Cho biểu thức: P=\(\dfrac{1}{\sqrt{x}+1}+\dfrac{x}{\sqrt{x}-x}\)
a.Rút gọn P
b.Tính P khi x=3 ;x=3 +\(2\sqrt{2}\)
c.Tìm x để P nguyên
d.Tìm x khi P=2
K
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AH
Akai Haruma
Giáo viên
3 tháng 4 2022
Lời giải:
a. ĐKXĐ: $x>0; x\neq 1$
\(P=\frac{1}{\sqrt{x}+1}+\frac{x}{\sqrt{x}(1-\sqrt{x})}=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}=\frac{1-\sqrt{x}+x+\sqrt{x}}{(1-\sqrt{x})(\sqrt{x}+1)}=\frac{x+1}{1-x}\)
b.
\(P=\frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{2}\)
14 tháng 2 2021
Sửa đề: \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{9}{4}\end{matrix}\right.\)
a) Ta có: \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2\cdot\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\cdot\dfrac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{2x+2\sqrt{x}+\sqrt{x}+1}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b) Ta có: \(x=\dfrac{3-2\sqrt{2}}{4}\)
\(\Leftrightarrow x=\dfrac{2-2\cdot\sqrt{2}\cdot1+1}{4}\)
\(\Leftrightarrow x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\)(thỏa ĐK)
Thay \(x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\) vào biểu thức \(P=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\), ta được:
\(P=\left(3\cdot\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{4}}-5\right):\left(2\cdot\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{4}}+1\right)\)
\(\Leftrightarrow P=\left(3\cdot\dfrac{\sqrt{2}-1}{2}-5\right):\left(2\cdot\dfrac{\sqrt{2}-1}{2}+1\right)\)
\(\Leftrightarrow P=\left(\dfrac{3\cdot\left(\sqrt{2}-1\right)}{2}-\dfrac{10}{2}\right):\left(\sqrt{2}-1+1\right)\)
\(\Leftrightarrow P=\dfrac{3\sqrt{2}-3-10}{2}:\sqrt{2}\)
\(\Leftrightarrow P=\dfrac{3\sqrt{2}-13}{2}\cdot\sqrt{2}\)
\(\Leftrightarrow P=\dfrac{6-13\sqrt{2}}{2}\)
Vậy: Khi \(x=\dfrac{3-2\sqrt{2}}{4}\) thì \(P=\dfrac{6-13\sqrt{2}}{2}\)
13 tháng 7 2021
a) Ta có: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
MH
26 tháng 4 2022
Điều kiện: \(x\ge0,x\ne1\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}+1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
Ta có \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0,\forall x\Rightarrow A>0\)
Lại có: \(A-2=\dfrac{2}{x+\sqrt{x}+1}-2=\dfrac{-2\left(x+\sqrt{x}\right)}{x+\sqrt{x}+1}\)
Mà \(x+\sqrt{x}+1>0;x+\sqrt{x}>0\) với mọi \(x\in TXĐ\)
\(\Rightarrow A-2< 0\Rightarrow A< 2\)
Vậy \(0< A< 2\)
12 tháng 8 2021
a)A=\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b) Thay x=3+2\(\sqrt{2}\)
A=\(\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}\)=\(\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2-2}}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)=\(\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}\)
A=\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
c)Ta có \(\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)>0
\(\Rightarrow\dfrac{2}{\sqrt{x}}\)<1\(\Rightarrow\sqrt{x}\)>2\(\Rightarrow x>4\)
H
12 tháng 11 2023
`a,`
\(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\\ =\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{4}{\sqrt{x}+1}\)
`b,` Để `A *B<0` ta có :
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{4}{\sqrt{x}+1}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-1}< 0\\ \Leftrightarrow\sqrt{x}-1< 0\left(vì.4>0\right)\\ \Leftrightarrow\sqrt{x}< 1\\ \Leftrightarrow0\le x< 1\)
Kết hợp với đkxđ ta có : \(0< x< 1\)
10 tháng 2 2021
ĐKXĐ: \(x>0\)
a) Ta có: \(M=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b) Vì x=16 thỏa mãn ĐKXĐ
nên Thay x=16 vào biểu thức \(M=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\), ta được:
\(M=\dfrac{16+\sqrt{16}+1}{\sqrt{16}}=\dfrac{16+4+1}{4}=\dfrac{21}{4}\)
Vậy: Khi x=16 thì \(M=\dfrac{21}{4}\)
c) Để \(M=\dfrac{13}{3}\) thì \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\dfrac{13}{3}\)
\(\Leftrightarrow3\left(x+\sqrt{x}+1\right)=13\sqrt{x}\)
\(\Leftrightarrow3x+3\sqrt{x}+3-13\sqrt{x}=0\)
\(\Leftrightarrow3x-10\sqrt{x}+3=0\)
\(\Leftrightarrow3x-\sqrt{x}-9\sqrt{x}+3=0\)
\(\Leftrightarrow\sqrt{x}\left(3\sqrt{x}-1\right)-3\left(3\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}=1\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{1}{3}\\x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{9}\left(nhận\right)\\x=9\left(nhận\right)\end{matrix}\right.\)
Vậy: Để \(M=\dfrac{13}{3}\) thì \(x\in\left\{\dfrac{1}{9};9\right\}\)