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1) ĐKXĐ: \(\left\{{}\begin{matrix}x^2+x+1\ge0\\x^2+1\ne0\end{matrix}\right.\)
Ta có:
+) \(x^2+x+1=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
+) \(x^2+1\ge1>0\forall x\)
Vậy biểu thức luôn xác định với mọi x
2) ĐKXĐ: \(\left\{{}\begin{matrix}x^2-2x+3>0\\x^2-x+1\ge0\end{matrix}\right.\)
Ta có:
+) \(x^2-2x+3=\left(x^2-2x+1\right)+2\)
\(=\left(x-1\right)^2+2\ge2>0\forall x\)
+) \(x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
Vậy biểu thức luôn xác định với mọi x
\(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{3x+\sqrt{x}+3-x+\sqrt{x}-1}{\sqrt{x}}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
a) Ta có: \(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x^2+\sqrt{x}}{x\sqrt{x}+x}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{x^2-2x+1}{2}\)
a)
Đkxđ:\(\left\{{}\begin{matrix}x-1\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ge0\end{matrix}\right.\)
\(=\)\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(x-1\right)^2}{2}\)
\(=\frac{x\sqrt{x}+2x+\sqrt{x}-2x-4\sqrt{x}-2-x\sqrt{x}+\sqrt{x}-2x+2}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}-2x}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}\left(1+\sqrt{x}\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}\left(x-1\right)}{2\left(\sqrt{x}+1\right)}=\frac{-2\sqrt{x}\left(x-1\right)}{2\sqrt{x}+2}\)
2:
a: Sửa đề: \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)
\(A=\dfrac{a^2+3}{\sqrt{a^2+2}}=\dfrac{a^2+2+1}{\sqrt{a^2+2}}=\sqrt{a^2+2}+\dfrac{1}{\sqrt{a^2+2}}\)
=>\(A>=2\cdot\sqrt{\sqrt{a^2+2}\cdot\dfrac{1}{\sqrt{a^2+2}}}=2\)
A=2 thì a^2+2=1
=>a^2=-1(loại)
=>A>2 với mọi a
b: \(\Leftrightarrow\sqrt{a}+\sqrt{b}< =\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}\)
=>\(a\sqrt{a}+b\sqrt{b}>=a\sqrt{b}+b\sqrt{a}\)
=>\(\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)>=0\)
=>(căn a+căn b)(a-2*căn ab+b)>=0
=>(căn a+căn b)(căn a-căn b)^2>=0(luôn đúng)
1
ĐK: `x>1`
PT trở thành:
\(\sqrt{\dfrac{2x-3}{x-1}}=2\\ \Leftrightarrow\dfrac{2x-3}{x-1}=2^2=4\\ \Leftrightarrow4x-4-2x+3=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\left(KTM\right)\)
Vậy PT vô nghiệm.
b
ĐK: \(x\ge2\)
Đặt \(t=\sqrt{x-2}\) (\(t\ge0\))
=> \(x=t^2+2\)
PT trở thành: \(t^2+2-5t+2=0\)
\(\Leftrightarrow t^2-5t+4=0\)
nhẩm nghiệm: `a+b+c=0` (`1+(-5)+4=0`)
\(\Rightarrow\left\{{}\begin{matrix}t=1\left(nhận\right)\\t=4\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{x-2}=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\left(TM\right)\\x=18\left(TM\right)\end{matrix}\right.\)
bạn cm các biểu thức trong căn > 0 ∀ x là xong =))