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\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
a: \(\left(x-2\right)^2+\left(x-y\right)^6+3\ge3\)
\(\Leftrightarrow A=\dfrac{2003}{\left(x-2\right)^2+\left(x-y\right)^6+3}\le\dfrac{2003}{3}\)
Dấu '=' xảy ra khi x=y=2
b: \(B=-\left(2x+\dfrac{1}{3}\right)^6+3\le3\forall x\)
Dấu '=' xảy ra khi x=-1/6
c: \(C=\dfrac{x^{2016}+2015+2}{x^{2016}+2015}=1+\dfrac{2}{x^{2016}+2015}\le\dfrac{2}{2015}+1=\dfrac{2017}{2015}\)
Dấu '=' xảy ra khi x=0
a) \(\left|x-\dfrac{5}{3}\right|< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{-1}{3}< x-\dfrac{5}{3}< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{-1}{3}+\dfrac{5}{3}< x-\dfrac{5}{3}+\dfrac{5}{3}< \dfrac{1}{3}+\dfrac{5}{3}\)
\(\Rightarrow\dfrac{4}{3}< x< 2\)
b) \(\left|x+\dfrac{11}{2}\right|>\left|-5,5\right|=5,5\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{11}{2}< 5,5\\x+\dfrac{11}{2}>5,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 5,5-\dfrac{11}{2}=0\\x>5,5-\dfrac{11}{2}=0\end{matrix}\right.\)
=> Với x khác 0 thì thõa mãn đề bài
c) \(\dfrac{2}{5}< \left|x-\dfrac{7}{5}\right|< \dfrac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2}{5}< x-\dfrac{7}{5}< \dfrac{3}{5}\\-\dfrac{2}{5}< x-\dfrac{7}{5}< -\dfrac{3}{5}\end{matrix}\right.\)
Ta thấy trường hợp 2 là trường hợp không thể xảy ra
=> Loại
Vậy \(\dfrac{2}{5}< x-\dfrac{7}{5}< \dfrac{3}{5}\)
\(\Rightarrow\dfrac{2}{5}+\dfrac{7}{5}< x< \dfrac{3}{5}+\dfrac{7}{5}\)
\(\Rightarrow\dfrac{9}{5}< x< 2\) (nhận)
p/s : làm đại nha , ko bik đúng sai
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
a) \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
⇔ \(\left(\dfrac{-8}{5}+x\right).\dfrac{13}{12}=\dfrac{13}{6}\)
⇔ \(-\dfrac{8}{5}+x=\dfrac{13}{6}:\dfrac{13}{12}\)
⇔ \(-\dfrac{8}{5}+x=2\)
⇔ \(x=2+\dfrac{8}{5}\)
⇔ \(x=\dfrac{18}{5}\)
b) \(\dfrac{-4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
⇔ \(-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
⇔ \(-\dfrac{4}{7}x=-\dfrac{3}{40}-\dfrac{7}{5}\)
⇔ \(-\dfrac{4}{7}x=-\dfrac{59}{40}\)
⇔ \(x=\left(-\dfrac{59}{40}\right):\left(-\dfrac{4}{7}\right)\)
⇔ \(x=\dfrac{413}{160}\)
a, \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
=> \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=\dfrac{13}{6}\)
=> \(\left(-1\dfrac{3}{5}+x\right)=\dfrac{13}{6}.\dfrac{12}{13}\)
=> \(\left(-1\dfrac{3}{5}+x\right)=2\)
=> \(\dfrac{-8}{5}+x=2\)
=> x= \(2+\dfrac{8}{5}=\dfrac{10}{5}+\dfrac{8}{5}\)
=> x= \(\dfrac{18}{5}\)
1.
\(\left(\dfrac{-2}{3}\right).0,75+1\dfrac{2}{3}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)^2\)
\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}.\left(\dfrac{9}{-4}\right)+\dfrac{1}{4}\)
\(=-\dfrac{1}{2}+\dfrac{45}{-12}+\dfrac{1}{4}\)
\(=-\dfrac{6}{12}+\dfrac{-45}{12}+\dfrac{3}{4}\)
\(=\dfrac{-48}{12}\)
\(=-4\)
2.
a) \(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{10}{20}\)
\(\Leftrightarrow x=\dfrac{-11}{20}\)
b) \(\left|x-\dfrac{2}{5}\right|+\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{11}{4}-\dfrac{3}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=-2\Rightarrow x=-2+\dfrac{2}{5}=\dfrac{-8}{5}\\x-\dfrac{2}{5}=2\Rightarrow x=2+\dfrac{2}{5}=\dfrac{12}{5}\end{matrix}\right.\)
3.
a) \(\dfrac{16}{2^n}=2\)
\(\Leftrightarrow2^n=16:2\)
\(\Leftrightarrow2^n=8\)
\(\Leftrightarrow2^n=2^3\)
\(\Leftrightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{81}=-27\)
\(\Leftrightarrow\left(-3\right)^n=\left(-27\right).81\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)
\(\Leftrightarrow n=7\)
4. Ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)
\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\) (2)
Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Vì \(x-y+x=-49\) ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)
Vậy \(\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)
Vì /2x+1/ ≥ 0
=> /2x+1/ + 2017 ≥ 2017
=> 2016/ /2x+1/ +2017 ≤ 2016/2017
Vậy Bmax = 2016/2017 khi /2x+1/ = 0 => 2x+1 =0 => 2x=-1
=> x = -1/2
\(0< \left(\dfrac{1}{2}\right)^{2016}< 1\Rightarrow\left\{{}\begin{matrix}0< 1-\left(\dfrac{1}{2}\right)^{2016}< 1\\1< 1+\left(\dfrac{1}{2}\right)^{2016}< 2\end{matrix}\right.\)
\(\Rightarrow0< x< 2\)
(0,2) có duy nhất 1 là nguyên => đáp số : 1
nhận thấy 1 < ( 1+ \(\left(\dfrac{-1}{2}\right)^{2016}\)) < 2
0 < (1-\(\left(\dfrac{1}{2}\right)^{2016}\)) < 1
lại có x nguyên \(\Rightarrow0\le x\le1\)
vay x= 0 ; 1