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a/ Giải rồi
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)
Pt trở thành:
\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)
\(\Leftrightarrow...\)
e/ ĐKXD: \(x>0\)
\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)
\(\Rightarrow t^2=x+\frac{1}{4x}+1\)
Pt trở thành:
\(5t=2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)
\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)
a) + \(VT=\sqrt{x^2+2x+10}+x^2+2x+1+7\)
\(=\sqrt{x^2+2x+1}+\left(x+1\right)^2+7>0\forall x\)
=> ptvn
d) ĐK : \(x^2+7x+7\ge0\)
Đặt \(t=\sqrt{x^2+7x+7}\ge0\) \(\Rightarrow t^2=x^2+7x+7\)
\(pt\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0\)
\(\Leftrightarrow3t^2+2t-5=0\Leftrightarrow\left(3t+5\right)\left(t-1\right)=0\)
\(\Leftrightarrow t=1\) ( do \(3t+5>0\forall t\ge0\) )
\(\Leftrightarrow x^2+7x+1=0\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) ( TM )
f) ĐK : \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\) thì pt trở thành :
\(a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)
\(a,A=\frac{\sqrt{x}+1}{\sqrt{x}}\) ĐKXĐ: x> 0
Với x = 81 ta có:
\(A=\frac{\sqrt{81}+1}{\sqrt{81}}=\frac{9+1}{9}=\frac{10}{9}\)
b,
\(ĐKXĐ:\hept{\begin{cases}\sqrt{x}-1\ne0\\\sqrt{x}-2\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\\x\ne4\end{cases}}}\)
\(B=\frac{3x}{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\frac{3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3x-x+1+x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3x-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\frac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3\sqrt{x}+3}{\sqrt{x}-2}\)
ta có:\(\frac{x-2\sqrt{x}+1}{x-\sqrt{x}+1}=\frac{1}{2}\)
\(\Rightarrow x-3\sqrt{x}+1=0\)
\(\Rightarrow\hept{\begin{cases}x+1=3\sqrt{x}\\x-3\sqrt{x}=-1\end{cases}}\)
lại có \(B=\frac{3x\sqrt{x}+10x+19}{x^2+7x+15}\)
\(=\frac{3x\sqrt{x}-9x+19x+19}{x^2-9x+16x+15}\)
\(=\frac{3\sqrt{x}\left(x-3\sqrt{x}\right)+19\left(x+1\right)}{\left(x+3\sqrt{x}\right)\left(x-3\sqrt{x}\right)+16x+15}\)
\(=\frac{-3\sqrt{x}+19\times3\sqrt{x}}{-1\times\left(x+3\sqrt{x}\right)+16x+15}\)
\(=\frac{57\sqrt{x}-3\sqrt{x}}{15x+15-3\sqrt{x}}\)
\(=\frac{54\sqrt{x}}{15\left(x+1\right)-3\sqrt{x}}\)
\(=\frac{54\sqrt{x}}{45\sqrt{x}-3\sqrt{x}}\)
\(=\frac{54\sqrt{x}}{42\sqrt{x}}=\frac{27}{21}\)