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TH1: x< -160
|x+102| = -(x+102) (vì x+102 < 0)
|x+106| = -(x+106) (vì x+106 < 0)
A = |x+102| - |x+106| = -(x+102) - [-(x+106)] = -x -102+x+106 = 4
TH2: -160 =< x <102
|x+102| = -(x+102) (vì x+102 < 0)
|x+106| = (x+106) (vì x+106 > 0)
A = |x+102| - |x+106| = -(x+102) -(x+106) = -x - 102 - x - 106 = -2x - 208
TH3: x >= 102
|x+102| = x+102 (vì x+102 >= 0)
|x+106| = x+106 (vì x+106 > 0)
A = |x+102| - |x+106| = x+102 -(x+106) = x+102 - x - 106 = -4
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\(\frac{x+110}{100}+\frac{x+8}{102}+\frac{x+6}{104}=\frac{x+4}{106}+\frac{x+2}{108}\)
\(\Leftrightarrow\frac{x+110}{100}+\left(\frac{x+8}{102}+1\right)+\left(\frac{x+6}{104}+1\right)=\left(\frac{x+4}{106}+1\right)+\left(\frac{x+2}{108}+1\right)\)
\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}=\frac{x+110}{106}+\frac{x+110}{108}\)
\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}-\frac{x+110}{106}-\frac{x+110}{108}=0\)
\(\Leftrightarrow\left(x+110\right)\left(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}+\frac{1}{106}+\frac{1}{108}\right)=0\)
\(\Leftrightarrow x+110=0\) (vì \(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}-\frac{1}{106}-\frac{1}{108}>0\))
\(\Leftrightarrow x=-110\)
Vậy \(x=-110\)
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a) \(P\left(-1\right)=\left(-1\right)^2+\left(-1\right)^4+...+\left(-1\right)^{106}\)
\(=1+1+...+1\)
=53
b) \(Q\left(-1\right)=\left(-1\right)+\left(-1\right)^3+\left(-1\right)^5+...+\left(-1\right)^{107}\)
\(=-1\cdot54=-54\)
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a) |5/3 - x| - |-5/6| = |-5/9|
=> |5/3 - x| - 5/6 = 5/9
=> |5/3 - x| = 5/9 + 5/6
=> |5/3 - x| = 25/18
=> \(\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=\frac{55}{18}\end{cases}}\)
a, \(\left|\frac{5}{3}-x\right|-\left|-\frac{5}{6}\right|=\left|-\frac{5}{9}\right|\)
\(\Leftrightarrow\left|\frac{5}{3}-x\right|-\frac{5}{6}=\frac{5}{9}\Rightarrow\left|\frac{5}{3}-x\right|=\frac{5}{9}+\frac{5}{6}=\frac{25}{18}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}\Rightarrow}x.\)
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