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\(D=\dfrac{1}{\sqrt{h+2\sqrt{h-1}}}+\dfrac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\dfrac{1}{\sqrt{h-1+2\sqrt{h-1}+1}}+\dfrac{1}{\sqrt{h-1-2\sqrt{h-1}+1}}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{h-1}+1\right)^2}}+\dfrac{1}{\sqrt{\left(\sqrt{h-1}-1\right)^2}}\)
\(=\dfrac{1}{\sqrt{h-1}+1}+\dfrac{1}{\sqrt{h-1}-1}\)
\(=\dfrac{\sqrt{h-1}-1+\sqrt{h-1}+1}{\left(\sqrt{h-1}+1\right)\left(\sqrt{h-1}-1\right)}\)
\(=\dfrac{2\sqrt{h-1}}{\left(\sqrt{h-1}+1\right)\left(\sqrt{h-1}-1\right)}\)
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
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 : \(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{a-\sqrt{a}-6}{a+\sqrt{a}-6}\)
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:
\(D=\frac{1}{\left|\sqrt{3-1}+1\right|}+\frac{1}{\left|\sqrt{3-1}-1\right|}\)
\(D=\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{2}-1}=\frac{\sqrt{2}-1+\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{2\sqrt{2}}{2-1}=2\sqrt{2}\)
B1:
a)
\(H=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{x-\sqrt{x}}\\ H=\dfrac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ H=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\H=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b)
\(H< 0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}< 0\)
vì \(\sqrt{x}\ge0\)
nên \(\sqrt{x}-1< 0\\ \sqrt{x}< 1\Rightarrow x< 1\)
vậy khi x<1 thì H < 0
b)
1.
\(Q=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}+1\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-1\right)\\ Q=\left(\dfrac{x+\sqrt{x}+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\dfrac{x-\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\right)\\ Q=\left[\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right]:\left[\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\right]\\ Q=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)=x-1\)
2.
\(Q< 1\Leftrightarrow x-1< 1\Leftrightarrow x< 2\)
vậy khi x< 2 thì Q<1
\(D=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4x}{x-1}\)ĐK:x\(\ge\)0.x khác 0
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x}{x-1}\)
\(=\dfrac{4x+4\sqrt{x}}{x-1}\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)
a. H=\(\dfrac{1}{\sqrt{x-1}-\sqrt{x}}+\dfrac{1}{\sqrt{x-1}+1}\dfrac{\sqrt{x^3}-x}{\sqrt{x}-1}\left(đkxđ:x\ge1\right)\)
H=\(-2\sqrt{x-1}+x\)
b. Với x=\(\dfrac{53}{9-2\sqrt{7}}:\)
H=\(-2\sqrt{\dfrac{53}{9-2\sqrt{7}}-1}+\dfrac{53}{9-2\sqrt{7}}\)
H\(=7\)
c. \(-2\sqrt{x-1}+x=16\)
\(\sqrt{x-1}=\dfrac{x-16}{2}\)
\(4x-4=x^2-32x+256\)
\(x^2-36x+260=0\)
x=26
d. Để H>1 thì x>3
\(A=\dfrac{1}{\sqrt{h+2\sqrt{h-1}}}+\dfrac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{h-1}+1\right)^2}}+\dfrac{1}{\sqrt{\left(\sqrt{h-1}-1\right)^2}}\)
\(=\dfrac{1}{\sqrt{h-1}+1}+\dfrac{1}{\left|\sqrt{h-1}-1\right|}\)
\(D=\dfrac{1}{\sqrt{h+2\sqrt{h-1}}}+\dfrac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\dfrac{1}{\sqrt{h-1}+1}+\dfrac{1}{\sqrt{h-1}-1}\)
\(=\dfrac{2\cdot\sqrt{h-1}}{h}\)