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1.
ĐKXĐ: \(x< 5\)
\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)
\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)
\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)
\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
b.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=2\)
a: \(x^2\cdot2\sqrt{3}+x+1=\sqrt{3}\cdot\left(x+1\right)\)
=>\(x^2\cdot2\sqrt{3}+x\left(1-\sqrt{3}\right)+1-\sqrt{3}=0\)
\(\text{Δ}=\left(1-\sqrt{3}\right)^2-4\cdot2\sqrt{3}\left(1-\sqrt{3}\right)\)
\(=4-2\sqrt{3}-8\sqrt{3}\left(1-\sqrt{3}\right)\)
\(=4-2\sqrt{3}-8\sqrt{3}+24=28-10\sqrt{3}=\left(5-\sqrt{3}\right)^2>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(1-\sqrt{3}\right)-\left(5-\sqrt{3}\right)}{2\cdot2\sqrt{3}}=\dfrac{-1+\sqrt{3}-5+\sqrt{3}}{4\sqrt{3}}=\dfrac{1-\sqrt{3}}{2}\\x_2=\dfrac{-\left(1-\sqrt{3}\right)+5-\sqrt{3}}{2\cdot2\sqrt{3}}=\dfrac{4}{4\sqrt{3}}=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
b: \(5x^2-3x+1=2x+31\)
=>\(5x^2-3x+1-2x-31=0\)
=>\(5x^2-5x-30=0\)
=>\(x^2-x-6=0\)
=>(x-3)(x+2)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
c: \(x^2+2\sqrt{2}x+4=3\left(x+\sqrt{2}\right)\)
=>\(x^2+2\sqrt{2}x+4-3x-3\sqrt{2}=0\)
=>\(x^2+x\left(2\sqrt{2}-3\right)+4-3\sqrt{2}=0\)
\(\text{Δ}=\left(2\sqrt{2}-3\right)^2-4\left(4-3\sqrt{2}\right)\)
\(=17-12\sqrt{2}-16+12\sqrt{2}=1\)>0
Do đó, phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(2\sqrt{2}-3\right)-1}{2}=\dfrac{-2\sqrt{2}+3-1}{2}=-\sqrt{2}+1\\x_2=\dfrac{-\left(2\sqrt{2}-3\right)+1}{2}=\dfrac{-2\sqrt{2}+4}{2}=-\sqrt{2}+2\end{matrix}\right.\)
1)\(\sqrt{4x^2+12x+9}=2-x\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)
\(\Leftrightarrow\left|2x+3\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
\(\)
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
b) đặt \(\sqrt{3x+1}=a\)(\(a\ge0\))
\(PT\Leftrightarrow\dfrac{a^2-1}{\sqrt{a^2+9}}+1=a\)
\(\Leftrightarrow\left(a-1\right)\left(1-\dfrac{a+1}{\sqrt{a^2+9}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+1=\sqrt{a^2+9}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)(tm)
c) bunyalovsky:
\(VT^2\le2\left(7-x+x-5\right)=4\)
\(\Leftrightarrow VT\le2\)
\(VF=\left(x-6\right)^2+2\ge2\)
Dấu = xảy ra khi x=6