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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)
c) Đặt \(a=\sqrt{x-4},b=\sqrt{y-4}\)với \(a,b\ge0\)thì pt đã cho trở thành:
\(2\left(a^2+4\right)b+2\left(b^2+4\right)a=\left(a^2+4\right)\left(b^2+4\right)\). chia 2 vế cho \(\left(a^2+4\right)\left(b^2+4\right)\)thì pt trở thành :
\(\frac{2b}{b^2+4}+\frac{2a}{a^2+4}=1\). Để ý rằng a=0 hoặc b=0 không thỏa mãn pt.
Xét \(a,b>0\). Theo BĐT AM-GM ta có: \(b^2+4\ge2\sqrt{4b^2}=4b,a^2+4\ge4a\)
\(\Rightarrow VT\le\frac{2a}{4a}+\frac{2b}{4b}=1\), dấu đẳng thức xảy ra khi và chỉ khi \(\hept{\begin{cases}a^2=4\\b^2=4\end{cases}\Leftrightarrow a=b=2\Leftrightarrow x=y=8}\)
Vậy x=8,y=8 là nghiệm của pt
e/ \(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)
\(\Leftrightarrow4+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)
\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)
Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\)thì pt thành
\(2a=-a^2+8\)
\(\Leftrightarrow a^2+2a-8=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-4\left(l\right)\\a=2\end{cases}}\)
\(\Leftrightarrow\sqrt{-x^2+8x-12}=2\)
\(\Leftrightarrow-x^2+8x-12=4\)
\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)
a/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)
\(1.x^2-4x-2\sqrt{2x-5}+5=0\left(x>=\dfrac{5}{2}\right)\)
\(\text{⇔}2x-5-2\sqrt{2x-5}+1+x^2-6x+9=0\)
\(\text{⇔}\left(\sqrt{2x-5}-1\right)^2+\left(x-3\right)^2=0\)
\(\text{⇔}\sqrt{2x-5}-1=0\) 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...........
\(d.Tuong-tự-nhé-bn\)