Rút gọn biểu thức
P = \(\dfrac{x+\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{x-6\sqrt{x}+4}{x-4}\)
và tìm giá trị của P khi x=9 + 4\(\sqrt{5}\)
K
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Những câu hỏi liên quan
19 tháng 10 2021
\(a,P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ P=\dfrac{x+16}{\sqrt{x}+3}\\ b,P=4\Leftrightarrow\dfrac{x+16}{\sqrt{x}+3}=4\\ \Leftrightarrow x+16=4\sqrt{x}+12\\ \Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\\ \Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
\(c,P=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\\ P=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6=2\cdot5-6=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\Leftrightarrow\sqrt{x}+3=5\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,x=3-2\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{2}-1\\ \Leftrightarrow P=\dfrac{3-2\sqrt{2}+16}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{\sqrt{2}+2}\\ P=\dfrac{\left(19-2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}=\dfrac{42-23\sqrt{2}}{2}\)
22 tháng 10 2023
giải phương trình
a)\(\sqrt{x^8}=256\) b)\(\sqrt{x^2-2x+1}=x-1\)
DH
2 tháng 3 2021
\(P=\left(\dfrac{x}{x\sqrt{x}-4\sqrt{x}}-\dfrac{6}{3\sqrt{x}-6}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
\(P=\left(\dfrac{\sqrt{x}}{x-4}-\dfrac{2\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}-2}{x-4}\right):\left(\dfrac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(P=\left(\dfrac{-6}{x-4}\right):\left(\dfrac{6}{\sqrt{x}+2}\right)=\dfrac{-1}{\sqrt{x}-2}\)
H9
HT.Phong (9A5)
CTVHS
5 tháng 8 2023
a) Thay x=25 vào B ta có:
\(B=\dfrac{\sqrt{25}+2}{\sqrt{25}-2}=\dfrac{7}{3}\)
b) \(A=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-1}{x-5\sqrt{x}+6}\)
\(A=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{x-9-x+4+2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{2\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{2}{\sqrt{x}-2}\)
c) Ta có: \(A>B\) Khi:
\(\dfrac{2}{\sqrt{x}-2}>\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\dfrac{2-\sqrt{x}-2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\dfrac{-\sqrt{x}}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}-\sqrt{x}< 0\\\sqrt{x}-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}-\sqrt{x}>0\\\sqrt{x}-2>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x>4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow0< x< 4\)
LD
17 tháng 5 2021
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
ĐKXĐ: \(x\ge0;x\ne4\)
\(P=\dfrac{x+\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{x-6\sqrt{x}+4}{x-4}\)
\(=\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-\left(2x-4\sqrt{x}-\sqrt{x}+2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-2x+4\sqrt{x}+\sqrt{x}-2+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(x+1\right)+2\left(x+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+1}{\sqrt{x}-2}\)
Khi \(x=9+4\sqrt{5}\)
Ta có: \(4+4\sqrt{5}+5=2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=\left(2+\sqrt{5}\right)^2\)
\(\Rightarrow\sqrt{x}=2+\sqrt{5}\)
\(\Rightarrow P=\dfrac{\left(2+\sqrt{5}\right)^2+1}{2+\sqrt{5}-2}=\dfrac{9+4\sqrt{5}+1}{\sqrt{5}}=\dfrac{10+4\sqrt{5}}{\sqrt{5}}=4+2\sqrt{5}\)
Vậy \(P=4+2\sqrt{5}\) khi \(x=9+4\sqrt{5}\).
\(D=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-2x+4\sqrt{x}+\sqrt{x}-2+x-6\sqrt{x}+4}{x-4}\)
\(=\dfrac{x\sqrt{x}+2x+2}{x-4}\)
Khi x=9+4căn 5 thì \(D=\dfrac{\left(9+4\sqrt{5}\right)\left(\sqrt{5}+2\right)+2\sqrt{5}+4+2}{\sqrt{5}-2}\)
\(=\dfrac{9\sqrt{5}+18+20+8\sqrt{5}+2\sqrt{5}+6}{\sqrt{5}-2}\)
=(44+19căn 5)*(căn 5+2)
=44căn 5+88+95+38căn 5
=82căn 5+183