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a) \(\text{ }x^4+y^4\ge x^3y+xy^3\)
\(\Leftrightarrow x^4+y^4-x^3y-xy^3\ge0\)
\(\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)(ĐPCM)
*NOTE: chứng minh đc vì (x-y)^2 >= 0 ; x^2 +xy +y^2 > 0
mình cũng làm đến nơi rồi nhưng sợ x^2+xy+y^2 chưa chắc lớn hơn 0 thanks bạn nhé
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\(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)\)
\(=1+x^2+y^2+x^2y^2+4xy+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x^2+y^2+2xy\right)+\left(x^2y^2+2xy+1\right)+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x+y\right)^2+\left(1+xy\right)^2+2\left(x+y\right)\left(1+xy\right)\)
\(=\left(x+y+1+xy\right)^2\) là SCP
(1+x2)(1+y2)+4xy+2(x+y)(1+xy)
= 1+y2+x2+x2y2+2xy+2xy+2(x+y)(1+xy)
=(x2+2xy+y2)+(x2y2+2xy+1)+2(x+y)(1+xy)
=(x+y)2+(xy+1)2+2(x+y)(1+xy)
=(x+y+xy+1)2
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c) \(\left(x+\dfrac{y}{x}\right)^3\)
\(=\left(\dfrac{x^2}{x}+\dfrac{y}{x}\right)^3\)
\(=\left(\dfrac{x^2+y}{x}\right)^3\)
\(=\dfrac{x^6+3x^4y+3x^2y^3+y^3}{x^3}\)
f) \(\left(x-\dfrac{1}{2}\right)^3\)
\(=x^3-3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3\)
\(=x^3-\dfrac{3}{2}x^2+\dfrac{3}{4}x-\dfrac{1}{8}\)
h) \(\left(x+\dfrac{y^2}{2}\right)^3\)
\(=\left(\dfrac{2x}{2}+\dfrac{y^2}{2}\right)^3\)
\(=\left(\dfrac{2x+y^2}{2}\right)^3\)
\(=\dfrac{8x^3+12x^2y^2+6xy^4+y^6}{8}\)
k) \(\left(x-\dfrac{1}{3}\right)^3\)
\(=x^3-3\cdot x^2\cdot\dfrac{1}{3}+3\cdot x\cdot\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3\)
\(=x^3-x^2+\dfrac{x}{3}-\dfrac{1}{27}\)
m) \(\left(x+\dfrac{y^2}{3}\right)^3\)
\(=\left(\dfrac{3x}{3}+\dfrac{y^2}{3}\right)^3\)
\(=\left(\dfrac{3x+y^2}{3}\right)^3\)
\(=\dfrac{27x^3+27x^2y^2+9xy^4+y^6}{27}\)
Q) \(2\left(x^2+\dfrac{1}{2}y\right)\left(2x^2-y\right)\)
\(=2\left(2x^4-x^2y+x^2y-\dfrac{1}{2}y^2\right)\)
\(=2\left(2x^4-\dfrac{1}{2}y^2\right)\)
\(=4x^4-y^2\)
rút gọn P=2/x-(x2/(x2-xy)+(x2-y2)/xy-y2/(y2-xy)):(x2-xy+y2)/(x-y)
r tìm gt P với |2x-1|=1 ; |y+1|=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+y=1\)
Áp dụng BĐT AM-GM, ta có:
\(\dfrac{x^2}{1}+\dfrac{y^2}{1}\ge\dfrac{\left(x+y\right)^2}{2}=\dfrac{1^2}{2}=\dfrac{1}{2}\)
--> \(x^2+y^2\ge\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left[\left(x+y\right)^2-2xy\right]\)
\(=2\left(x^2-xy+y^2\right)-3\left(1-2xy\right)\)
\(=2x^2-2xy+2y^2-3+6xy\)
\(=2x^2+4xy+2y^2-3\)
ko đúng bạn ơi
thanks bạn