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x^3+y^3+z^3-3xyz = 0
<=> (x+y+z).(x^2+y^2+z^2-xy-yz-zx) = 0
Mà x+y+z > 0 => x^2+y^2+z^2-xy-yz-zx = 0
<=> 2x^2+2y^2+2z^2-2xy-2yz-2zx = 0
<=> (x-y)^2+(y-z)^2+(z-x)^2 = 0
=> x-y=0;y-z=0;z-x=0
=> P = 0
k mk nha
\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz.\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(=\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2xz-2yz\right)\)
\(=\frac{1}{2}\left(x+y+z\right)\text{[}\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)\text{]}\)
\(=\frac{1}{2}\left(x+y+z\right)\text{[}\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\text{]}\left(\text{đ}pcm\right)\)
Dùng biến đổi sau: \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(VT=z^3+\left(x+y\right)^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(z+x+y\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3zx\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(=\frac{1}{2}\left(x+y+z\right)\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]\)
\(=\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)
\(=VP\)
x³ + y³ + z³ - 3xyz = (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz).
~~~~~~~~
Bài làm trên mình đã sử dụng hằng đẳng thức đáng nhớ sau:
(a+b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a-b)
=> a³ + b³ = (a+b)³ - 3ab(a-b).
Chúc bạn học giỏi!
VT=\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy.\left(x+y+z\right)\)
\(=\left(x+y\right)^2-\left(x+y\right).z+z^2-3xy\left(\text{vì }x+y+z=1\right)\)
\(=x^2+2xy+y^2-xz-yz+z^3-3xy\)
\(=x^2+y^2+z^2-xy-yz-xz\)
\(=\frac{1}{2}.\left(2x^2+2y^2+2z^2-2xy-2yz-2xz\right)\)
\(=\frac{1}{2}.\left[\left(x^2-2xy-y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)\right]\)
\(=\frac{1}{2}.\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)=VP
=>dpcm
Ta có : \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=x+y+z\left(x^2+y^2+z^2+2xy+xz+yz\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(=x^2+y^2+z^2-xy-yz-xz=\frac{\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)}{2}=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)
Cho \(x^4+y^4+z^4=3\). Tìm MAX của \(A=x^2\left(x+y\right)+y^2\left(y+z\right)+z^2\left(z+x\right)\)
Ta có:
\(3x^4+1=x^4+x^4+x^4+1\ge4\sqrt[4]{x^4.x^4.x^4.1}=4x^3\)
Tương tự: \(3y^4+1\ge4y^3\) ; \(3z^4+1\ge4z^3\)
=> \(3\left(x^4+y^4+z^4\right)+3\ge4\left(x^3+y^3+z^3\right)\) (1)
Thay vào:
\(A=x^2\left(x+y\right)+y^2\left(y+z\right)+z^2\left(z+x\right)\)
\(A=x^3+x^2y+y^3+y^2z+z^3+z^2x\)
\(A=x^3+y^3+z^3+\left(x^2y+y^2z+z^2x\right)\)
\(\le x^3+y^3+z^3+\left(\frac{x^3+x^3+y^3+y^3+y^3+z^3+z^3+z^3+x^3}{3}\right)\)
\(=2\left(x^3+y^3+z^3\right)\)
\(=\frac{1}{2}\left[4\left(x^3+y^3+z^3\right)\right]\le\frac{1}{2}\left[3\left(x^4+y^4+z^4\right)+3\right]\)
\(=\frac{1}{2}\left[3.3+3\right]=\frac{12}{2}=6\)
Dấu "=" xảy ra khi: \(x=y=z=1\)
Vậy Max(A) = 6 khi x = y = z = 1
Bài 2: Ta có: x, y, z không âm và \(x+y+z=\frac{3}{2}\)nên \(0\le x\le\frac{3}{2}\Rightarrow2-x>0\)
Áp dụng bất đẳng thức AM - GM dạng \(ab\le\frac{\left(a+b\right)^2}{4}\), ta được: \(x+2xy+4xyz=x+4xy\left(z+\frac{1}{2}\right)\le x+4x.\frac{\left(y+z+\frac{1}{2}\right)^2}{4}=x+x\left(2-x\right)^2\)
Ta cần chứng minh \(x+x\left(2-x\right)^2\le2\Leftrightarrow\left(2-x\right)\left(x-1\right)^2\ge0\)*đúng*
Đẳng thức xảy ra khi \(\left(x,y,z\right)=\left(1,\frac{1}{2},0\right)\)
Bài 3: Áp dụng đánh giá quen thuộc \(4ab\le\left(a+b\right)^2\), ta có: \(2\le\left(x+y\right)^3+4xy\le\left(x+y\right)^3+\left(x+y\right)^2\)
Đặt x + y = t thì ta được: \(t^3+t^2-2\ge0\Leftrightarrow\left(t-1\right)\left(t^2+2t+2\right)\ge0\Rightarrow t\ge1\)(dễ thấy \(t^2+2t+2>0\forall t\))
\(\Rightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\ge\frac{1}{2}\)
\(P=3\left(x^4+y^4+x^2y^2\right)-2\left(x^2+y^2\right)+1=3\left[\frac{3}{4}\left(x^2+y^2\right)^2+\frac{1}{4}\left(x^2-y^2\right)^2\right]-2\left(x^2+y^2\right)+1\ge\frac{9}{4}\left(x^2+y^2\right)^2-2\left(x^2+y^2\right)+1\)\(=\frac{9}{4}\left[\left(x^2+y^2\right)^2+\frac{1}{4}\right]-2\left(x^2+y^2\right)+\frac{7}{16}\ge\frac{9}{4}.2\sqrt{\left(x^2+y^2\right)^2.\frac{1}{4}}-2\left(x^2+y^2\right)+\frac{7}{16}=\frac{9}{4}\left(x^2+y^2\right)-2\left(x^2+y^2\right)+\frac{7}{16}=\frac{1}{4}\left(x^2+y^2\right)+\frac{7}{16}\ge\frac{1}{8}+\frac{7}{16}=\frac{9}{16}\)Đẳng thức xảy ra khi x = y = 1/2