Rút gọn biểu thức :
\(\sqrt{\frac{28\left(a-2\right)^2}{7}}\)với \(a>2\)
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YY
26 tháng 5 2018
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(a-1\right)^2}{4a}\left(\frac{\left(\sqrt{a}-1-\sqrt{a}-1\right)\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)
\(=\frac{\left(a-1\right)\left(-2\right)2\sqrt{a}}{4a}=-\frac{\left(a-1\right)}{\sqrt{a}}\)
13 tháng 7 2019
a) ĐKXĐ : \(a>0;a\ne1\)
\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\right)\)
\(Q=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right)\)
\(Q=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{3}\)
\(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}\)
b) \(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}>2\Rightarrow\sqrt{a}-6\sqrt{a}+2>0\Rightarrow-5\sqrt{a}>-2\Rightarrow0< \sqrt{a}< \frac{2}{5}\)
\(\Rightarrow0< a< \frac{4}{25}\)
11 tháng 8 2020
Có: \(A=\sqrt{\frac{1}{1^2}+\frac{1}{a^2}+\frac{1}{\left(-a-1\right)^2}}\)
Có: \(1+a+\left(-a-1\right)=1+a-1-a=0\)
=> \(\sqrt{\frac{1}{1^2}+\frac{1}{a^2}+\frac{1}{\left(-a-1\right)^2}}=\sqrt{\left(\frac{1}{1}+\frac{1}{a}+\frac{1}{-a-1}\right)^2}=\frac{1}{1}+\frac{1}{a}+\frac{1}{-a-1}\)
=> \(A=1+\frac{1}{a}-\frac{1}{a+1}=1+\frac{1}{a\left(a+1\right)}\)
VẬY \(A=1+\frac{1}{a\left(a+1\right)}\)
11 tháng 8 2020
\(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\)
\(=\sqrt{\left(\frac{1}{a}-\frac{1}{a+1}\right)^2+\frac{2}{a\left(a+1\right)}+1}\)
\(=\sqrt{\left[\frac{1}{a\left(a+1\right)}+1\right]^2}=\left|\frac{1}{a}-\frac{1}{a+1}+1\right|\)
3 tháng 2 2022
Bài 1:
a: \(=\sqrt{\dfrac{7-4\sqrt{3}}{2-\sqrt{3}}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
Bài 2:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
CT
24 tháng 4 2019
=(\(\frac{\sqrt{a-b}\left(\sqrt{a+b}-\sqrt{a-b}\right)}{\left(\sqrt{a+b}+\sqrt{a-b}\right)\left(\sqrt{a+b}-\sqrt{a-b}\right)}\)+\(\frac{a-b}{\sqrt{a-b}\left(\sqrt{a+b}-\sqrt{a-b}\right)}\)):\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)
=(\(\frac{\sqrt{a^2-b^2}-\left(a-b\right)}{a+b-a+b}+\frac{\sqrt{a^2-b^2}+a-b}{a+b-a+b}\)):\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)
=\(\frac{2\sqrt{a^2-b^2}}{2b}\):\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)
=\(\frac{\sqrt{a^2-b^2}}{b}\)*\(\frac{a^2+b^2}{\sqrt{a^2-b^2}}\)
=\(\frac{a^2+b^2}{b}\)
25 tháng 4 2019
b/ Thế \(b=a-1\)thì ta có
\(P=\frac{a^2+\left(a-1\right)^2}{a-1}=\frac{2a^2-2a+1}{a-1}\)
\(\Leftrightarrow2a^2-\left(2+P\right)a+1+P=0\)
\(\Rightarrow\Delta_a=\left(2+P\right)^2-4.2.\left(1+P\right)\ge0\)
\(\Leftrightarrow P\ge2+2\sqrt{2}\)
21 tháng 10 2023
1:
a: \(\sqrt{25}+\sqrt{49}=5+7=12\)
b: \(\sqrt{121}-\sqrt{81}=11-9=2\)
2: x>-2
=>2x>-4
=>2x+1>-3
=>Với x>-2 thì \(\sqrt{2x+1}\) chưa chắc có nghĩa
3:
a: \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)
\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)
\(=\sqrt{3}-1-\sqrt{3}=-1\)
b: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)
\(=21-14\sqrt{2}+14\sqrt{2}=21\)
c:
\(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}-6\sqrt{3}+2\sqrt{3}}{\sqrt{3}}=3+2-6=-1\)
Answer:
\(\sqrt{\frac{28\left(a-2\right)^2}{7}}\)
\(=\sqrt{4\left(a-2\right)^2}\)
\(=2\left|a-2\right|\)
\(=2\left(a-2\right)\) vì a > 2
\(=2a-4\)