\(B=\text{}\)\(\ sum _ { i = 1 } ^ { 2020 }\) \(\frac{1}{\left(\sqrt{i}+\sqrt{i+1}\right)\left(\sqrt{i\left(i+1\right)}+\sqrt{i}+\sqrt{i+1}+1\right)}\)
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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:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
18 tháng 3 2020
a) \(A=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\cdot\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(A=\left(\frac{a-1}{2\sqrt{a}}\right)^2\cdot\left[\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(A=\frac{\left(a-1\right)^2\cdot\left(-4\sqrt{a}\right)}{4a\cdot\left(a-1\right)}\)
\(A=\frac{-\left(a-1\right)}{\sqrt{a}}=\frac{-a+1}{\sqrt{a}}\)
b) \(A< 0\Leftrightarrow\frac{-a+1}{\sqrt{a}}< 0\Leftrightarrow-a+1< 0\Leftrightarrow a>1\)
c) \(A=-2\Leftrightarrow\frac{-a+1}{\sqrt{a}}=-2\)
\(\Leftrightarrow-a+1=-2\sqrt{a}\)
\(\Leftrightarrow a-2\sqrt{a}-1=0\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)^2-2=0\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)^2=2\)
Vì \(\sqrt{a}-1\ge-1\Rightarrow\sqrt{a}-1=\sqrt{2}\Leftrightarrow a=3+2\sqrt{2}\) (t/m)
Vậy...
TD
19 tháng 11 2016
a,\(\left(\sqrt{6}-\sqrt{10}\right)\sqrt{4+\sqrt{15}}=\sqrt{6}.\sqrt{4-\sqrt{15}}-\sqrt{10}.\sqrt{4+\sqrt{15}}\)
=\(\sqrt{24+6\sqrt{15}}-\sqrt{40+10\sqrt{15}}=\sqrt{\left(\sqrt{15}+3\right)^2}-\sqrt{\left(\sqrt{15}+5\right)^2}\)
=\(\sqrt{15}+3-\sqrt{15}-5=-2\)
b,\(\left(\sqrt{3}+\sqrt{30}\right)\sqrt{10-\sqrt{41-4\sqrt{10}}}\)
=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{40-2\sqrt{40}+1}}\)
=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{\left(\sqrt{40}-1\right)^2}}\)
=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{40}+1}\)
=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{11-2\sqrt{10}}=\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{\left(\sqrt{10}-1\right)^2}\)
=\(\sqrt{3}\left(1+\sqrt{10}\right)\left(\sqrt{10}-1\right)=9\sqrt{3}\)
2,\(A=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)-a-2}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}\left(1-\sqrt{a}\right)-\sqrt{a}+4}{1-a}\right)\)
\(A=\left(\frac{a+\sqrt{a}-a-2}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}-a-\sqrt{a}+4}{1-a}\right)=\left(\frac{\sqrt{a}+2}{\sqrt{a}+1}\right).\left(\frac{1-a}{4-a}\right)\)
\(A=\frac{\sqrt{a}-2}{\sqrt{a}+1}.\frac{a-1}{a-4}=\frac{\sqrt{a}-1}{\sqrt{a}+2}\)
b, ̣để \(A=\frac{1}{2}\Rightarrow\frac{\sqrt{a}-1}{\sqrt{a}+2}=\frac{1}{2}\Leftrightarrow2\sqrt{a}-2=\sqrt{a}+2\Leftrightarrow\sqrt{a}=4\Leftrightarrow a=16\left(t.m\right)\)
TD
19 tháng 11 2016
Bạn oi bài 2 hàng A thú 2 phải là \(\frac{\sqrt{a}-2}{\sqrt{a}+1}\) mình nhầm
26 tháng 9 2020
bình phương lên
\(Q^2=\frac{x-2\sqrt{\left(x-1\right)}+x+2\sqrt{\left(x-1\right)}+2\sqrt{\left(x-2\right)^2}}{x^2-4\left(x-1\right)}.\left(\frac{x-2}{x-1}\right)^2\)
\(=\frac{2x+2\left(x-2\right)}{\left(x-2\right)^2}.\frac{\left(x-2\right)^2}{\left(x-1\right)^2}=\frac{2\left(x+x-2\right)}{\left(x-1\right)^2}=\frac{4\left(x-1\right)}{\left(x-1\right)^2}=\frac{4}{x-1}\)
\(\Rightarrow Q=\frac{2}{\sqrt{x-1}}\)
7 tháng 8 2023
\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)
\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)
Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)
\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)
Không hiện tex ;-; Bổ sung: \(i=\overline{1,2020}\)
i là số gì vậy ?