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Bạn xem lại đề câu b và c nhé !
a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)
\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ
\(\Rightarrow x\ge2\) thỏa mãn đề.
d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)
Pt tương đương :
\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )
e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)
\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)
Phương trình (1) tương đương :
\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )
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)
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)
\(1.x^2-4x-2\sqrt{2x-5}+5=0\left(x>=\dfrac{5}{2}\right)\)
\(\text{⇔}2x-5-2\sqrt{2x-5}+1+x^2-6x+9=0\)
\(\text{⇔}\left(\sqrt{2x-5}-1\right)^2+\left(x-3\right)^2=0\)
\(\text{⇔}\sqrt{2x-5}-1=0\) hoặc \(x-3=0\)
\(\text{⇔}x=3\left(TM\right)\)
KL...........
\(2.x+y+4=2\sqrt{x}+4\sqrt{y-1}\)
\(\text{⇔}x-2\sqrt{x}+1+y-1-4\sqrt{y-1}+4=0\)
\(\text{⇔}\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-2\right)^2=0\)
\(\text{⇔}x=1;y=5\)
KL..........
\(3.\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-5}=\dfrac{1}{2}\left(x+y+z-7\right)\)
\(\text{⇔}2\sqrt{x-2}+2\sqrt{y-3}+2\sqrt{z-5}=x+y+z-7\)
\(\text{⇔}x-2-2\sqrt{x-2}+1+y-3-2\sqrt{y-3}+1+z-5-2\sqrt{z-5}+1=0\)
\(\text{⇔}\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-5}-1\right)^2=0\)
\(\text{⇔}x=1;y=4;z=6\)
KL...........
\(d.Tuong-tự-nhé-bn\)