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21 tháng 5 2016
- Ta chứng minh bất đẳng thức phụ dưới đây: \(\frac{1}{\sqrt{x}\left(x+1\right)}=\frac{\sqrt{x}}{x\left(x+1\right)}=\sqrt{x}\left(\frac{1}{x}-\frac{1}{x+1}\right)=\sqrt{x}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x+1}}\right)\)\(=\left(1+\frac{\sqrt{x}}{\sqrt{x+1}}\right)\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)< 2\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)\)
Áp dụng : \(\frac{1}{\sqrt{1}.2}< 2.\left(1-\frac{1}{\sqrt{2}}\right)\)
\(\frac{1}{\sqrt{2}.3}< 2.\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)\)
...................................
\(\frac{1}{\sqrt{2015}.2016}< 2.\left(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)\)
Cộng các BĐT trên với nhau được : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}}< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)=2\left(1-\frac{1}{\sqrt{2016}}\right)< 2\left(1-\frac{1}{\sqrt{2025}}\right)=\frac{88}{45}\)
Từ đó suy ra đpcm
21 tháng 5 2016
Cái ............... là gì vậy bn
9 tháng 3 2016
\(\frac{\left(x+1\right)^2-\frac{x}{2}}{4}=\frac{\left(2x-3\right)^2}{3}-\frac{\frac{x+1}{4}-\frac{x\left(3-2x\right)}{3}}{4}\)
\(\Rightarrow3\left[\left(x+1\right)^2-\frac{x}{2}\right]=4\left(2x-3\right)^2-3\left[\frac{x+1}{4}-\frac{x\left(3-2x\right)}{3}\right]\)
\(\Rightarrow3\left(x+1\right)^2-\frac{3x}{2}=4\left(2x-3\right)^2-\frac{3\left(x+1\right)}{4}+\frac{3x\left(3-2x\right)}{3}\)
\(\Rightarrow36\left(x+1\right)^2-18x=48\left(2x-3\right)^2-9\left(x+1\right)+12x\left(3-2x\right)\)
=> 36.(x2 + 2x + 1) - 18x = 48.(4x2 - 12x + 9) - 9(x + 1) + 12x(3 - 2x)
=> 36x2 + 72x + 36 - 18x - 192x2 + 576x - 432 + 9x + 9 - 36x + 24x2 = 0
=> -132x2 + 603x - 387 = 0
Có: \(\Delta=603^2-4.\left(-387\right)\left(-132\right)=159273\Rightarrow\sqrt{\Delta}=\sqrt{159273}\)
\(\Rightarrow x=\frac{-603+\sqrt{159273}}{-264}\) hoặc \(x=\frac{-603-\sqrt{159273}}{-264}\)
Vậy phương trình có 2 nghiệm : x = \(\left\{\frac{-603+\sqrt{159273}}{-264};\frac{-603-\sqrt{159273}}{-264}\right\}\)
Câu này không có nghiệm nguyên nha bạn.
16 tháng 8 2019
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
16 tháng 8 2019
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
NG
13 tháng 10 2020
\(\text{a, }\frac{-2}{5}+x=\left(\frac{-1}{3}\right)^2+\frac{2}{3}\)
\(\Leftrightarrow\frac{-2}{5}+x=\frac{1}{9}+\frac{6}{9}\)
\(\Leftrightarrow\text{ }\frac{-2}{5}+x=\frac{7}{9}\)
\(\Leftrightarrow\text{ }x=\frac{7}{9}-\frac{-2}{5}\)
\(\Leftrightarrow\text{ }x=\frac{53}{45}\)
\(\text{Vậy }x=\frac{53}{45}\)
\(\text{Chia hay cộng mình không biết nên mình làm 2 TH, cái nào đúng đề thì bạn nhìn nha:}\)
\(\text{TH 1: Dấu chia}\)
\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2:\frac{9}{25}\)
\(\text{ }\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}:\frac{9}{25}\)
\(\text{ }\Leftrightarrow\frac{3}{5}-2x=1\)
\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)
\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)
\(\text{ }\Leftrightarrow2x=\frac{-2}{5}\)
\(\text{ }\Leftrightarrow x=\frac{-2}{5}:2\)
\(\text{ }\Leftrightarrow x=\frac{-1}{5}\)
\(\text{Vậy }\text{}x=\frac{-1}{5}\)
\(\text{TH 2:Dấu cộng}\)
\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2+\frac{9}{25}\)
\(\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}+\frac{9}{25}\)
\(\Leftrightarrow\frac{3}{5}-2x=\frac{18}{25}\)
\(\Leftrightarrow2x=\frac{3}{5}-\frac{18}{25}\)
\(\Leftrightarrow2x=\frac{-3}{25}\)
\(\Leftrightarrow x=\frac{-3}{25}:2\)
\(\Leftrightarrow x=\frac{-3}{50}\)
\(\text{Vậy }x=\frac{-3}{50}\)
\(\text{c, }\left|2x-1\right|=\frac{1}{2}-\frac{-2}{3}\)
\(\Leftrightarrow\left|2x-1\right|=\frac{7}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{7}{6}\\2x-1=\frac{-7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{13}{6}\\2x=\frac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{12}\\x=\frac{-1}{12}\end{matrix}\right.\)
\(\text{Vậy }x\in\left\{\frac{13}{12};\frac{-1}{12}\right\}\)
\(\text{d, }\left(x-\frac{3}{4}\right).\frac{1}{2}=\left(\frac{-1}{2}\right)^2\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right).\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{4}:\frac{1}{2}\)
\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{5}{4}\)
\(\text{Vậy }x=\frac{5}{4}\)
6 tháng 7 2016
mk ko biết cứ bấm đại thui, bn có thể giúp mk ko ???
\(\left|2x-\frac{1}{2}\right|-2x=\frac{1}{2}\)
\(\Rightarrow\left|2x-\frac{1}{2}\right|=\frac{1}{2}+2x\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=\frac{1}{2}+2x\\2x-\frac{1}{2}=-\left(\frac{1}{2}+2x\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x-2x=\frac{1}{2}+\frac{1}{2}\\2x-\frac{1}{2}=-\frac{1}{2}-2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x+2x=\frac{-1}{2}+\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\4x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=0\end{cases}}\)