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a) vì | x + \(\frac{5}{3}\)| \(\ge\)0 nên A = | x + \(\frac{5}{3}\)| + 112 \(\ge\)112
dấu " = " xảy ra khi | x + \(\frac{5}{3}\)| = 0 hay x = \(\frac{-5}{3}\)
\(\Rightarrow\)GTNN của A là 112 khi | x + \(\frac{5}{3}\) | = 0 hay x = \(\frac{-5}{3}\)
b) B = | x - 2,7 | + | x + 8,5 |
B = | 2,7 - x | + | x + 8,5 | \(\ge\)| 2,7 - x + x + 8,5 | = 11,2
\(\Rightarrow\)GTNN của B là 11,2 khi ( 2,7 - x ) . ( x + 8,5 ) \(\ge\)0 hay -8,5 \(\le\)x \(\le\)2,7
c) C = \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|2x+\frac{1}{4}\right|\)
C = \(\left|x+\frac{1}{2}\right|+\left|-\frac{1}{3}-x\right|+\left|2x+\frac{1}{4}\right|\)\(\ge\)\(\left|x+\frac{1}{2}-\frac{1}{3}-x\right|+\left|2x+\frac{1}{4}\right|=\frac{1}{6}+\left|2x+\frac{1}{4}\right|\ge\frac{1}{6}\)
\(\Rightarrow\)GTNN của C là \(\frac{1}{6}\)khi \(\hept{\begin{cases}2x+\frac{1}{4}=0\Leftrightarrow x=\frac{-1}{8}\\\left(x+\frac{1}{2}\right).\left(-\frac{1}{3}-x\right)\ge0\Leftrightarrow\frac{-1}{2}\le x\le\frac{-1}{3}\end{cases}}\)
1 :\(\frac{7}{20}\)
2 \(\frac{1}{4}\)
3 \(\frac{23}{2}\)
4 2187
5 64
6 x=16
7 x=\(\frac{-1}{243}\)
8 mϵ∅
cho mình hỏi cài này là j vậy
Đề 2
1) \(\frac{7}{20}.\)
2) \(\frac{1}{4}.\)
3) \(\frac{23}{2}.\)
4) \(2187.\)
5) \(64.\)
6) \(x=16.\)
7) \(x=\left(-\frac{1}{3}\right)^5\)
8) \(m\in\varnothing.\)
Chúc bạn học tốt!
a/ \(\left|x-\frac{1}{3}\right|=2\frac{1}{5}\)
\(\left|x-\frac{1}{3}\right|=\frac{11}{5}\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{11}{5}\\x-\frac{1}{3}=\frac{-11}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{11}{5}+\frac{1}{3}\\x=\frac{-11}{5}+\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{33}{15}+\frac{5}{15}\\x=\frac{-33}{15}+\frac{5}{15}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{38}{15}\\x=\frac{28}{15}\end{matrix}\right.\)
Vậy:
c/
\(\left|x+\frac{1}{4}\right|-\frac{3}{4}=5\%\)
\(\left|x+\frac{1}{4}\right|-\frac{3}{4}=\frac{1}{20}\)
\(\left|x+\frac{1}{4}\right|=\frac{1}{20}+\frac{15}{20}\)
\(\left|x+\frac{1}{4}\right|=\frac{4}{5}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{4}{5}\\x+\frac{1}{4}=\frac{-4}{5}\end{matrix}\right.\)
Bạn tự tính tiếp nhé!
a) \(5x-7=3x+9\)
\(\Rightarrow5x-3x=9+7\)
\(\Rightarrow2x=16\)
\(\Rightarrow x=16:2\)
\(\Rightarrow x=8\)
Vậy \(x=8.\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{2}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{2}{5}-\frac{1}{2}\\x=\left(-\frac{2}{5}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}.\)
c) \(5x-\left|9-7x\right|=3\)
\(\Rightarrow\left|9-7x\right|=5x-3\)
\(\Rightarrow\left[{}\begin{matrix}9-7x=5x-3\\9-7x=3-5x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}9+3=5x+7x\\9-3=-5x+7x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}12=12x\\6=2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=12:12\\x=6:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{1;3\right\}.\)
d) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Rightarrow-5+\left|3x-1\right|+6=4\)
\(\Rightarrow-5+\left|3x-1\right|=4-6\)
\(\Rightarrow-5+\left|3x-1\right|=-2\)
\(\Rightarrow\left|3x-1\right|=\left(-2\right)+5\)
\(\Rightarrow\left|3x-1\right|=3.\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=3\\3x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=4\\3x=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4:3\\x=\left(-2\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{4}{3};-\frac{2}{3}\right\}.\)
Chúc bạn học tốt!
\(|x+1,5|=2\)
\(|x+\frac{3}{2}|=2\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{2}=2\\x+\frac{3}{2}=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2-\frac{3}{2}\\x=-2-\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-7}{2}\end{matrix}\right.\)
Vậy \(x=\left[{}\begin{matrix}\frac{1}{2}\\\frac{-7}{2}\end{matrix}\right.\)
Bài 1:
a) Ta có: \(\frac{-3}{4}< \frac{a}{12}< \frac{-5}{9}\)
\(\Leftrightarrow\frac{-27}{36}< \frac{3a}{36}< \frac{-20}{36}\)
Suy ra: \(-27< 3a< -20\)
\(\Leftrightarrow3a\in\left\{-26;-25;-24;-23;-22;-21\right\}\)
\(\Leftrightarrow a\in\left\{\frac{-26}{3};\frac{-25}{3};-8;-\frac{23}{3};-\frac{22}{3};-7\right\}\)
mà \(a\in Z\)
nên \(a\in\left\{-8;-7\right\}\)
a) 2-|3/2x-1/4|=|-5/4|
=> |3/2x-1/4| = 2-|-5/4| = 2-5/4 = 3/4
=> 3/2x-1/4 = 3/4 hoac -3/4
Khi 3/2x-1/4=3/4 => x=2/3
Khi 3/2x-1/4 = -3/4 => -1/3
Vay x la { 2/3 ; -1/3 }
b) tu la