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\(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)
\(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^x.\left(\frac{1}{2}\right)^4=17\)
\(\left(\frac{1}{2}\right)^x.\left[1+\left(\frac{1}{2}\right)^4\right]=17\)
\(\left(\frac{1}{2}\right)^x.\frac{17}{16}=17\)
\(\left(\frac{1}{2}\right)^x=\frac{17.16}{17}=16\)
\(\left(\frac{1}{2}\right)^x=16=\left(\frac{1}{2}\right)^{-4}\)
=> x = -4
\(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^x\left[1+\left(\frac{1}{2}\right)^4\right]=17\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^x\left(1+\frac{1}{16}\right)=17\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^x.\frac{17}{16}=17\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^x=16\)
\(\Leftrightarrow\frac{1}{2^x}=\frac{1}{2^{-4}}\)
\(\Rightarrow x=-4\)
Bài 1:
a) (2x-3). (x+1) < 0
=>2x-3 và x+1 ngược dấu
Mà 2x-3<x+1 với mọi x
\(\Rightarrow\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Rightarrow\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)\(\Rightarrow-1< x< \frac{3}{2}\)
b)\(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Rightarrow x-\frac{1}{2}\) và x+3 cùng dấu
Xét \(\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\)\(\Rightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\)
Xét \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)\(\Rightarrow\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
=>....
Bài 2:
\(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\cdot\frac{998}{3003}\)
\(=\frac{499}{3003}\)
Ta có : \(\dfrac{x}{4}-\dfrac{1}{y}=\dfrac{1}{2}\left(y\ne0\right)\)
\(\Leftrightarrow\dfrac{xy-4}{4y}=\dfrac{1}{2}\)
\(\Leftrightarrow2xy-8=4y\)
\(\Leftrightarrow xy-2y-4=0\)
\(\Leftrightarrow y\left(x-2\right)=4\)
\(\Leftrightarrow x-2=\dfrac{4}{y}\left(1\right)\)
Mà x, y là các số nguyên .
\(\Rightarrow y\inƯ_{\left(4\right)}\)
\(\Rightarrow y\in\left\{1;-1;2;-2;4;-4\right\}\)
- Thay lần lượt y vào ( 1 ) ta được x lần lượt là : \(\left\{6;-2;4;0;3;1\right\}\)
Vậy ...
a, \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)
\(\Rightarrow\frac{1}{2^x}+\frac{1}{2^x}\cdot\frac{1}{16}=17\)
\(\Rightarrow\frac{1}{2^x}\left(1+\frac{1}{16}\right)=17\)
\(\Rightarrow\frac{1}{2^x}\cdot\frac{17}{16}=17\)
\(\Rightarrow\frac{1}{2^x}=17:\frac{17}{16}=\frac{1}{16}=\frac{1}{2^4}\)
=> x = 4
b, Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;....;\left|x+\frac{1}{99.100}\right|\ge0\)
\(\Rightarrow\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+...+x+\frac{1}{99.100}=100x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)
\(\Rightarrow99x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100x\)
\(\Rightarrow100x-99x=1-\frac{1}{100}\)
\(\Rightarrow x=\frac{99}{100}\)
Giúp mk với
Ta có \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^x\left(1+\frac{1^4}{2^4}\right)=17\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^x.\frac{17}{16}=17\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^x=17:\frac{17}{16}=16\)
\(\Leftrightarrow\frac{1}{2^x}=16\Leftrightarrow1=2^{4+x}\Leftrightarrow4+x=0\Leftrightarrow x=-4\)