Cho \(\ A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right) :\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a) Rút gọn A
b) Tìm a để \(A>\frac{1}{6}\)
Bai 1)A=\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right),\)
a)rút gọn A
2) B=1+\(\left(\frac{2a+\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}-1}{1-a\sqrt{a}}\right).\left(\frac{a-\sqrt{a}}{2\sqrt{a}-1}\right),\)
a) rút gọn B
b) tìm a để B=\(\frac{\sqrt{6}}{1+\sqrt{6}}\)
c) chứng minh rằng B>2/3
Giúp mk với nếu đúng mk tick cho nha ,cảm ơn
cho biểu thức A =\(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a/ Rút gọn A
b/ Tìm giá trị của a để A>1/6
\(đkxđ\Leftrightarrow x\ge0;x\ne1;x\ne4\)
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)
\(=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\)\(\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
\(A< \frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
\(\Rightarrow\frac{2\left(\sqrt{a}-2\right)}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\Rightarrow\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
Vì \(6\sqrt{a}>0\Rightarrow\sqrt{a}-4>0\Rightarrow\sqrt{a}>4\Rightarrow a>16\)
Vậy \(P>\frac{1}{6}\Leftrightarrow a>16\)
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)....\)
a) Rút gọn A, b) tìm a để A<0.c) Tìm A để A=-2
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\).\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
= \(\left[\left(\frac{\sqrt{a}}{2}\right)^2-2\frac{\sqrt{a}}{2}\frac{1}{2\sqrt{a}}+\left(\frac{1}{2\sqrt{a}}\right)^2\right]\).\(\left[\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{a-1}\right]\)
=\(\left(\frac{a}{4}-\frac{1}{2}+\frac{1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1\right)^2}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)^2}{a-1}\right]\)
=\(\left(\frac{a^2}{4a}-\frac{2a}{4a}+\frac{1}{4a}\right)\).\(\left[\frac{\left[\left(\sqrt{a}-1\right)-\left(\sqrt{a}+1\right)\right]\cdot\left[\left(\sqrt{a}-1\right)+\left(\sqrt{a}+1\right)\right]}{a-1}\right]\)
=\(\left(\frac{a^2-2a+1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1-\sqrt{a}+1\right).\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right]\)
=\(\frac{\left(a-1\right)^2}{1}\).\(\frac{-4\sqrt{a}}{a-1}\)
=\(\frac{-\left(a-1\right)}{1}\)= - a + 1
hok tốt
\(ChoQ=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
a, rút gọn
b, chứng minh nếu 0<x<1 thì Q>0
c, tìm GTLN của Q
\(ChoA=\frac{1}{2\left(1+\sqrt{x}+2\right)}+\frac{1}{2\left(1-\sqrt{x}+2\right)}\)
a, tìm x để a có nghĩa
b, rút gon A
c, tìm X nguyên để A nguyên
\(ChoA=\left(\frac{\sqrt{a}}{\sqrt{a-1}}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2}{a-1}\right)\)
a, Rút gọn A
b, tính A Khi a=3+\(2\sqrt{2}\)
Cho biểu thức :
\(A=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a-1}}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
a, Rút gọn A
b, Tìm a để A=7
c, Tìm a để A>6
Cho biểu thức P=\(\left[\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right]:\left(\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}-1}\right)\)
a/ Rút gọn P
b/ Tìm a để \(\frac{1}{P}-\frac{\sqrt{a}+1}{8}\ge1\)
cho biểu thức P=\(\left[\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right]:\left(\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}-1}\right)\)
a/ Rút gọn P
b/ Tìm a để \(\frac{1}{P}-\frac{\sqrt{a}+1}{8}\ge1\)
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{a}\right):\left(\frac{\sqrt{a}-1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a, Rút gọn
b, Tìm a để A dương
1) cho biểu thức P=\(\frac{\sqrt{a}+2}{\sqrt{a}+3-}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{2-\sqrt{a}}\)
a/ rút gọn P
b/ tìm giá trị của a để P<1
2) cho biểu thức P=\(\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
a/ rút gọn P
b/ tìm giá trị của P<0
Cho biểu thức: P = \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a-1}}\right)\)
a, Rút gọn P
b, Tìm giá trị của a để \(P>\frac{1}{6}\)