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Ta thấy:\(\left|3x+\frac{1}{7}\right|\ge0\)
\(\Rightarrow-\left|3x+\frac{1}{7}\right|\le0\)
\(\Rightarrow-\left|3x+\frac{1}{7}\right|+\frac{5}{3}\le\frac{5}{3}\)
\(\Rightarrow C\le\frac{5}{3}\)
Dấu= khi \(x=-\frac{1}{7}\)
Vậy MinC=\(\frac{5}{3}\) khi \(x=-\frac{1}{7}\)
\(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2\)
\(=\left(-\frac{11}{4}+\frac{1}{2}\right)^2\)
\(=\left(-\frac{11}{4}+\frac{2}{4}\right)^2\)
\(=\left(-\frac{9}{4}\right)^2\)
\(=\frac{81}{16}\)
\(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2\)
\(=\left(\frac{-11}{4}+\frac{1}{2}\right)^2\)
\(=\left(\frac{-11}{4}+\frac{2}{4}\right)^2\)
\(=\left(\frac{-9}{4}\right)^2\)
\(=\frac{81}{16}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow\frac{a}{b}.\frac{c}{d}=\frac{a+c}{b+d}.\frac{a+c}{b+d}\)
\(\Rightarrow\frac{ac}{bd}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(đpcm\right)\)
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
Ta có:
\(\frac{ac}{bd}=\frac{bkdk}{bd}=k^2\) (1)
\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{\left[k.\left(b+d\right)\right]^2}{\left(b+d\right)^2}=\frac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\) (2)
Từ (1) và (2) suy ra \(\frac{ac}{bd}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(đpcm\right)\)
Ta có:A=\(\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+....+\left(\frac{1}{2}\right)^{99}\right]\)
\(\frac{1}{2}A\)=\(\frac{1}{2}\)\(\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{4}\right)^4+....+\left(\frac{1}{2}\right)^{99}\right]\)
\(\frac{1}{2}A\)=\(\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^5+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(\frac{1}{2}A-A\)=\(\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^5+...+\left(\frac{1}{2}\right)^{100}\right]\)-\(\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+....+\left(\frac{1}{2}\right)^{99}\right]\)
\(-\frac{1}{2}A\)=\(\left(\frac{1}{2}^{100}\right)-\frac{1}{2}\)
\(-\frac{1}{2}A\)=\(-\frac{1}{2}\)
A=\(-\frac{1}{2}:\left(-\frac{1}{2}\right)\)
A=1
Chúc bạn học tốt!
a) \(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right).\left(x+8\right)}+\frac{6}{\left(x+8\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14}{\left(x+2\right).\left(x+14\right)}-\frac{x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14-x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{16}{\left(x+2\right).\left(x+4\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow x=16\)
Vậy x = 16
\(b,\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(vì\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
\(\text{Vậy }x=-1\)
b. (x+1)(1/10+1/11+1/12-1/13-1/14)=0
x+1=0 (vì : 1/10+1/11+1/12-1/13-1/14>0)
x=-1
\(\left(x+2\right)\left(x+\frac{2}{3}\right)>0\)
(+) \(\begin{cases}x+2>0\\x+\frac{2}{3}>0\end{cases}\)\(\Rightarrow\begin{cases}x>-2\\x>-\frac{2}{3}\end{cases}\)\(\Rightarrow x>-\frac{2}{3}\)
(+) \(\begin{cases}x+2< 0\\x+\frac{2}{3}< 0\end{cases}\)\(\Rightarrow\begin{cases}x< -2\\x< -\frac{2}{3}\end{cases}\)\(\Rightarrow x< -2\)
Vậy \(x>-\frac{2}{3}\) ; \(x< -2\)
\(\frac{64}{\left(-2\right)^x}=\left(-16\right)^2:4^3\)
<=> \(\frac{64}{\left(-2\right)^x}=4\)
<=> \(\frac{64}{\left(-2\right)^x}=\frac{64}{16}\)
<=> (-2)x = 16
<=> x = 4
Ta có : \(E=\left|x+5\right|+\left|x+2\right|+\left|x-7\right|+\left|x-8\right|=\left(\left|x+5\right|+\left|8-x\right|\right)+\left(\left|7-x\right|+\left|x+2\right|\right)\)
\(\ge\left|x+5+8-x\right|+\left|7-x+x+2\right|=22\)
Dấu "=" xảy ra khi \(\begin{cases}-5\le x\le8\\-2\le x\le7\end{cases}\) \(\Rightarrow-2\le x\le7\)
Vậy MIN E = 22 khi \(-2\le x\le7\)
\(\left(x-\frac{3}{4}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)
\(\Leftrightarrow x-\frac{3}{4}=\left(\frac{2}{3}\right)^2\)
\(\Leftrightarrow x=\frac{43}{36}\)