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A= \(\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{-\frac{131}{100}}{\frac{13}{5}}-\frac{5}{6}:2\)
=\(-\frac{131}{260}-\frac{5}{12}=-\frac{359}{390}\)
B= \(\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}.\frac{26}{75}=\frac{13}{12}\)
b) ta có : A=\(-\frac{359}{390}\approx-0,9\)
B= \(\frac{13}{12}\approx1,08\)
=> A<x<B mà x nguyên => x=0 hoặc x=1
- 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
Cái ............... là gì vậy bn
mk ko biết cứ bấm đại thui, bn có thể giúp mk ko ???
\(\left[\frac{x}{\left(x+4\right)\left(x-4\right)}-\frac{x-4}{x\left(x+4\right)}\right]:\frac{2\left(x-2\right)}{x\left(x+4\right)}\)\(=\left[\frac{x^2-\left(x-4\right)^2}{x\left(x+4\right)\left(x-4\right)}\right].\left[\frac{x\left(x+4\right)}{2\left(x-2\right)}\right]\)\(=\left(\frac{x^2-x^2+8x-16}{x\left(x+4\right)\left(X-4\right)}\right).\frac{x\left(x+4\right)}{2\left(x-2\right)}=\frac{8\left(x-2\right).x\left(x+4\right)}{x\left(x+4\right)\left(x-4\right).2\left(x-2\right)}=\frac{4}{x-4}\)
a)\(\left(2x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\) hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại) hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)
\(\Leftrightarrow-1< x< \frac{3}{2}\)
b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)
c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)
Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow x=4\)
