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Câu a : Ta có : \(x^3+x^2z+y^2z-xyz+y^3=0\)
\(\Leftrightarrow\left(x^3+y^3\right)+\left(x^2z+y^2z-xyz\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow\left(x^2-xy+y^2\right)\left(x+y+z\right)=0\)
\(\Leftrightarrow x+y+z=0\) ( đpcm )
Câu b : \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
Câu c : Ta có : \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow a+b+c=0\) ( đúng )
Ta có :
\(x^3+x^2z+y^2z-xyz+y^3\)
\(=x^3+y^3+x^2z+y^2z-xyz\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2+y^2-xy\right)\)
\(=\left(x+y+z\right)\left(x^2-xy+y^2\right)\)
\(=0\left(x^2-xy+y^2\right)\)
\(=0\left(ĐPCM\right)\)
\(\left(\frac{2x+2y-z}{3}\right)^2+\left(\frac{2y+2z-x}{3}\right)^2+\left(\frac{2z+2x-y}{3}\right)^2\\ =\frac{4x^2+4y^2+z^2+8xy-4xz-4yz}{9}+\frac{4y^2+4z^2+x^2+8yz-4xy-4xz}{9}+\frac{4z^2+4x^2+y^2+8xz-4yz-4xy}{9}\\ =\frac{9x^2+9y^2+9z^2}{9}=x^2+y^2+z^2\)
- Ta có : \(\left(\frac{2x+2y-z}{3}\right)^2+\left(\frac{2y+2z-x}{3}\right)^2+\left(\frac{2x+2z-y}{3}\right)^2\)
\(=\frac{\left(2x+2y-z\right)^2}{9}+\frac{\left(2y+2z-x\right)^2}{9}+\frac{\left(2x+2z-y\right)^2}{9}\)
\(=\frac{\left(2x+2y-z\right)^2+\left(2y+2z-x\right)^2+\left(2x+2z-y\right)^2}{9}\)
\(=\frac{4x^2+4y^2+z^2+8xy-4yz-4xz+4y^2+4z^2+x^2+8yz-4xy-4xz+4x^2+4z^2+y^2+8xz-4xy-4yz}{9}\)
\(=\frac{9x^2+9y^2+9z^2}{9}=\frac{9\left(x^2+y^2+z^2\right)}{9}=x^2+y^2+z^2\)
Có : (a-b)^2>=0
<=> a^2+b^2-2ab >=0
<=>a^2+b^2 >= 2ab
<=>a^2+b^2+2ab >= 4ab
<=> (a+b)^2 >= 4ab
Với a,b >0 thì chia cả 2 vế cho (a+b).ab thì :
a+b/ab >= 4/a+b
<=>4/a+b <= 1/a+1/b
<=> 1/a+b <= 1/4.(1/a+1/b) ( với mọi a,b > 0 )
Áp dụng bđt trên cho x;y;z > 0 thì : x/2x+y+z = x. 1/(x+y)+(z+x) <= x/4 .( 1/x+y+1/x+z) = x/4.(x+y) + x/4.(x+z)
Tương tự : y/x+2y+z <= y/4.(y+x) + y/4.(y+z)
z/x+y+2z <= z/4.(z+x) + z/4.(z+y)
=> VT <= [ x/4.(x+y) + y/4.(y+x) ] + [ y/4.(y+z) + z/4.(z+y) ] + [ z/4.(z+x) + x/4.(x+z) ] = 1/4 + 1/4 + 1/4 = 3/4
=> ĐPCM
Dấu "=" xảy ra <=> x=y=z > 0
k mk nha
áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) với mọi a,b >0
Thì \(\frac{x}{x+y}+\frac{x}{x+z}\ge\frac{4x}{2x+y+z}\)
Tương tự thì đpcm
Cách này nhanh này thành đơ
\(\left(\dfrac{2x+2y-z}{3}\right)^2+\left(\dfrac{2y+2z-x}{3}\right)^2+\left(\dfrac{2z+2x-y}{3}\right)^2\)
\(=\dfrac{4y^2+4x^2+z^2+8xy-4xz-4yz+4y^2+4z^2+x^2+8yz-4xy-4xz}{9}+\dfrac{\left(2z+2x-y\right)^2}{9}\)
\(=\dfrac{8y^2+5x^2+5z^2+4xy-8xz+4yz+4z^2+4x^2+y^2+8xz-4yz-4xy}{9}\)
\(=\dfrac{9y^2+9z^2+9x^2}{9}=x^2+y^2+z^2\)
Ta đặt: \(\left\{{}\begin{matrix}\dfrac{1}{x^2}=a\\\dfrac{1}{y^2}=b\\\dfrac{1}{z^2}=c\end{matrix}\right.\)\(\Rightarrow\sqrt{abc}=abc=1\)
Ta có: \(\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{1}{\sqrt{b}+\sqrt{bc}+1}+\dfrac{1}{\sqrt{c}+\sqrt{ca}+1}\)
\(=\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{1}{\sqrt{b}+\dfrac{1}{\sqrt{a}}+1}+\dfrac{1}{\dfrac{1}{\sqrt{ab}}+\sqrt{ca}+1}\)
\(=\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{\sqrt{a}}{\sqrt{ba}+1+\sqrt{a}}+\dfrac{1}{1+\sqrt{ab}+\sqrt{a}}=1\)
Quay lại bài toán, sau khi đặt bài toán trở thành:
\(P=\dfrac{1}{2b+a+3}+\dfrac{1}{2c+b+3}+\dfrac{1}{2a+c+3}\)
\(=\dfrac{1}{\left(a+b\right)+\left(b+1\right)+2}+\dfrac{1}{\left(b+c\right)+\left(c+1\right)+2}+\dfrac{1}{\left(c+a\right)+\left(a+1\right)+2}\)
\(\le\dfrac{1}{2}\left(\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{1}{\sqrt{b}+\sqrt{bc}+1}+\dfrac{1}{\sqrt{c}+\sqrt{ca}+1}\right)=\dfrac{1}{2}\)
Cái đó t cố tình bỏ đấy. B phải tự làm chứ chẳng lẽ t làm hết??
Lời giải:
Ta có:
\(x^3+x^2z+y^2z-xyz+y^3=(x^3+y^3)+(x^2z+y^2z-xyz)\)
\(=(x+y)(x^2-xy+y^2)+z(x^2+y^2-xy)\)
\(=(x^2-xy+y^2)(x+y+z)=(x^2-xy+y^2).0=0\)
Ta có đpcm.