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\(x^2+y^2+z^2-2x+4y-6z=15\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2=29\)
Đặt \(P=\left|2x-3y+4z-20\right|=\left|2\left(x-1\right)-3\left(y+2\right)+4\left(z-3\right)\right|\)
\(P^2=\left[2\left(x-1\right)-3\left(y+2\right)+4\left(z-3\right)\right]^2\)
\(P^2\le\left(2^2+3^2+4^2\right)\left[\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2\right]=29^2\)
\(\Rightarrow P\le29\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2=29\\\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{4}\end{matrix}\right.\)
1) \(21x^2+21y^2+z^2\)
\(=18\left(x^2+y^2\right)+z^2+3\left(x^2+y^2\right)\)
\(\ge9\left(x+y\right)^2+z^2+3.2xy\)
\(\ge2.3\left(x+y\right).z+6xy\)
\(=6\left(xy+yz+zx\right)=6.13=78\)
Dấu "=" xảy ra <=> x = y ; 3(x+y) = z; xy + yz + zx= 13 <=> x = y = 1; z= 6
2) \(x+y+z=3xyz\)
<=> \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=3\)
Đặt: \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)=> ab + bc + ca = 3
Ta cần chứng minh: \(3a^2+b^2+3c^2\ge6\)
Ta có: \(3a^2+b^2+3c^2=\left(a^2+c^2\right)+2\left(a^2+c^2\right)+b^2\)
\(\ge2ac+\left(a+c\right)^2+b^2\ge2ac+2\left(a+c\right).b=2\left(ac+ab+bc\right)=6\)
Vậy: \(\frac{3}{x^2}+\frac{1}{y^2}+\frac{3}{z^2}\ge6\)
Dấu "=" xảy ra <=> a = c = \(\sqrt{\frac{3}{5}}\); \(b=2\sqrt{\frac{3}{5}}\)
khi đó: \(x=z=\sqrt{\frac{5}{3}};y=\sqrt{\frac{5}{3}}\)
\(a,4x+4y\)
\(=4\left(x+y\right)\)
b,\(2xy-y^2+2x-y\)
\(=\left(2xy+2x\right)-\left(y^2+y\right)\)
\(=2x\left(y+1\right)-y\left(y+1\right)\)
\(=\left(y+1\right)\left(2x-y\right)\)
\(c,2x^3y-8x^2y^2+8xy^3\)
\(=2xy\left(x^2-4xy+4y^2\right)\)
\(=2xy\left(x^2-2.x.2y+\left(2y\right)^2\right)\)
\(=2xy\left(x-2y\right)^2\)
Bai 2:
\(a,x^2-81=0\)
\(\Rightarrow x^2-9^2=0\)
\(\Rightarrow\left(x-9\right)\left(x+9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x+9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
\(b,x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
a) 5x2 + 10y2 - 6xy - 4x - 2y + 3
= ( x2 - 6xy + 9y2 ) + ( 4x2 - 4x + 1 ) + ( y2 - 2y + 1 ) + 1
= ( x - 3y )2 + ( 2x - 1 )2 + ( y - 1 )2 + 1 ≥ 1 > 0 ∀ x, y, z
=> đpcm
b) x2 + 4y2 + z2 - 2x - 6z + 8y + 15
= ( x2 - 2x + 1 ) + ( 4y2 + 8y + 4 ) + ( z2 - 6z + 9 ) + 1
= ( x - 1 )2 + ( 2y + 2 )2 + ( z - 3 )2 + 1 ≥ 1 > 0 ∀ x, y, z
=> đpcm
a,
\(x^2+4y^2-x+4y+2=\left(x^2-x+\dfrac{1}{4}\right)+4\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+4\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x,y\)
b,
\(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0-3\left(-c\right)\left(-a\right)\left(-b\right)=0-3\left(-abc\right)=3abc\left(dpcm\right)\)
Bài 1:
\(x^2+y^2-2x-4y+5=0\)
\(\Leftrightarrow (x^2-2x+1)+(y^2-4y+4)=0\)
\(\Leftrightarrow (x-1)^2+(y-2)^2=0\)
Vì $(x-1)^2; (y-2)^2\geq 0$ với mọi $x,y\in\mathbb{R}$ nên để tổng của chúng bằng $0$ thì $(x-1)^2=(y-2)^2=0$
$\Rightarrow x=1; y=2$
Vậy...........
Bài 2:
Ta có:
\(a(a-b)+b(b-c)+c(c-a)=0\)
\(\Leftrightarrow 2a(a-b)+2b(b-c)+2c(c-a)=0\)
\(\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0\)
\(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\)
Lập luận tương tự bài 1, ta suy ra :
\((a-b)^2=(b-c)^2=(c-a)^2=0\Rightarrow a=b=c\)
Khi đó, thay $b=c=a$ ta có:
\(P=a^3+b^3+c^3-3abc+3ab-3c+5\)
\(=3a^3-3a^3+3a^2-3a+5=3a^2-3a+5\)
\(=3(a^2-a+\frac{1}{4})+\frac{17}{4}=3(a-\frac{1}{2})^2+\frac{17}{4}\geq \frac{17}{4}\)
Vậy $P_{\min}=\frac{17}{4}$
Giá trị này đạt được tại $b=c=a=\frac{1}{2}$
\(19^2\le\left(2x+y\right)^2=\left(\frac{2}{\sqrt{3}}.\sqrt{3}x+\frac{1}{2}.2y\right)^2\le\left(\frac{4}{3}+\frac{1}{4}\right)\left(3x^2+4y^2\right)\)
\(\Rightarrow3x^2+4y^2\ge\frac{19^2}{\frac{4}{3}+\frac{1}{4}}=228\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{x}{2}=\frac{y}{4}\\2x+y=19\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{19}{4}\\y=\frac{19}{2}\end{matrix}\right.\)
Hình như bạn ghi ko đúng đề, nếu đề là \(2x+y=\sqrt{19}\) thì \(3x^2+4y^2\ge12\) mới đúng