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Bài 1a/
\(\frac{1}{1+x+xy}=\frac{xyz}{xyz+x+xy}=\frac{yz}{1+y+yz}\)
\(\frac{1}{1+z+xz}=\frac{y}{y+yz+xyz}=\frac{y}{1+y+yz}\)
Vậy \(M=\frac{1}{1+y+yz}+\frac{y}{1+y+yz}+\frac{yz}{1+y+yz}=1\)
Chiều về làm tiếp
Bài 1b:Lời giải này chủ yếu nhờ dự đoán trước Min là 2011/2012 đạt được khi x=2012
Ta có \(P=\frac{2012x^2-2.2012x+2012^2}{2012x^2}=\frac{\left(x-2012\right)^2+2011x^2}{2012x^2}\ge\frac{2011x^2}{2012x^2}=\frac{2011}{2012}\)
Bài 2: Dùng phân tích thành bình phương
\(10x^2+y^2+4z^2+6x-4y-4xz+5=\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)\)
\(=\left(3x+1\right)^2+\left(y-2\right)^2+\left(x-2z\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}}\)
Bài 3:
a/\(pt\Leftrightarrow\left(x+6\right)\left(x-5\right)\left(x^2-x+1\right)=0\Leftrightarrow x=-6,x=5\)
b/ta phân tích vế trái thành:\(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
9x2 + y2 + 2z2 - 18x + 4z - 6y + 20 = 0
<=> 9x2 - 18x + 9 + y2 - 6y + 9 + 2x2 + 4z + 2 = 0
<=> 9(x2 - 2x + 1) + (y - 3)2 + 2(z2 + 2z + 1) = 0
<=> 9(x - 1)2 + (y - 3)2 + 2(z + 1)2 = 0
<=> \(\left\{\begin{matrix}x-1=0\\y-3=0\\z+1=0\end{matrix}\right.\)
<=> \(\left\{\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(9x^2+y^2+2z^2-18x+4z-6z+20=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\\z+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
a) Ta có :
\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Ta thấy : \(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\forall x,y,z\)
Do đó : \(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(3x-3\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\) ( thỏa mãn )
Vậy : \(\left(x,y,z\right)=\left(1,3,-1\right)\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\\z+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
9x2 + y2 + 2z2 - 18x + 4z - 6y + 20 = 0
( 9x2 -18x + 9) +( y2 - 6y + 9) +2(z2+2z +1) = 0
( 3x-3)2 + ( y-3)2 + 2( z+1)2 = 0
vì ( 3x-3)^2 , (y-3)^2 , 2( z+1)^2 >0 \(\Rightarrow\left(3x-3\right)^2=\left(y-3\right)^2=2\left(z+1\right)^2\))^2
\(\Leftrightarrow\hept{\begin{cases}3x-3=0\\y-3=0\\2\left(z+1\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
1) \(9x^2+y^2-2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
mà: \(9\left(x-1\right)^2\ge0;\left(y-3\right)^2\ge0;2\left(z+1\right)^2\ge0\)
nên \(_{\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}}\)
2) Ta có: \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Leftrightarrow\left(\frac{ayz+bxz+cxy}{xyz}\right)=0\Leftrightarrow ayz+bxz+cxy=0\)
Lại có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Rightarrow\left(\frac{x^2}{a^2}\right)+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1\)
mà : \(\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=\frac{2xyabc^2+2yzbca^2+2xzacb^2}{a^2b^2c^2}=\frac{2abc\left(cxy+ayz+bxz\right)}{a^2b^2c^2}=\frac{2abc\cdot0}{a^2b^2c^2}=0\)
Vậy \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)
1 ) \(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Vì \(\hept{\begin{cases}9\left(x-1\right)^2\ge0\\\left(y-3\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{cases}}\)
\(\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\)
Để \(9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\) thì \(\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}}\)
2 ) Ta có : \(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{2xy}{ab}+\frac{y^2}{b^2}+\frac{2xz}{ac}+\frac{z^2}{c^2}+\frac{2yz}{bc}=1\)
\(\Leftrightarrow\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\left(\frac{2xy}{ab}+\frac{2xz}{ac}+\frac{2yz}{bc}\right)=1\)
\(\Leftrightarrow\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)=1\)
\(\Leftrightarrow\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}.0=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\) (đpcm(
Giải các phương trình:
\(a,\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(b,x^4-30x^2+31x-30=0\)
a, Đặt \(x^2-5x=a\)
\(\Rightarrow\)\(a^2+10a+24=0\)
\(\Rightarrow a^2+4a+6a+24=0\)
\(\Rightarrow\left(a+4\right)\left(a+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+4=0\\a+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2-5x+4=0\left(1\right)\\x^2-5x+6=0\left(2\right)\end{cases}}}\)
Giải pt (1) ta có : \(x^2-5x+4=0\)
\(\Rightarrow x^2-4x-x+4=0\)
\(\Rightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
Giải pt (2) ta có : \(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Vậy \(S=\left\{1;2;3;4\right\}\)
\(x^4-30x^2+31x-30=0\)
\(\Rightarrow x^4-30x^2+x+30x-30=0\)
\(\Rightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
\(\Rightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)\)
\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)\)
\(\Rightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
Mà \(x^2-x+1>0\)với \(\forall\)\(x\)
\(\Rightarrow x^2+x-30=0\)
\(\Rightarrow x^2-5x+6x-30=0\)
\(\Rightarrow x\left(x-5\right)+6\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)
Vậy \(S=\left\{5;-6\right\}\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y-3\right)^2+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
+ \(\left\{{}\begin{matrix}9\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\\2\left(z+1\right)^2\ge0\forall z\end{matrix}\right.\)
\(\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\forall x,y,z\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\left(TM\right)\)