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Bài làm:
Ta có:
\(D=\frac{1}{\sqrt{h+2\sqrt{h-1}}}+\frac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(D=\frac{1}{\sqrt{\left(h-1\right)+2\sqrt{h-1}+1}}+\frac{1}{\sqrt{\left(h-1\right)-2\sqrt{h-1}+1}}\)
\(D=\frac{1}{\sqrt{\left(\sqrt{h-1}+1\right)^2}}+\frac{1}{\sqrt{\left(\sqrt{h-1}-1\right)^2}}\)
\(D=\frac{1}{\left|\sqrt{h-1}+1\right|}+\frac{1}{\left|\sqrt{h-1}-1\right|}\)
Tại h = 3 thì giá trị của D là:
\(D=\frac{1}{\left|\sqrt{3-1}+1\right|}+\frac{1}{\left|\sqrt{3-1}-1\right|}\)
\(D=\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{2}-1}=\frac{\sqrt{2}-1+\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{2\sqrt{2}}{2-1}=2\sqrt{2}\)
a: \(H=\dfrac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{x-1-x}+x\)
\(=-2\sqrt{x-1}+x\)
b: \(x=\dfrac{53}{9-2\sqrt{7}}=9+2\sqrt{7}\)
Khi x=9+2 căn 7 thì \(H=-2\cdot\sqrt{8+2\sqrt{7}}+9+2\sqrt{7}\)
\(=-2\left(\sqrt{7}+1\right)+9+2\sqrt{7}\)
=-2+9=7
a) Rut gon H
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}-\dfrac{1}{\sqrt{a}-2}\)
DKXD : \(\left\{{}\begin{matrix}\sqrt{a}+3\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ne9\\a\ne4\end{matrix}\right.\)
Ta co : \(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{a-\sqrt{a}-6}{a+\sqrt{a}-6}\)
1,
\(D=\frac{1}{\sqrt{h+2\sqrt{h-1}}}+\frac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\frac{1}{\sqrt{h-1+2\sqrt{h-1}+1}}+\frac{1}{\sqrt{h-1-2\sqrt{h-1}+1}}\)
\(=\frac{1}{\sqrt{h-1}+1}+\frac{1}{\sqrt{h-1}-1}\)
\(=\frac{\sqrt{h-1}-1+\sqrt{h-1}+1}{h-1-1}\)
\(=\frac{2\sqrt{h-1}}{h-2}\)
Thay \(h=3\)vào D ta có:
\(D=\frac{2\sqrt{3-1}}{3-2}=2\sqrt{2}\)
Vậy với \(h=3\)thì \(D=2\sqrt{2}\)
2,
a, \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)(ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(TM\right)\)
Vậy PT có nghiệm là \(x=2\)
b, \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)(ĐK: \(-\sqrt{2}\le x\le\sqrt{2}\))
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}=-3\)
\(\Leftrightarrow0=-3\)(vô lí)
Vậy PT đã cho vô nghiệm.
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
a: \(A=\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}\cdot\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}=\sqrt{1-a}\)
b: \(a=\sqrt{3}\left(2-\sqrt{3}\right)=2\sqrt{3}-3\)
Khi \(a=2\sqrt{3}-3\) thì \(A=\sqrt{1-2\sqrt{3}+3}=\sqrt{3}-1\)
a. \(ĐKXĐ:a\ge0,a\ne2\)
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)
\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{a-4-8-\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\left(a-4\sqrt{a}\right)+\left(3\sqrt{a}-12\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{\sqrt{a}\left(\sqrt{a}-4\right)+3\left(\sqrt{a}-4\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b. Mk nghĩ là H < 2 chứ
\(H=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}< 2\)
\(\Leftrightarrow\dfrac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}=\dfrac{-\sqrt{a}}{\sqrt{a}-2}< 0\)
\(\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
c. \(a^2+3a=0\Leftrightarrow a\left(a+3\right)=0\Leftrightarrow\left[{}\begin{matrix}a=0\left(n\right)\\a=-3\left(l\right)\end{matrix}\right.\)
Thay \(a=0\) và H ta được:
\(\dfrac{0-4}{0-2}=2\)
d. \(H=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=5\Leftrightarrow\dfrac{\sqrt{a}-2-2}{\sqrt{a}-2}=5\Leftrightarrow1-\dfrac{2}{\sqrt{a}-2}=5\)
\(\Leftrightarrow\dfrac{2}{\sqrt{a}-2}=-4\Leftrightarrow-4\sqrt{a}+8=2\Leftrightarrow-4\sqrt{a}=-6\Leftrightarrow\sqrt{a}=\dfrac{3}{2}\Leftrightarrow a=\dfrac{9}{4}\)
\(D=\dfrac{1}{\sqrt{h+2\sqrt{h-1}}}+\dfrac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\dfrac{1}{\sqrt{h-1+2\sqrt{h-1}+1}}+\dfrac{1}{\sqrt{h-1-2\sqrt{h-1}+1}}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{h-1}+1\right)^2}}+\dfrac{1}{\sqrt{\left(\sqrt{h-1}-1\right)^2}}\)
\(=\dfrac{1}{\sqrt{h-1}+1}+\dfrac{1}{\sqrt{h-1}-1}\)
\(=\dfrac{\sqrt{h-1}-1+\sqrt{h-1}+1}{\left(\sqrt{h-1}+1\right)\left(\sqrt{h-1}-1\right)}\)
\(=\dfrac{2\sqrt{h-1}}{\left(\sqrt{h-1}+1\right)\left(\sqrt{h-1}-1\right)}\)
Thay \(h=3\) vào biểu thức ta được :
\(\dfrac{2\sqrt{3-1}}{\left(\sqrt{3-1}+1\right)\left(\sqrt{3-1}-1\right)}=\dfrac{2\sqrt{2}}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{2\sqrt{2}}{1}=2\sqrt{2}\)
Chúc bạn học tốt