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a)
\(P=a\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}+\frac{a}{b}=a\sqrt{\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}\)
=\(a\sqrt{\frac{a^2\left(a+1\right)^2+2a\left(a+1\right)+1}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}=a\sqrt{\frac{\left[a\left(a+1\right)+1\right]^2}{\left[a\left(a+1\right)\right]^2}}+\frac{a}{a+1}\)
\(=a.\frac{a\left(a+1\right)+1}{a\left(a+1\right)}+\frac{a}{a+1}=a+\frac{1}{a+1}+\frac{a}{a+1}=a+1\)
Vay P=a+1
phan b,c ap dung phan a la ra
CM bài toán phụ: \(x+y+z=0\)
CM: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) với x,y,z dương
Ta có: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}\)
\(=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\cdot\frac{x+y+z}{xyz}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)
\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Áp dụng vào ta được: \(Q=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)
\(Q=2021-\frac{1}{2021}=...\)
a, Ta có: \(x=4-2\sqrt{3}\)\(=3-2\sqrt{3}+1\)\(=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}\)\(=\sqrt{3}-1\)
Thay \(\sqrt{x}=\sqrt{3}-1\) vào biểu thức P ta có:
\(P=\frac{\sqrt{3}-1+1}{\sqrt{3}-1-4}\)\(=\frac{\sqrt{3}}{\sqrt{3}-5}\)\(=\frac{\sqrt{3}.\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right).\left(\sqrt{3}+5\right)}\)\(=\frac{3-5\sqrt{3}}{3-25}\)\(=\frac{5\sqrt{3}-3}{22}\)
Vậy \(P=\frac{5\sqrt{3}-3}{22}\)khi \(x=4-2\sqrt{3}\)
b, \(E=\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\)\(=\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right).\left(\sqrt{3}+1\right)}\)\(-\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)}\)
\(=\frac{\sqrt{3}+1-\sqrt{3}+1}{3-1}\) \(=\frac{2}{2}=1\)
a, Ta có : \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
Thay vào P ta được : \(P=\frac{\sqrt{3}-1+1}{\sqrt{3}-1-4}=\frac{\sqrt{3}}{\sqrt{3}-5}=\frac{\sqrt{3}\left(\sqrt{3}+5\right)}{-22}=-\frac{3+5\sqrt{3}}{22}\)
b, \(E=\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}+1-\sqrt{3}+1}{2}=1\)
a)ĐKXĐ : x > 0
P = \(\left(\frac{x-1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(1+\sqrt{x}\right)}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{1}{\sqrt{x}}.\left(\sqrt{x}-1+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\sqrt{x}-1}{\sqrt{x}}.\left(1-\frac{1}{\sqrt{x}+1}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right).\sqrt{x}}{\sqrt{x}}\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy P = \(\frac{\sqrt{x}+1}{\sqrt{x}}\)
b) x = \(\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{3}-1\)
=> P = \(\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}-1+1}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\)
= \(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3+1}\right)}=\frac{3+\sqrt{3}}{3-1}=\frac{3+\sqrt{3}}{2}\)
c)\(P\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{\sqrt{x}}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x-4}=5\sqrt{x-4}\)
Đặt \(\hept{\begin{cases}a=\sqrt{x}\\b=\sqrt{x-4}\end{cases}\Rightarrow a^2+b^2=x-\left(x-4\right)=4}\)
\(\Rightarrow\hept{\begin{cases}a^2-b^2=4\\b=5a-4\end{cases}\Rightarrow\hept{\begin{cases}a^2-\left(5a-4\right)^2=4\left(^∗\right)\\b=5a-4\end{cases}}}\)
Từ (*) <=> a2 -(25a2 -40a + 16 ) =4
<=> -24a2 + 40a - 20 = 0
=> \(\Delta'=-80< 0\)
=> PT vô nghiệm
=> ko tồn tại x thỏa mãn
mọi ng ơi mk viết thiếu dấu ngoặc nha.thiếu ngoặc lownns nha. đóng ngoắc ở trước dấu chia
\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2020}+\sqrt{2021}}\)
\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}+\sqrt{1}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{2021}-\sqrt{2020}}{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}\)
\(=\dfrac{\sqrt{2}-\sqrt{1}}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{2021}-\sqrt{2020}}{2021-2020}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2021}-\sqrt{2020}\)
\(=\sqrt{2021}-1\)