Tính P =\(\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\)
nếu x=\(\frac{\sqrt{3}}{4}\)
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2 tháng 7 2018
Ta có: \(Q=\frac{1-2x}{1-\sqrt{1-2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\)
\(=\frac{1-2x}{1-\sqrt{1^2-2.x.1+\left(\sqrt{x}\right)^2}}+\frac{1-2x}{1-\sqrt{1-2.x.1+\left(\sqrt{x}\right)^2}}\)
\(=\frac{1-2x}{1-\sqrt{\left(1-\sqrt{x}\right)^2}}+\frac{1-2x}{1-\sqrt{\left(1-\sqrt{x}\right)^2}}\)
\(=\frac{1-2x}{1-\left|1-\sqrt{x}\right|}+\frac{1-2x}{1-\left|1-\sqrt{x}\right|}\)
\(=\frac{1-2x}{\sqrt{x}}+\frac{1-2x}{\sqrt{x}}=\left(\frac{1-2x}{\sqrt{x}}\right)^2\)
Thế \(x=\frac{\sqrt{3}}{4}\) ta được: \(Q=\left(\frac{1-2.\frac{\sqrt{3}}{4}}{\sqrt{\frac{\sqrt{3}}{4}}}\right)^2=0,04145188433\)
19 tháng 7 2016
Ta có: M= \(\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\)= \(\frac{\left(1+2x\right)\left(1-\sqrt{1+2x}\right)+\left(1-2x\right)\left(1+\sqrt{1+2x}\right)}{1-\left(1-2x\right)}\)=\(\frac{1-\sqrt{1+2x}+2x-2x\sqrt{1+2x}+1+\sqrt{1+2x}-2x-2x\sqrt{1+2x}}{2x}\)
=\(\frac{2}{2x}=\frac{1}{x}\)
Với x=\(\frac{\sqrt{3}}{4}\)=> M=\(\frac{4}{\sqrt{3}}\)
19 tháng 7 2016
Ta có \(\sqrt{\left(1+2x\right)^2}\)= 1 + 2x (1)
+ \(\sqrt{\left(1-2x\right)^2}\)= 1 - 2x (2)
(1) +(2) = 2
Có \(\sqrt{1+2x}.\sqrt{1-2x}\)= \(\sqrt{1-4x^2}=\frac{1}{2}\) (3)
Từ (1),(2),(3) \(\Rightarrow\)\(\left(\sqrt{1+2x}+\sqrt{1-2x}\right)^2\)= 3 \(\Rightarrow\)\(\sqrt{1+2x}+\sqrt{1-2x}\)=\(\sqrt{3}\) (4)
\(\left(\sqrt{1+2x}-\sqrt{1-2x}\right)^2\)= 1 \(\Rightarrow\) \(\sqrt{1+2x}-\sqrt{1-2x}\)= 1 (5)
Có M= \(\frac{\left(1+2x\right).\left(1-\sqrt{1-2x}\right)+\left(1-2x\right).\left(1+\sqrt{1+2x}\right)}{\left(1+\sqrt{1+2x}\right).\left(1-\sqrt{1-2x}\right)}\)
Xét TS= \(1-\sqrt{1-2x}+2x-2x.\sqrt{1-2x}+1+\sqrt{1+2x}-2x-2x.\sqrt{1+2x}\)
= 2+ \(\sqrt{1+2x}-\sqrt{1-2x}\)- 2x\(\left(\sqrt{1+2x}+\sqrt{1-2x}\right)\)
Thay (4), (5) và x vào TS ta có TS= \(2+1-2.\frac{\sqrt{3}}{4}.\sqrt{3}=\frac{3}{2}\) (6)
Xét MS=\(1-\sqrt{1-2x}+\sqrt{1+2x}-\sqrt{1-4x^2}\)
Thay (5) và x vào MS ta có MS= \(1+1-\frac{1}{2}\)=\(\frac{3}{2}\) (7)
Từ (6),(7) ta có giá trị của M= 1
8 tháng 10 2020
\(\Leftrightarrow\frac{7x+4}{\sqrt{2\left(x-1\right)\left(x+1\right)}}+\frac{2\sqrt{2x+1}}{\sqrt{2\left(x+1\right)}}=3+\frac{3\sqrt{2x+1}}{\sqrt{x-1}}\)
\(\Leftrightarrow7x+4+2\sqrt{\left(2x+1\right)\left(x-1\right)}=3\sqrt{2\left(x-1\right)\left(x+1\right)}+3\sqrt{2\left(2x+1\right)\left(x+1\right)}\)
\(\Leftrightarrow\left(7x+4+\sqrt{8x^2-4x-4}\right)^2=\left(\sqrt{18x^2-18}+\sqrt{36^2+54x+18}\right)^2\)
\(\Leftrightarrow\left(7x+4\right)^2+8x^2-4x-4+2\left(7x+4\right)\sqrt{8x^2-4x-4}\)\(=18x^2-18+36x^2+54x+18+2\sqrt{\left(18x^2-18\right)\left(36x^2+54x+18\right)}\)
\(\Leftrightarrow3x^2-2x+12+4\left(7x+4\right)\sqrt{\left(x-1\right)\left(2x+1\right)}=36\left(x+1\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow3x^2-2x+12=4\left(2x+5\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\left(3x^2-2x+12\right)^2=16\left(2x+5\right)^2\left(x-1\right)\left(2x+1\right)\)
\(\Leftrightarrow119x^4+588x^3+1940x^2-672x-544=0\left(1\right)\)
Ta thấy x>1 => Vế trái (1) \(>119.1^4+588.1^3+1940.1^2-672.1-544=1431>0\)
=> pt vô nghiệm.
LH
24 tháng 3 2020
a) \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}=2\)
Ta có: \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}\ge2\sqrt{\sqrt{\frac{2x-1}{x+1}}\cdot\sqrt{\frac{x+1}{2x-1}}}=2\) (BĐT Cô-si)
Mà \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}=2\) (theo đề bài)
Suy ra dấu bằng phải xảy ra \(\Rightarrow\sqrt{\frac{2x-1}{x+1}}=\sqrt{\frac{x+1}{2x-1}}\) \(\Leftrightarrow\frac{2x-1}{x+1}=\frac{x+1}{2x-1}\) \(\Leftrightarrow\left(2x-1\right)^2=\left(x+1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+1\\2x-1=-x-1\end{matrix}\right.\Leftrightarrow\) \(x=2\) (tmđkxđ) hoặc \(x=0\) (không tmđkxđ)
Vậy \(S=\left\{2\right\}\).
Bạn đừng quên tự tìm ĐKXĐ cho câu a nhé bạn.
c) \(x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6=0\) ĐKXĐ: \(x>0\)
Vì \(x>0\Rightarrow x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6>0\)
Vậy \(S=\varnothing\).
\(P=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
\(=\frac{1}{\sqrt{2}}\left[\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\right]\)
\(=\frac{1}{\sqrt{2}}\left[\frac{2+\sqrt{3}}{\sqrt{2}+\frac{\sqrt{3}+1}{\sqrt{2}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\frac{\sqrt{3}-1}{\sqrt{2}}}\right]\)
Vì \(\sqrt{2\pm\sqrt{3}}=\sqrt{\frac{\left(1\pm\sqrt{3}\right)^2}{2}}=\frac{\sqrt{3}\pm1}{\sqrt{2}}\)
\(P=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}=\frac{3+\sqrt{3}}{6}+\frac{3-\sqrt{3}}{6}=1\)