Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(\sqrt{x-1}=a\), khi đó ta có:
\(P=\left(\dfrac{\sqrt{x-1}}{3+\sqrt{x-1}}+\dfrac{x+8}{10-x}\right):\left(\dfrac{3\sqrt{x-1}+1}{x-3\sqrt{x-1}-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
\(=\left[\dfrac{\sqrt{x-1}}{\sqrt{x-1}+3}+\dfrac{\left(x-1\right)+9}{9-\left(x-1\right)}\right]:\left[\dfrac{3\sqrt{x-1}+1}{\left(x-1\right)-3\sqrt{x-1}}-\dfrac{1}{\sqrt{x-1}}\right]\)
\(=\left(\dfrac{a}{a+3}+\dfrac{a^2+9}{9-a^2}\right):\left(\dfrac{3a+1}{a^2-3a}-\dfrac{1}{a}\right)\)
\(=\dfrac{a\left(3-a\right)+\left(a^2+9\right)}{\left(3+a\right)\left(3-a\right)}:\dfrac{\left(3a-1\right)-\left(a-3\right)}{a\left(a-3\right)}\)
\(=\dfrac{3a-a^2+a^2+9}{\left(3+a\right)\left(3-a\right)}:\dfrac{3a-1-a+3}{a\left(a-3\right)}\)
\(=\dfrac{3a+9}{\left(3+a\right)\left(3-a\right)}:\dfrac{2a+4}{a\left(a-3\right)}\)
\(=\dfrac{3\left(a+3\right)}{\left(a+3\right)\left(a-3\right)}.\dfrac{a\left(a-3\right)}{2\left(a+2\right)}\)
\(=\dfrac{-3a}{2\left(a+2\right)}\).
Suy ra: P \(=\dfrac{-3\sqrt{x-1}}{2\left(\sqrt{x-1}+2\right)}\).
Ta lại có: \(x=\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}\)
\(=\sqrt[4]{\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)^2}}-\sqrt[4]{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}-\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}\)
\(=\sqrt{\dfrac{\left(\sqrt{2}+1\right)^2}{2-1}}-\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{2-1}}\)
\(=\left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)\)
\(=2\).
Suy ra: \(P=\dfrac{-3\sqrt{2-1}}{2\left(\sqrt{2-1}+2\right)}=\dfrac{-3}{2.3}=-\dfrac{1}{2}\).
\(Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8}{2-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\dfrac{4x-8\sqrt{x}-8x-16\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-4x-20\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}-1}\)
\(=\dfrac{4x+20\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
Để Q=-1 thì \(4x+20\sqrt{x}=-x-3\sqrt{x}-2\)
=>\(5x+23\sqrt{x}+2=0\)
hay \(x\in\varnothing\)
\(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-8\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-x-3-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}+1+\sqrt{x}-1\right)\left(\sqrt{x}+1-\sqrt{x}+1\right)-8\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{-x-4}\right)\)
\(\Leftrightarrow P=\dfrac{-4\sqrt{x}}{-x-4}=\dfrac{4\sqrt{x}}{x+4}\)
Thay x = \(3+2\sqrt{2}\) ta được :
\(P=\dfrac{4\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}+4}=\dfrac{4\left(\sqrt{2}+1\right)}{7+2\sqrt{2}}=\dfrac{4\sqrt{2}+4}{7+2\sqrt{2}}\)
\(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{x-1}:\dfrac{\sqrt{x}-x-3-\sqrt{x}-1}{x-1}=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{x-1}{-x-4}=\dfrac{4\sqrt{x}}{x+4}\left(x\ne4;x\ge0;x\ne1\right)\)
Ta có : \(x=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\left(TMĐKXĐ\right)\)
\(P=\dfrac{4\left(\sqrt{2}+1\right)}{3+2\sqrt{2}+4}=\dfrac{4+4\sqrt{2}}{7+2\sqrt{2}}\)
a: ĐKXĐ: x>=0; x<>1
b: \(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+\sqrt{x}+1}\)
c: Khi \(x=8-2\sqrt{7}\) vào A, ta được:
\(A=\dfrac{2}{8-2\sqrt{7}+\sqrt{7}-1+1}=\dfrac{2}{8-\sqrt{7}}\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)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, \(P=\left(1-\dfrac{2\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}+x+1}\right)\)(ĐK: \(x\ge0,x\ne-1\))
\(=\left(\dfrac{x-2\sqrt{x}+1}{x+1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)+x+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{x+1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{x+1}:\left(\dfrac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{x+1}:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}+1\right)}\)
\(=\sqrt{x}+1\)
b, ĐK: \(x\ge0,x\ne-1\)
\(x=2019-2\sqrt{2018}=\left(\sqrt{2018}-1\right)^2\)
Thay \(x=\left(\sqrt{2018}-1\right)^2\)(TMĐK) vào P ta có:
\(P=\sqrt{2018}-1+1=\sqrt{2018}\)
Vậy với \(x=2019-2\sqrt{2018}\) thì \(P=\sqrt{2018}\)
2.
\(P=\dfrac{\sqrt{x-2018}}{x+2}+\dfrac{\sqrt{x-2019}}{x}\)\(P=\dfrac{\sqrt{\left(x-2018\right).2020}}{\left(x+2\right)\sqrt{2020}}+\dfrac{\sqrt{\left(x-2019\right).2019}}{\sqrt{2019}.x}\)
Áp dụng BĐT AM-GM:
\(\sqrt{\left(x-2018\right).2020}\le\dfrac{1}{2}\left(x-2018+2020\right)=\dfrac{1}{2}\left(x+2\right)\)
\(\sqrt{\left(x-2019\right).2019}\le\dfrac{1}{2}\left(x-2019+2019\right)=\dfrac{1}{2}x\)
\(\Rightarrow P\le\dfrac{x+2}{2\sqrt{2020}\left(x+2\right)}+\dfrac{x}{2\sqrt{2019}.x}=\dfrac{1}{2\sqrt{2020}}+\dfrac{1}{2\sqrt{2019}}\)
\("="\Leftrightarrow x=4038\)
không phải bơ đâu, oan cho tớ quá :>
27/11 thi nên ít lên, với cả chị tớ cũng không cho chat :>
lấy mật khẩu của tớ vô đọc góc ib là biết mà :>