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20 tháng 1 2020
\(\hept{\begin{cases}\sqrt{2x+3}+x^2+2x=\sqrt{2y-1}+y^2-2y\left(1\right)\\\sqrt{x-2}+\sqrt{y-1}=3\left(2\right)\end{cases}}\)
\(Đkxđ:x\ge2;y\ge1\)
\(\left(1\right)\Leftrightarrow\sqrt{2x+3}-\sqrt{2y-1}=y^2-x^2-2\left(y+x\right)\)
\(\frac{2x-2y+4}{\sqrt{2x+3}+\sqrt{2y-1}}=\left(x+y\right)\left(y-x\right)-2\left(y+x\right)\)
\(\Leftrightarrow\frac{2\left(x-y+2\right)}{\sqrt{2x+3}+\sqrt{2y-1}}+\left(x+y\right)\left(x-y+2\right)=0\)
\(\Leftrightarrow\left(x-y+2\right)\left(\frac{2}{\sqrt{2x+3}+\sqrt{2y-1}}+x+y\right)=0\)
\(\Leftrightarrow x-y+2=0\)
\(\Leftrightarrow x=y-2\)
Thay vào \(\left(2\right)\) ...................................................................
6 tháng 10 2019
b,ĐK:\(-3\le x\le\frac{3}{2}\)
\(PT\Leftrightarrow x-1+4\left(\sqrt{x+3}-2\right)+2\left(\sqrt{3-2x}-1\right)=0\)
\(\Leftrightarrow x-1+\frac{4\left(x-1\right)}{\sqrt{x+3}+2}+\frac{2\left(2-2x\right)}{\sqrt{3-2x}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(1+\frac{4}{\sqrt{x+3}+2}-\frac{4}{\sqrt{3-2x}+1}\right)=0\)
Với \(x\ge-3\) \(\Rightarrow\frac{4}{\sqrt{x+3}+2}>0\) và \(3-2x\le9\Rightarrow-\frac{4}{\sqrt{3-2x}+1}\ge-1\)
\(\Rightarrow1+\frac{4}{\sqrt{x+3}+2}-\frac{4}{\sqrt{3-2x}+1}>0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)(tm)
6 tháng 10 2019
c,Đk: \(x\ge2,y\ge3,z\ge5\)
pt <=> \(x-2\sqrt{x-2}+y-4\sqrt{y-3}+z-6\sqrt{z-5}+4=0\)
<=> \(\left(x-2\right)-2\sqrt{x-2}+1+\left(y-3\right)-4\sqrt{y-3}+4+\left(z-5\right)-6\sqrt{z-5}+9=0\)
<=>\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=\)0
=>\(\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y-3}-2=0\\\sqrt{z-5}-3=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=3\\y=7\\z=14\end{matrix}\right.\)(t/m)
d, \(2x+2y+2z=\sqrt{4x-1}+\sqrt{4y-1}+\sqrt{4z-1}\left(đk:x,y,z\ge\frac{1}{4}\right)\)
<=> \(4x+4y+4z=2\sqrt{4x-1}+2\sqrt{4y-1}+2\sqrt{4z-1}\)
<=> \(\left(4x-1\right)-2\sqrt{4x-1}+1+\left(4y-1\right)-2\sqrt{4y-1}+1+\left(4z-1\right)-2\sqrt{4z-1}+1=0\)
<=>\(\left(\sqrt{4x-1}-1\right)^2+\left(\sqrt{4y-1}-1\right)^2+\left(\sqrt{4z-1}-1\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{4x-1}-1=0\\\sqrt{4y-1}-1=0\\\sqrt{4z-1}-1=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\\z=\frac{1}{2}\end{matrix}\right.\)(tm)
B
8 tháng 7 2019
\(A=\sqrt{2x^2-4x+3}+3\)
Ta có: \(2x^2-4x+3\)
\(=2\left(x^2-2x+\frac{3}{2}\right)\)
\(=2\left(x^2-2.x.1+1^2+\frac{1}{2}\right)\)
\(=2[\left(x-1\right)^2+\frac{1}{2}]\)
\(=2\left(x-1\right)^2+1\ge1\)
\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}\)
\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}+3\ge3+\sqrt{1}=4\)
\(\Rightarrow MinA=4\Leftrightarrow x=1\)
BH
1 tháng 3 2020
\(\hept{\begin{cases}x^2-2x\sqrt{y}+2y=x\\y^2-2y\sqrt{z}+2z=y\\z^2-2z\sqrt{x}+2x=z\end{cases}}\)
\(\Leftrightarrow x^2-2x\sqrt{y}+2y+y^2-2y\sqrt{z}+2z+z^2-2z\sqrt{x}+2x=x+y+z\)
\(\Leftrightarrow\left(x-\sqrt{y}\right)^2+\left(y-\sqrt{z}\right)^2+\left(z-\sqrt{x}\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-\sqrt{y}=0\\y-\sqrt{z}=0\\z-\sqrt{x}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\sqrt{y}\\y=\sqrt{z}\\z=\sqrt{x}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=y=z=0\\x=y=z=1\end{cases}}\)