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Bài 1:
b: \(\Leftrightarrow2+\sqrt{3x-5}=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1=3x-5\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+6=0\\x>=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
c: \(\Leftrightarrow5x+7=16\left(x+3\right)\)
=>16x+48=5x+7
=>11x=-41
hay x=-41/11
\(\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}=6-\dfrac{1}{\sqrt{x-1}}-\dfrac{1}{\sqrt{y-2}}-\dfrac{1}{\sqrt{z-3}}\Leftrightarrow\left(\sqrt{x-1}+\dfrac{1}{\sqrt{x-1}}\right)+\left(\sqrt{y-2}+\dfrac{1}{\sqrt{y-2}}\right)+\left(\sqrt{z-3}+\dfrac{1}{\sqrt{z-3}}\right)=6\)Áp dụng bất đẳng thức cô si ta có :
\(\sqrt{x-1}+\dfrac{1}{\sqrt{x-1}}\ge2\sqrt{\sqrt{x-1}.\dfrac{1}{\sqrt{x-1}}}=2\)
Tương tự :\(\sqrt{y-2}+\dfrac{1}{\sqrt{y-2}}\ge2\)
\(\sqrt{z-3}+\dfrac{1}{\sqrt{z-3}}\ge2\)
Do đó :\(\left(\sqrt{x-1}+\dfrac{1}{\sqrt{x-1}}\right)+\left(\sqrt{y-2}+\dfrac{1}{\sqrt{y-2}}\right)+\left(\sqrt{z-3}+\dfrac{1}{\sqrt{z-3}}\right)\ge6\)Dấu "=+ xảy ra khi :\(\left\{{}\begin{matrix}\sqrt{x-1}=\dfrac{1}{\sqrt{x-1}}\\\sqrt{y-2}=\dfrac{1}{\sqrt{y-2}}\\\sqrt{z-3}=\dfrac{1}{\sqrt{z-3}}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=1\\z-3=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\\z=4\end{matrix}\right.\)
Vậy \(x=2,y=3,z=4\)
4.a)\(x-2\sqrt{x}+3\)
\(=x-2\sqrt{x}+1+2\)
\(=\left(\sqrt{x}-1\right)^2+2\)
Vì \(\left(\sqrt{x}-1\right)^2\ge0,\forall x\)
\(\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow Min_{bt}=2\) khi \(\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
b)Ta có:
\(x-4\sqrt{y}+13\ge0\)
\(\Leftrightarrow x-4\sqrt{y}\ge-13\)
Dấu "=" xảy ra khi \(x-4\sqrt{y}=0\Leftrightarrow x=4\sqrt{y}\)
Vậy \(min_{bt}=0\) khi \(x=4\sqrt{y}\)
c)Ta có:
\(2x-4\sqrt{y}+6\ge0\)
\(\Leftrightarrow x-2\sqrt{y}+3\ge0\)
\(\Leftrightarrow x-2\sqrt{y}\ge-3\)
Dấu "=" xảy ra khi \(x-2\sqrt{y}=0\Leftrightarrow x=2\sqrt{y}\)
Vậy \(Min_{bt}=0\) khi \(x=2\sqrt{y}\)
d)Ta có:
\(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\)
Vì \(\left(x+1\right)^2\ge0,\forall x\)
\(\Leftrightarrow\left(x+1\right)^2+4\ge4\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)^2+4}\le\frac{1}{4}\)
\(\Leftrightarrow-\frac{1}{\left(x+1\right)^2+4}\ge-\frac{1}{4}\)
\(\Leftrightarrow-\frac{4}{\left(x+1\right)^2+4}\ge-1\)
Vậy \(Min_{bt}=-1\) khi \(x+1=0\Leftrightarrow x=-1\)
ĐK:\(\left\{{}\begin{matrix}x\ge2\\y\ge3\\z\ge5\end{matrix}\right.\)
\(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\Leftrightarrow x-2\sqrt{x-2}+y-4\sqrt{y-3}+z-6\sqrt{z-5}+4=0\Leftrightarrow x-2-2\sqrt{x-2}+1+y-3-4\sqrt{y-3}+4+z-5-6\sqrt{z-5}+9=0\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y-3}-2=0\\\sqrt{z-5}-3=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=3\\y=7\\z=14\end{matrix}\right.\)(tm)
Vậy (x;y;z)=(3;7;14)
ĐKXĐ:\(\left\{{}\begin{matrix}x\ge2\\y\ge3\\z\ge5\end{matrix}\right.\)
Ta có x+y+z+4=\(2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
\(\Leftrightarrow\)\(x-2\sqrt{x-2}+y-4\sqrt{y-3}+z-6\sqrt{z-5}+4=0\)
\(\Leftrightarrow\)\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5+6\sqrt{z-5}+9\right)=0\)
\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
mà 3 biểu thức trên đều \(\ge\)0 nên để =0 thì
\(\)\(\sqrt{x-2}=1;\sqrt{y-3}=2;\sqrt{z-5=3}\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=14\end{matrix}\right.\)
Lời giải:
a/ ĐKXĐ: $x\geq 0; y\geq 1$
PT $\Leftrightarrow (x-2\sqrt{x}+1)+[(y-1)-4\sqrt{y-1}+4]=0$
$\Leftrightarrow (\sqrt{x}-1)^2+(\sqrt{y-1}-2)^2=0$
Vì $(\sqrt{x}-1)^2\geq 0; (\sqrt{y-1}-2)^2\geq 0$ với mọi $x,y$ thuộc đkxđ
Do đó để tổng của chúng bằng $0$ thì:
$\sqrt{x}-1=\sqrt{y-1}-2=0$
$\Leftrightarrow x=1; y=5$
b. ĐKXĐ: $x\geq 0; y\geq 1; z\geq 2$
PT $\Leftrightarrow 2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z$
$\Leftrightarrow (x-2\sqrt{x}+1)+[(y-1)-2\sqrt{y-1}+1]+[(z-2)-2\sqrt{z-2}+1]=0$
$\Leftrightarrow (\sqrt{x}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z-2}-1)^2=0$
$\Rightarrow \sqrt{x}-1=\sqrt{y-1}-1=\sqrt{z-2}-1=0$
$\Leftrightarrow x=1; y=2; z=3$