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\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
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c) l x - 5 l = 2x
\(\Leftrightarrow\orbr{\begin{cases}x-5=2x\\x-5=-2x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2x=5\\x+2x=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=5\\3x=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}\)
Hok tốt!!!!!!!
Tìm x, biết:
a) |2x + 1| = 17
<=>\(\orbr{\begin{cases}2x+1=17\\2x+1=-17\end{cases}}\)
<=>\(\orbr{\begin{cases}2x=16\\2x=-18\end{cases}}\)
<=> \(\hept{\begin{cases}x=8\\x=-9\end{cases}}\)
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<=>|2x-7|+|2x+10|=|2x-7|+2|x+5|
=>|2x-7|+2|x+5|=17
=>x=-5 hoặc \(\frac{7}{2}\)
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| 2x - 7 | + | 2x + 10 | = 17 (*)
Ta có :
| 2x - 7 | + | 2x + 10 |
= | 2x - 7 | + | -( 2x + 10 ) |
= | 2x - 7 | + | -2x - 10 |
Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :
| 2x - 7 | + | -2x - 10 | ≥ | 2x - 7 - 2x - 10 | = | -17 | = 17 ( đúng với (*) )
Đẳng thức xảy ra ( tức (*) ) khi ab ≥ 0
=> ( 2x - 7 )( -2x - 10 ) ≥ 0
Xét hai trường hợp
1. \(\hept{\begin{cases}2x-7\ge0\\-2x-10\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ge7\\-2x\ge10\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{7}{2}\\x\le-\frac{10}{2}\end{cases}}\)( loại )
2. \(\hept{\begin{cases}2x-7\le0\\-2x-10\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\le7\\-2x\le10\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{7}{2}\\x\ge-\frac{10}{2}\end{cases}\Leftrightarrow}-5\le x\le3,5\)
Vì x nguyên => x ∈ { -5 ; -4 ; -3 ; -2 ; -1 ; 0 ; 1 ; 2 ; 3 }
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a ) \(\left|2x-3\right|=\left|x+5\right|\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=x+5\\2x-3=-x+5\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-x=5+3\\2x+x=5+3\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\3x=8\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=\frac{8}{3}\left(loại\right)\end{array}\right.\)
Vậy ...........
b ) \(\left|x\right|+2x=7\)
\(\Rightarrow\left[\begin{array}{nghiempt}-x+2x=7\\x+2x=7\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=7\\3x=7\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=7\left(loại\right)\\x=\frac{7}{3}\end{array}\right.\)
Vậy .........................
c ) \(\left|x\right|+2x=7\) ( giống câu b )