chứng minh
\(\sqrt{x+2\sqrt{x-1}}\)+\(\sqrt{x-2\sqrt{x-1}}\)=2\(\sqrt{x-1}\)
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AT
4 tháng 7 2021
\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1+1}}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)
\(=\sqrt{x-1}-1+\sqrt{x-1}+1\left(x\ge2\right)=2\sqrt{x-1}\)
a) \(\dfrac{1}{\sqrt{5}+\sqrt{7}}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\dfrac{\sqrt{7}-\sqrt{5}}{2}\)
c) \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{5}}=\dfrac{7}{2\sqrt{5}-\sqrt{3}}=\dfrac{7\left(2\sqrt{5}+\sqrt{3}\right)}{\left(2\sqrt{5}+\sqrt{3}\right)\left(2\sqrt{5}-\sqrt{3}\right)}\)
\(=\dfrac{14\sqrt{5}+7\sqrt{3}}{17}\)
AT
22 tháng 6 2021
\(P=\left(\dfrac{\sqrt[4]{x^2}-\sqrt[4]{x}}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{1+\dfrac{2}{\sqrt{x}}+\dfrac{1}{x}}}{1+\sqrt{x}}\)
\(=\left(\dfrac{\sqrt[4]{x}\left(\sqrt[4]{x}-1\right)}{1-\sqrt[4]{x}}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\sqrt{\left(\dfrac{1}{\sqrt{x}}+1\right)^2}}{1+\sqrt{x}}\)
\(=\left(-\sqrt[4]{x}+\dfrac{1+\sqrt{x}}{\sqrt[4]{x}}\right)^2-\dfrac{\dfrac{1}{\sqrt{x}}+1}{1+\sqrt{x}}\)
\(=\left(\dfrac{1}{\sqrt[4]{x}}\right)^2-\dfrac{\dfrac{\sqrt{x}+1}{\sqrt{x}}}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
\(A-1=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-1=\dfrac{\sqrt{x}-2-\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+1}\)
Do \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1>0;\forall x\in D\)
\(\Rightarrow\dfrac{-3}{\sqrt{x}+1}< 0\)
\(\Rightarrow A-1< 0\Rightarrow A< 1\)
Y
12 tháng 6 2023
\(B=\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{3-\sqrt{x}}{x-1}\left(dkxd:x\ne1,x\ge0\right)\)
\(=\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2\sqrt{x}-3\sqrt{x}-3+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\left(dpcm\right)\)
12 tháng 6 2023
\(B=\dfrac{2x+2\sqrt{x}-3\sqrt{x}-3+3-\sqrt{x}}{x-1}=\dfrac{2x-2\sqrt{x}}{x-1}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
HP
22 tháng 10 2021
a.
A = \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\dfrac{\left(x-2+\sqrt{x}\right).\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{\left(x-2+\sqrt{x}\right)\left(\sqrt{x}-1\right)+\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-\left(x\sqrt{x}+2x+x+2\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-x\sqrt{x}-2x-x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-3x-4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-\left(3x+4\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-\sqrt{x}\left(3\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{3\sqrt{x}+4}{x-\sqrt{x}+2\sqrt{x}-2}\)
A = \(\dfrac{3\sqrt{x}+4}{x-\sqrt{x}-2}\)
H
11 tháng 7 2023
\(A+B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\left(\text{đ}pcm\right)\)
11 tháng 7 2023
A+B
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}\)
\(=\dfrac{2x-3\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
VH
10 tháng 10 2019
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6+\sqrt{3}-3+6-\sqrt{3}-3}{9-3}=\frac{6}{6}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-1+2x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\frac{2}{\sqrt{x}}\)
Đặt \(\sqrt{x-1}=t\)
Đẳng thức đã cho trở thành:
\(\sqrt{t^2+1+2t}+\sqrt{t^2+1-2t}\)
\(=\sqrt{\left(t+1\right)^2}+\sqrt{\left(t-1\right)^2}\)
= t+1 + t - 1
= 2t
Vậy đẳng thức đã được chứng minh