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b) \(\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}\) (*)
Đk: \(\left\{{}\begin{matrix}x>3\\y>1\\z>665\end{matrix}\right.\)
(*) \(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\dfrac{x-3}{\sqrt{x-3}}-\dfrac{y-1}{\sqrt{y-1}}-\dfrac{z-665}{\sqrt{z-665}}\)
\(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}-82+\dfrac{x-3}{\sqrt{x-3}}+\dfrac{y-1}{\sqrt{y-1}}+\dfrac{z-665}{\sqrt{z-665}}=0\)
\(\Leftrightarrow\left(\dfrac{x-3}{\sqrt{x-3}}-\dfrac{8\sqrt{x-3}}{\sqrt{x-3}}+\dfrac{16}{\sqrt{x-3}}\right)+\left(\dfrac{y-1}{\sqrt{y-1}}-\dfrac{4\sqrt{y-1}}{\sqrt{y-1}}+\dfrac{4}{\sqrt{y-1}}\right)+\left(\dfrac{z-665}{\sqrt{z-665}}-\dfrac{70\sqrt{z-665}}{\sqrt{z-665}}+\dfrac{1225}{\sqrt{z-665}}\right)=0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x-3}-4\right)^2}{\sqrt{x-3}}+\dfrac{\left(\sqrt{y-1}-2\right)^2}{\sqrt{y-1}}+\dfrac{\left(\sqrt{z-665}-35\right)^2}{\sqrt{z-665}}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-3}-4=0\\\sqrt{y-1}-2=0\\\sqrt{z-665}-35=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=19\\y=5\\z=1890\end{matrix}\right.\)
Kl: x=19, y= 5, z=1890
\(\frac{x+\sqrt{x}}{3\sqrt{x}-1}=6+2\sqrt{5}\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{3\left(\sqrt{x}-1\right)}=2\left(3+\sqrt{5}\right)\)
\(\Leftrightarrow\frac{\sqrt{x}}{3}=2\left(3+\sqrt{5}\right)\)
\(\Leftrightarrow\sqrt{x}=6\left(3+\sqrt{5}\right)\)
\(\Leftrightarrow x=36\left(3+\sqrt{5}\right)^2\)
\(\Leftrightarrow x=36\left(14+6\sqrt{5}\right)\)
a,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\) (*)(đk \(x\ge-2\))
<=> \(\sqrt{\left(x+2\right)-4\sqrt{x+2}+4}+\sqrt{\left(x+2\right)-6\sqrt{x+2}+9}\)=1
<=> \(\sqrt{\left(\sqrt{x+2}-2\right)^2}+\sqrt{\left(\sqrt{x+2}-3\right)^2}=1\)
<=> \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|\)=1 (1)
TH1: \(0\le\sqrt{x+2}< 2\)
Từ (1) =>\(2-\sqrt{x+2}+3-\sqrt{x+2}=1\)
<=> \(5-2\sqrt{x+2}=1\) <=> \(2\sqrt{x+1}=4\) <=> \(\sqrt{x+1}=2\)
<=> \(x+1=4\) <=> x=3(không t/m \(\sqrt{x+2}\le2\))
TH2 : \(2\le\sqrt{x+2}\le3\)
Từ (1) =>\(\sqrt{x+2}-2+3-\sqrt{x+2}=1\)
<=> \(1=1\) (luôn đúng)
Từ TH2 <=> 4\(\le x+2\le9\) <=> \(2\le x\le7\)
TH3 \(\sqrt{x+2}>3\)
Từ (1) => \(\sqrt{x+2}-2+\sqrt{x+2}-3=1\)
<=> \(2\sqrt{x+2}=6\) <=> \(\sqrt{x+2}=3\) <=> \(x+2=9\) <=> x=7 (không t/m \(\sqrt{x+2}>3\))
Vậy pt (*) có tập nghiệm S=\(\left\{2\le x\le7\right\}\)
b, \(x^2-10x+27=\sqrt{6-x}+\sqrt{x-4}\) (*) (đk :\(4\le x\le6\))
Vs a,b \(\ge0\) ta có \(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a^2+b^2\right)}\)(tự CM nha)
Dấu "=" xảy ra <=> a=b
Áp dụng bđt trên ta có: \(\sqrt{6-x}+\sqrt{x-4}\le\sqrt{2\left(6-x+x-4\right)}=\sqrt{2.2}=2\)
<=> \(\sqrt{6-x}+\sqrt{x-4}\le2\)(1)
Lại có: \(x^2-10x+27=x^2-10x+25+2=\left(x-5\right)^2+2\ge2\)
<=> \(x^2-10x+27\ge2\) (2)
Từ (1),(2) => Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}6-x=x-4\\x-5=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}6+4=2x\\x=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=5\\x=5\end{matrix}\right.\left(tm\right)\)
Vậy pt (*) có tập nghiệm S=\(\left\{5\right\}\)
c, \(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\)(*) (đk: x\(\ge0\))
<=> \(x\left(x-2\right)-\sqrt{x}\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)
<=> \(\left(x-\sqrt{x}\right)\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)
<=> \(\sqrt{x}\left(\sqrt{x}-1\right)\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)
<=> \(\left(\sqrt{x}-1\right)\left[\sqrt{x}\left(x-2\right)-4\right]=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}\left(x-2\right)-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}\left(x-2\right)=4\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x\left(x-2\right)^2=16\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=1\\x\left(x^2-4x+4\right)-16=0\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=1\\x^3-4x^2+4x-16=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=1\\x^2\left(x-4\right)+4\left(x-4\right)=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=1\\\left(x^2+4\right)\left(x-4\right)=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\left(tm\right)\)
Vậy pt (*) có tập nghiệm S=\(\left\{1;4\right\}\)
d) x2+3x+1=(x+3)\(\sqrt{x^2+1}\)
<=>(\(\sqrt{x^2+1}-3x+3\sqrt{x^2+1}-\left(x^2+1\right)=0\)
<=>\(\left(\sqrt{x^2+1}-3\right)\left(x-\sqrt{x^2+1}\right)=0\)
<=>\(\sqrt{x^2+1}=3\) hoặc \(x=\sqrt{x^2+1}\)
=>x=\(2\sqrt{2}\)
\(\sqrt{x+6-2\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\)
\(\Rightarrow\sqrt{\left(\sqrt{x+2}-1\right)^2+3}+\sqrt{\left(\sqrt{x+2}-3\right)^2}=1\)
\(\Rightarrow\sqrt{\left(\sqrt{x+2}-1\right)^2+3}+\sqrt{x+2}=4\)
\(\Rightarrow\left(\sqrt{x+2}-1\right)^2+3=\left(4-\sqrt{x}+2\right)^2\)
\(\Rightarrow x+2-2\sqrt{x+2}+1+3=16-8\sqrt{x+2}+x+2\)
\(\Rightarrow x-2\sqrt{x+2}-x+8\sqrt{x+2}=12\)
\(\Rightarrow6\sqrt{x+2}=12\)
\(\Rightarrow\sqrt{x+2}=2\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=2\)
Vậy x=2
\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\) ( ĐK : \(x\ge-2\) )
\(\Leftrightarrow\sqrt{x+2-4\sqrt{x+2}+4}+\sqrt{x+2-6\sqrt{x+2}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+2}-2\right)^2}+\sqrt{\left(\sqrt{x+2}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|=1\)
Ta có : \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|=\left|\sqrt{x+2}-2\right|+\left|3-\sqrt{x+2}\right|\)
Áp dụng BĐT : \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow\left|\sqrt{x+2}-2\right|+\left|3-\sqrt{x+2}\right|\ge\left|\sqrt{x+2}-2+3-\sqrt{x+2}\right|=1\)
Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x+2}-2\ge0\\3-\sqrt{x+2}\ge0\end{matrix}\right.\Leftrightarrow2\le x\le7\)