1. Cho A = \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{1}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
a) Rút gọn A.
b) Tìm a để A = 4; A >-6.
c) Tính A khi \(a^3-3=0\)
Cho biểu thức \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\cdot\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
a. Rút gọn A
b. Tìm \(x\) để \(A>-6\)
c. Tính A khi \(a^2-3=0\)
a) \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\left(đk:a>0,x\ne1\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}.\dfrac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{a-1}\)
\(=\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{2\sqrt{a}}\)
\(=\dfrac{-4a}{2\sqrt{a}}=-2\sqrt{a}\)
b) \(A=-2\sqrt{a}>-6\)
\(\Leftrightarrow\sqrt{a}< 3\Leftrightarrow0\le a< 9\) và \(a\ne1\)
c) \(a^2-3=0\Leftrightarrow a^2=3\Leftrightarrow\sqrt{a}=\sqrt[4]{3}\)
\(\Rightarrow A=-2\sqrt{a}=-2\sqrt[4]{3}\)
B=\(\left(\dfrac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{3}{\sqrt{1-a^2}}+1\right)\)
a) Rút gọn
b) Tìm B khi a=\(\dfrac{\sqrt{3}}{2+\sqrt{3}}\)
c) Tìm a để \(\sqrt{B}>B\)
a) ĐKXĐ: \(a>1;a\ne-1\)
\(B=\left(\dfrac{3}{\sqrt{1+a}}+\dfrac{\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right):\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)
\(\Leftrightarrow B=\dfrac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}.\dfrac{\sqrt{1+a}.\sqrt{1-a}}{3+\sqrt{1+a}.\sqrt{1-a}}\)
\(\Leftrightarrow B=\sqrt{1-a}\)
b) Thay a=\(\dfrac{\sqrt{3}}{2+\sqrt{3}}\) vào B ta được:
\(B=\sqrt{1-\dfrac{\sqrt{3}}{2+\sqrt{3}}}\)
\(\Leftrightarrow B\) \(=\sqrt{\dfrac{2+\sqrt{3}-\sqrt{3}}{2+\sqrt{3}}}\)
\(\Leftrightarrow B\) \(=\sqrt{\dfrac{2}{2+\sqrt{3}}}\)
\(\Leftrightarrow B\)\(=\sqrt{\dfrac{4}{4+2\sqrt{3}}}\) \(\Leftrightarrow B\) \(=\dfrac{\sqrt{4}}{\sqrt{3+2\sqrt{3}+1}}\)
\(\Leftrightarrow B=\dfrac{2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\) \(\Leftrightarrow B=\dfrac{2}{\sqrt{3}+1}=\dfrac{2.\left(\sqrt{3}-1\right)}{3-1}=\sqrt{3}-1\)
c) Có \(\sqrt{B}>B\) \(\Leftrightarrow\sqrt{\sqrt{1-a}}>\sqrt{1-a}\)
\(\Leftrightarrow\sqrt{1-a}>1-a\)
\(\Leftrightarrow\sqrt{1-a}-\left(1-a\right)>0\)
\(\Leftrightarrow\sqrt{1-a}.\left(1-\sqrt{1-a}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{1-a}>0\\1-\sqrt{1-a}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{1-a}< 0\\1-\sqrt{1-a}< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a< 1\\a>0\end{matrix}\right.\\\left\{{}\begin{matrix}a>1\\a< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}0< a< 1\\a>1;a< 0\end{matrix}\right.\)
A=\(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)với a>0,a khác 1
a)rút gọn A
b)tính giá trị của A biết a=4+2\(\sqrt{3}\)
c)tìm a để A<0
a) \(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
\(A=\left[\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
\(A=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right]\)
\(A=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left[\dfrac{\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(A=\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(A=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\left(\sqrt{a}-1\right)\)
\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
\(A=\dfrac{a-1}{\sqrt{a}}\)
b) Ta có:
\(a=4+2\sqrt{3}=\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2=\left(\sqrt{3}+1\right)^2\)
Thay vào A ta có:
\(A=\dfrac{\left(\sqrt{3}+1\right)^2-1}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\dfrac{4+2\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{3+2\sqrt{3}}{\sqrt{3}+1}\)
c) \(A< 0\) khi:
\(\dfrac{a-1}{\sqrt{a}}< 0\)
Mà: \(\sqrt{a}\ge0\forall x\) (xác định)
\(\Leftrightarrow a-1< 0\)
\(\Leftrightarrow a< 1\)
Kết hợp với đk:
\(0< a< 1\)
cho C= \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right)\div\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a) rút gọn C
b) tìm a để C= \(\dfrac{1}{4}\)
lm nhanh giúp mk nhé mk đang cần gấp
a) \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(a>0.a\ne1\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1-\sqrt{a}-2}{\sqrt{a}-1}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{-1}{\sqrt{a}-1}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(1-\sqrt{a}\right)=-\dfrac{1}{\sqrt{a}}\)
b) \(C=\dfrac{1}{4}\Rightarrow-\dfrac{1}{\sqrt{a}}=\dfrac{1}{4}\Rightarrow\sqrt{a}=-4\) (vô lý) \(\Rightarrow\) không có a thỏa đề
Cho biểu thức: Q= \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
a) Tìm ĐKXĐ và rút gọn P.
b) Tìm a để Q dương.
c) Tính giá trị của biểu thức biết a= \(9-4\sqrt{5}\).
Sửa đề: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a+3\sqrt{a}+2-a+3\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{6\sqrt{a}}\)
\(=\dfrac{a-4}{6a\left(\sqrt{a}-1\right)}\)
c) Thay \(a=9-4\sqrt{5}\) vào Q, ta được:
\(Q=\dfrac{5-4\sqrt{5}}{6\left(9-4\sqrt{5}\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(9\sqrt{5}-27-20+12\sqrt{5}\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(21\sqrt{5}-47\right)}\)
\(=\dfrac{\left(5-4\sqrt{5}\right)\left(21\sqrt{5}+47\right)}{-24}\)
\(=\dfrac{105\sqrt{5}+235-420-188\sqrt{5}}{-24}\)
\(=\dfrac{-83\sqrt{5}-185}{-24}=\dfrac{83\sqrt{5}+185}{24}\)
Rút gọn các biểu thức:
\(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{x-4}{3\sqrt{x}}\)
\(B=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}+\dfrac{6-7\sqrt{a}}{a-4}\right).\left(\sqrt{a}+2\right)\)
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
rút gọn biểu thức a
A= \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a/ rút gọn A
b/ tìm giá trị để A dương
a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
C=(\(B=\left(\dfrac{1}{\sqrt{x-1}}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\) a Tìm đkxd của B
b rút gọn B
c tìm a sao cho B \(\le\)\(\dfrac{1}{3}\)
Cho \(B=\left(1+\dfrac{\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
a, Rút gọn B
b, Tìm a để B<1
c, Cho \(a=19-8\sqrt{3}\). Tính B
d, Tìm a ∈ Z để b ∈ Z
e, Tìm giá trị lớn nhất của M
Cho biểu thức: M= \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\).
a) Tìm ĐKXĐ của M.
b) Rút gọn M.
c) Tìm giá trị của a để M=-4.
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
b) Ta có: \(M=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}\left[\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right]}{2\sqrt{a}}\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}\)
\(=\dfrac{-4\sqrt{a}}{2}=-2\sqrt{a}\)
c) Để M=-4 thì \(-2\sqrt{a}=-4\)
\(\Leftrightarrow\sqrt{a}=2\)
hay a=4(thỏa ĐK)