^2-\left(2xy\right)^2=x^4+2x^2y^2+y^4-4x^2y^2\)
\(=x^4-2x^2y^2+y^4=\left(x^2-y^2\right)^2=\left[\left(x-y\right)\left(x+y\right)\right]^2\)
\(=\left(x-y\right)^2\left(x+y\right)^2=VP\)
c) \(VT=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2=3xy\left(x+y\right)=VP\)
d) \(VT=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3=2y\left(y^2+3x^2\right)=VP\)
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Lời giải:
$A=(9+1)(9^2+1)(9^4+1)(9^8+1)$
$8A=(9-1)(9+1)(9^2+1)(9^4+1)(9^8+1)$
$=(9^2-1)(9^2+1)(9^4+1)(9^8+1)$
$=(9^4-1)(9^4+1)(9^8+1)$
$=(9^8-1)(9^8+1)=9^{16}-1$
$\Rightarrow A=\frac{9^{16}-1}{8}$
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