help me please nhanh tk với tym cho
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2.| 2x - 3 | = \(\frac{1}{2}\)
| 2x - 3 | = \(\frac{1}{2}:2\)
| 2x - 3 | = \(\frac{1}{4}\)
Th 1 : 2x - 3 = \(\frac{1}{4}\)
2x = \(\frac{1}{4}+3\)
2x = \(\frac{13}{4}\)
x = \(\frac{13}{4}:2\)
x = \(\frac{13}{8}\)
2 . | 2x - 3 | = 1/2
<=> | 2x - 3 | = 1/4
<=> 2x - 3 = 1/4
hoặc 2x - 3 = -1/4
<=> x = 13/8
hoặc x = 11/8
7,5 - 3 . | 5- xx | = - 4,5
<=> - 3 | 5 - x | = -12
<=> | 5 - x | = 4
<=> 5 - x = 4
hoặc 5 -x = -4
<=> x = 1 hoặc x = 9
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Vì AOB là góc bẹt => AOB =180*
Có BOC + COA=AOB
=>BOC=180-50=130*
b,Từ O kẻ 1 tia OM vuông góc vơi AB
Ta có BOD+DOM+MOC+COA=AOB
<=> 40* + DOC +50* = 180* (VÌ DOM+MOC=DOC)
=> DOC =90* hay OC vuông góc với OD
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a) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
b) \(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{97.99}\)
\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=2.\frac{32}{99}=\frac{64}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 7 :
a) ( 0,25 )3 . 32 = \(\left(\frac{1}{4}\right)^3.32\)\(=\frac{1}{64}.32=\frac{1}{2}\)
b) ( -0,125 )3 . 804 = \(\left(\frac{-1}{8}\right)^3.8^4.10^4\)\(=\frac{-1}{8^3}.8^4.10000=-8.10000=-80000\)
c) \(\frac{81^{11}.3^{17}}{27^{10}.9^{15}}=\frac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{44}.3^{17}}{3^3^0.3^3^0}=\frac{3^{61}}{3^{60}}=3\)
Bài 8 :
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\Rightarrow\frac{1}{2^m}=\frac{1}{2^5}\Rightarrow m=5\)
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\Rightarrow\frac{7^3}{5^3}=\frac{7^n}{5^n}\Rightarrow n=3\)