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Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
\(\dfrac{2}{xy}-\dfrac{2}{y\left(x+y\right)}-\dfrac{2}{x\left(x+y\right)}=\dfrac{2\left(x+y\right)-2x-2y}{xy\left(x+y\right)}=0\)
\(A=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{\left(x+y\right)^2}}\)
\(=\sqrt{\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{y}\right)^2+\left(\dfrac{1}{x+y}\right)^2+2\times\dfrac{1}{x}\times\dfrac{1}{y}-2\times\dfrac{1}{y}\times\dfrac{1}{x+y}-2\times\dfrac{1}{x}\times\dfrac{1}{x+y}}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{x+y}\right)}\)
\(=\left|\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{x+y}\right|\left(\text{đ}pcm\right)\)
ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1+\frac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\frac{\sqrt{a}+1+a+\sqrt{a}}{\sqrt{a}+1}.\frac{1-\sqrt{a}+a-\sqrt{a}}{1-\sqrt{a}}\)
\(=\frac{1+2\sqrt{a}+a}{\sqrt{a}+1}.\frac{1-2\sqrt{a}+a}{1-\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.\frac{\left(1-\sqrt{a}\right)^2}{1-\sqrt{a}}\)
\(=\left(\sqrt{a}+1\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1+\frac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\frac{\sqrt{a}+1+a+\sqrt{a}}{\sqrt{a}+1}\right)\left(\frac{1-\sqrt{a}+a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\frac{a+2\sqrt{a}+1}{\sqrt{a}+1}.\frac{a-2\sqrt{a}+1}{1-\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.-\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}\)
\(=-\left(\sqrt{a}+1\right).\left(\sqrt{a}-1\right)\)
\(=1-a\)
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)
Áp dụng BĐT cô si cho 2 số ko âm \(\sqrt{a}\) và \(\sqrt{b}\) ta được:
\(\sqrt{a}+\sqrt{b}\ge2\sqrt{\sqrt{ab}}\)
Suy ta: \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{ab}}}=\sqrt{\sqrt{ab}}=\sqrt[4]{ab}\)
=>điều cần chứng minh
Thay \(x=\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)\) vào chỗ \(\sqrt{x^2-1}\)
\(\Leftrightarrow2\sqrt{a\left(a+1\right)}-2a< 1\)
Lại có:\(2\sqrt{a\left(a+1\right)}\le a+a+1=2a+1\)
\(\Rightarrow2\sqrt{a\left(a+1\right)}-2a\le2a+1-2a=1\)
Dấu "=" không xảy ra
\(\Rightarrow\sqrt{a+1}-\sqrt{a}< \dfrac{1}{2\sqrt{a}}\)(đpcm)
Tại sao dấu "=" không xảy ra ?