Cho biểu thức
A= \(\left(\dfrac{\sqrt{a}-1}{3\sqrt{a}-1}-\dfrac{1}{1+3\sqrt{a}}+\dfrac{8\sqrt{a}}{9a-1}\right):\left(1-\dfrac{3\sqrt{a}-2}{3\sqrt{a}+1}\right)\)
a) Rút gọn A
b) Tìm giá trị của a để A= \(\dfrac{6}{5}\)
K
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H9
HT.Phong (9A5)
CTVHS
28 tháng 7 2023
a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)
\(M=\sqrt{3}-\sqrt{3}+1\)
\(M=1\)
b) Ta có:
\(N=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(N=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)
\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Theo đề ta có: \(M=2N\)
Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)
\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)
\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=2\)
\(\Leftrightarrow a=4\left(tm\right)\)
24 tháng 11 2021
\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
27 tháng 7 2021
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}+3}\)
b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow1-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 1\)
hay x<1
Kết hợp ĐKXĐ, ta được: 0<x<1
20 tháng 10 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)
\(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{a-1}\)
\(=\dfrac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)}{\sqrt{a}}+\dfrac{3a+3\sqrt{a}-a-\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
b: \(P=\sqrt{a}+7\)
=>\(2\left(a+2\sqrt{a}+1\right)=a+7\sqrt{a}\)
=>\(2a+4\sqrt{a}+2-a-7\sqrt{a}=0\)
=>\(a-3\sqrt{a}+2=0\)
=>\(\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\)
=>\(\left[{}\begin{matrix}a=1\left(loại\right)\\a=4\left(nhận\right)\end{matrix}\right.\)
c: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}=\dfrac{2a-2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\left(a-\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}\right)}{\sqrt{a}}=\dfrac{2\left[\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}{\sqrt{a}}>0\)
=>P>6
5 tháng 4 2022
a: \(P=\dfrac{a+\sqrt{a}+1}{a+1}:\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{a+\sqrt{a}+1}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+\sqrt{a}+1}{\sqrt{a}-1}\)
b: Để P<1 thì P-1<0
\(\Leftrightarrow\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)
hay 0<a<1
26 tháng 12 2021
a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
27 tháng 12 2021
\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
AH
Akai Haruma
Giáo viên
12 tháng 6 2021
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
a)
\(A=\frac{x+\sqrt{x}+1}{x+1}:\left[\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{(\sqrt{x}-1)(x+1)}\right]\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{(\sqrt{x}-1)(x+1)}=\frac{x+\sqrt{x}+1}{x+1}.\frac{(\sqrt{x}-1)(x+1)}{(\sqrt{x}-1)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b)
\(A=7\Leftrightarrow x+\sqrt{x}+1=7(\sqrt{x}-1)\)
\(\Leftrightarrow x-6\sqrt{x}+8=0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x=4\\ x=16\end{matrix}\right.\) (đều thỏa mãn)
c)
\(x=2(2+\sqrt{3})=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\Rightarrow \sqrt{x}=\sqrt{3}+1\)
\(\Rightarrow A=\frac{4+2\sqrt{3}+\sqrt{3}+1+1}{\sqrt{3}}=\frac{6+3\sqrt{3}}{\sqrt{3}}=3+2\sqrt{3}\)
d)
\(A< 1\Leftrightarrow \frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1<0\Leftrightarrow \frac{x-2\sqrt{x}+2}{\sqrt{x}-1}<0\)
\(\Leftrightarrow \frac{(\sqrt{x}-1)^2+1}{\sqrt{x}-1}<0\Leftrightarrow \sqrt{x}-1< 0\Leftrightarrow 0\leq x< 1\)
11 tháng 9 2023
a: \(A=3+\left(-2\right)\cdot\sqrt{3}+3\cdot\sqrt{3}-2-\sqrt{3}\)
\(=3-2=1\)
\(B=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b: B<A
=>B-1<0
=>\(\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}}< 0\)
=>-1/căn x<0
=>căn x>0
=>x>0 và x<>1
a.(ĐKXĐ: \(a\ge0,a\ne\dfrac{1}{9}\))
=> \(A=\left(\dfrac{\sqrt{a}-1}{3\sqrt{a}-1}-\dfrac{1}{1+3\sqrt{a}}+\dfrac{8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\right):\dfrac{3\sqrt{a}+1-3\sqrt{a}+2}{3\sqrt{a}+1}=\dfrac{\left(\sqrt{a}-1\right)\left(1+3\sqrt{a}\right)-3\sqrt{a}+1+8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}.\dfrac{3\sqrt{a}+1}{3}\)
\(=\dfrac{\sqrt{a}+3a-1-3\sqrt{a}-3\sqrt{a}+1+8\sqrt{a}}{3\left(3\sqrt{a}-1\right)}=\dfrac{3\sqrt{a}+3a}{3\left(3\sqrt{a}-1\right)}=\dfrac{3\left(\sqrt{a}+a\right)}{3\left(3\sqrt{a}-1\right)}=\dfrac{\sqrt{a}+a}{3\sqrt{a}-1}\)
b. Để A \(=\dfrac{6}{5}\Leftrightarrow\dfrac{\sqrt{a}+a}{3\sqrt{a}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(\sqrt{a}+a\right)=6\left(3\sqrt{a}-1\right)\)
\(\Leftrightarrow5\sqrt{a}+5a-18\sqrt{a}+6=0\)
\(\Leftrightarrow5a-13\sqrt{a}+6=0\)
\(\Leftrightarrow5a-10\sqrt{a}-3\sqrt{a}+6=0\)
\(\Leftrightarrow5\sqrt{a}\left(\sqrt{a}-2\right)-3\left(\sqrt{a}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(5\sqrt{a}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}-2=0\\5\sqrt{a}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=2\\\sqrt{a}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4\\a=\dfrac{9}{25}\end{matrix}\right.\)(nhận)
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