Question

[Dưa Trong Cúc]

Giải phương trình :

a,\(13x-2\sqrt{x}.\left(3+2y\right)+y^2+1=0\)

b,\(x+4\sqrt{x+3}+2\sqrt{3-2x}=11\)

c,\(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

d,\(2x+2y+2z=\sqrt{4x-1}+\sqrt{4y-1}+\sqrt{4z-1}\)

Answer 1

[Nguyễn Thị Ngọc Thơ]

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)

Answer 2

[Lê Thị Thục Hiền]

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)