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Trả lời:
a, \(\left|x\right|=5\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
Vậy x = 5; x = - 5
b, \(\left|x\right|< 2\) ( vô lí )
Vậy không tìm được x thỏa mãn đề bài.
c, \(\left|x\right|=-1\)( vô lí )
Vậy không tìm được x thỏa mãn đề bài.
d, \(\left|x\right|=\left|-5\right|\)
\(\Rightarrow\left|x\right|=5\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
Vậy x = 5; x = - 5
e, \(\left|x+3\right|=0\)
\(\Rightarrow x+3=0\)
\(\Rightarrow x=-3\)
Vậy x = - 3
f, \(\left|x-1\right|=4\)
\(\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
Vậy x = 5; x = - 3
g, \(\left|x-5\right|=10\)
\(\Rightarrow\orbr{\begin{cases}x-5=10\\x-5=-10\end{cases}\Rightarrow\orbr{\begin{cases}x=15\\x=-5\end{cases}}}\)
Vậy x = 15; x = - 5
h, \(\left|x+1\right|=-2\) ( vô lí )
Vậy không tìm được x thỏa mãn đề bài.
i, \(\left|x+4\right|=5-\left(-1\right)\)
\(\Rightarrow\left|x+4\right|=6\)
\(\Rightarrow\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}}\)
Vậy x = 2; x = - 10
k, \(\left|x-1\right|=-10-3\)
\(\Rightarrow\left|x-1\right|=-13\) ( vô lí )
Vậy không tìm được x thỏa mãn đề bài.
l, \(\left|x+2\right|=12+\left(-3\right)+\left|-4\right|\)
\(\Rightarrow\left|x+2\right|=12-3+4\)
\(\Rightarrow\left|x+2\right|=13\)
\(\Rightarrow\orbr{\begin{cases}x+2=13\\x+2=-13\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-15\end{cases}}}\)
Vậy x = 11; x = - 15
m, \(\left|x+2\right|-12=-1\)
\(\Rightarrow\left|x+2\right|=11\)
\(\Rightarrow\orbr{\begin{cases}x+2=11\\x+2=-11\end{cases}\Rightarrow\orbr{\begin{cases}x=9\\x=-13\end{cases}}}\)
Vậy x = 9; x = - 13
n, \(135-\left|9-x\right|=-1\)
\(\Rightarrow\left|9-x\right|=136\)
\(\Rightarrow\orbr{\begin{cases}9-x=136\\9-x=-136\end{cases}\Rightarrow\orbr{\begin{cases}x=-127\\x=145\end{cases}}}\)
Vậy x = - 127; x = 145
o, \(\left|2x+3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}2x+3=5\\2x+3=-5\end{cases}\Rightarrow\orbr{\begin{cases}2x=2\\x=-8\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-4\end{cases}}}\)
Vậy x = 1; x = - 4
4, Q = |x+\(\frac{1}{5}\) | -x +\(\frac{4}{7}\)
xét x \(\ge\) \(-\frac{1}{5}\)
Ta Có Q = |x+\(\frac{1}{5}\) | -x + \(\frac{4}{7}\) = x+\(\frac{1}{5}\) - x +\(\frac{4}{7}\) = \(\frac{27}{35}\) (1)
xét x \(< -\frac{1}{5}\)
Ta có Q = | x +\(\frac{1}{5}\) | - x + \(\frac{4}{7}\) = -x - \(\frac{1}{5}\) - x + \(\frac{4}{7}\) = -2x + \(\frac{13}{35}\)
với x \(< -\frac{1}{5}\)
=> -2x \(>\) \(\frac{2}{5}\)
=> -2x + \(\frac{13}{35}\) \(>\frac{27}{35}\) (2)
Từ (1) và (2) => MinQ = \(\frac{27}{35}\) khi \(x\ge-\frac{1}{5}\)
5 , D = |x| + |8-x|
D = |x| + |8-x| \(\ge\) |x+8-x| = |8| = 8
Dấu ''='' xảy ra khi x(8-x) \(\ge\) 0 <=> 0\(\le\)x\(\le\) 8
Vậy MinD = 8 khi \(0\le x\le8\)
6,L= |x - 2012| + |2011 - x|
L = |x-2012| + |2011-x| \(\ge\) | x-2012 + 2011 - x | = |-1| = 1
Dấu ''= '' xảy ra khi ( x-2012)(2011-x) \(\ge\) 0
làm nốt câu 6 nãy ấn nhầm
<=> 2011\(\le\) x \(\le\) 2012
Vậy MinL = 1 khi \(2011\le x\le2012\)
7 , E = | x- \(\frac{2006}{2007}\) | + |x-1|
Ta có :
E = |x-\(\frac{2006}{2007}\) | + |1-x|
E = | x - \(\frac{2006}{2007}\) | + |1-x| \(\ge\) | x - \(\frac{2006}{2007}\) + 1 - x | = \(\frac{1}{2007}\)
Dấu ''='' xảy ra khi (x- \(\frac{2006}{2007}\) ) ( 1-x ) \(\ge0\) <=> \(\frac{2006}{2007}\le x\le1\)
Vậy MinE = \(\frac{1}{2007}\) khi \(\frac{2006}{2007}\le x\le1\)
8 ,F = | x -\(\frac{1}{4}\) | + | \(x-\frac{3}{4}\) |
Ta có :
F = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) - x |
F = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) -x | \(\ge\) | x - \(\frac{1}{4}\) + \(\frac{3}{4}\) -x | = \(\frac{1}{2}\)
Dấu ''='' xảy ra khi ( x-\(\frac{1}{4}\) ) ( \(\frac{3}{4}-x\) ) \(\ge\) 0 <=> \(\frac{1}{4}\le x\le\frac{3}{4}\)
Vậy MinF = \(\frac{1}{2}\) khi \(\frac{1}{4}\le x\le\frac{3}{4}\)
a) -4/5 + 5/2x = -3/10
5/2x = -3/10 + 4/5
5/2x = 1/5
5/2x = 1/2
x = 1/2 : 5/2
x = 1/5
b) 4/3 + 5/8 : x = 1/12
5/8x = 1/12 - 4/3
5/8x = -5/4
5 = -5/4.8x
5 = -10x
5/-10 = x
-1/2 = x
x = -1/2
c) (x - 1/3)(x - 2/5) = 0
x - 1/3 = 0 hoặc x - 2/5 = 0
x = 0 + 1/3 x = 0 + 2/5
x = 1/3 x = 2/5
a: \(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)
\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)=\left(\dfrac{x-9}{10}+1\right)+\left(\dfrac{x-10}{11}+1\right)+\left(\dfrac{x-11}{12}+1\right)\)
=>x+1=0
hay x=-1
c: |x-2|=13
=>x-2=13 hoặc x-2=-13
=>x=15 hoặc x=-11
d: \(\Leftrightarrow3\left|x-2\right|+4\left|x-2\right|=2-\dfrac{1}{3}=\dfrac{5}{3}\)
=>7|x-2|=5/3
=>|x-2|=5/21
=>x-2=5/21 hoặc x-2=-5/21
=>x=47/21 hoặc x=37/21
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)