Giải phương trình, hệ phương trình:
a) \(\frac{\sqrt{x-2013}-1}{x-2013}+\frac{\sqrt{y-2014}-1}{y-2014}+\frac{\sqrt{z-2015}-1}{z-2015}=\frac{3}{4}\)
b) \(\left\{{}\begin{matrix}x^3+1=2y\\y^3+1=2x\end{matrix}\right.\)
c)\(\sqrt{x^2-3x+2}+\sqrt{x-3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
d)\(5x-2\sqrt{x}\left(2+y\right)+y^2+1=0\)
c/ ĐKXĐ: \(x\ge3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x-3}-\sqrt{x-2}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}\right)-\left(\sqrt{\left(x-1\right)\left(x+3\right)}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}-\sqrt{x+3}=0\\\sqrt{x-1}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2< 3\left(ktm\right)\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x>2013\\y>2014\\z>2015\end{matrix}\right.\)
\(\Leftrightarrow\frac{1}{4}-\frac{\sqrt{x-2013}-1}{x-2013}+\frac{1}{4}-\frac{\sqrt{y-2014}-1}{y-2014}+\frac{1}{4}-\frac{\sqrt{z-2015}-1}{z-2015}=0\)
\(\Leftrightarrow\frac{x-2013-4\sqrt{x-2013}+4}{4\left(x-2013\right)}+\frac{y-2014-4\sqrt{y-2014}+4}{4\left(y-2014\right)}+\frac{z-2015-4\sqrt{z-2015}+4}{4\left(z-2015\right)}=0\)
\(\Leftrightarrow\left(\frac{\sqrt{x-2013}-2}{2\sqrt{x-2013}}\right)^2+\left(\frac{\sqrt{y-2014}-2}{2\sqrt{y-2014}}\right)^2+\left(\frac{\sqrt{z-2015}-2}{2\sqrt{z-2015}}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2013}-2=0\\\sqrt{y-2014}-2=0\\\sqrt{z-2015}-2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)
b/ Trừ vế cho vế 2 pt ta được:
\(x^3-y^3=2\left(y-x\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+y^2-xy\right)+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+y^2-xy+2\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}+2\right]=0\)
\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)
Thay vào pt đầu:
\(x^3+1=2x\Leftrightarrow x^3-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow...\)
c/ ĐKXĐ: \(x\ge3\)
Câu này có vẻ ghi đề ko đúng, pt này vô nghiệm
\(\Leftrightarrow\sqrt{x^2-3x+2}-\sqrt{x^2+2x-3}+\sqrt{x-3}-\sqrt{x-2}=0\)
\(\Leftrightarrow\frac{\left(x^2-3x+2\right)-\left(x^2+2x-3\right)}{\sqrt{x^2-3x+2}+\sqrt{x^2+2x-3}}+\frac{\left(x-3\right)-\left(x-2\right)}{\sqrt{x-3}+\sqrt{x-2}}=0\)
\(\Leftrightarrow\frac{-5x+1}{\sqrt{x^2-3x+2}+\sqrt{x^2+2x-3}}+\frac{-1}{\sqrt{x-3}+\sqrt{x-2}}=0\)
Với \(x\ge3\Rightarrow-5x+1< 0\Rightarrow\) vế trái luôn âm
Do đó pt đã cho vô nghiệm
Kiểm tra lại bên vế trái là \(\sqrt{x-3}\) hay \(\sqrt{x+3}\)
d/ ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow5x-4\sqrt{x}-2y\sqrt{x}+y^2+1=0\)
\(\Leftrightarrow\left(4x-4\sqrt{x}+1\right)+\left(x-2y\sqrt{x}+y^2\right)=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)^2+\left(\sqrt{x}-y\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{4}\\y=\frac{1}{2}\end{matrix}\right.\)
