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a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
câu 1 bỏ dấu ngoặc rồi tính
( 36 + 79 ) + ( 145 _ 79 _ 36 )
\(=36+79+145-79-36\)
\(=\left(36-36\right)+\left(79-79\right)+145\)\
\(=0+0+145=145\)
10 _ [ 12 _ ( -9 _ 1 ) ]
\(=10-12-10\)
\(=10-10-12\)
\(=0-12=-12\)
( 38 _ 29 + 43) _ ( 43 + 38 )
\(=38-29+43-43-38\)
\(=\left(38-38\right)+\left(43-43\right)-29\)
\(=0+0-29=-29\)
271 _ [ ( -43 ) + 271 _ ( -17 ) ]
\(=271+43-271-17\)
\(=\left(271-271\right)+\left(43-17\right)\)
\(=0+26=26\)
- 144 _ [ 29 _ ( + 144 ) _ ( + 144 )]
\(=-144-19+144+144\)
\(=\left(-144+144+144\right)-19\)
\(=144-19=125\)
đợi mk lm tiếp câu 2 nha .
bài 2 tính tổng các số nguyên
- 18 < hoặc bằng x < hoặc bằng 17
\(\Rightarrow x\in\left\{-18;-17;-16;....;17\right\}\)
tổng \(x=-18+\left(-17\right)+\left(-16\right)+...+17=-18\)
- 27 < hoặc bằng x < hoặc bằng 27
\(\Rightarrow x\in\left\{-27;-26;-25;..;27\right\}\)
Tổng \(x=-27+\left(-26\right)+\left(-25\right)+...+27=0\)
a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)
=\(\frac{3}{8}+\frac{2}{13}\)
=\(\frac{55}{104}.\)
b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)
=\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)
=\(\frac{2}{7}+\frac{2}{7}\)
=\(\frac{4}{7}\)
c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)
=\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)
=\(-\frac{16}{55}--\frac{1}{50}\)
=\(-\frac{149}{550}.\)
d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)
=\(-\frac{417}{1564}\)
e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)
=\(\frac{380}{391}.\)
g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)
=\(-\frac{5}{28}+-\frac{132}{85}\)
= \(-1.731512605.\)
k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................
a: \(=\dfrac{-39+19+10}{12}=\dfrac{-10}{12}=\dfrac{-5}{6}\)
b: \(=\dfrac{2^{30}\cdot3^{16}\cdot7-2^{34}\cdot3^{15}}{2^{28}\cdot3^{21}-2^{28}\cdot3^{17}}\)
\(=\dfrac{2^{30}\cdot3^{15}\left(3\cdot7-2^4\right)}{2^{28}\cdot3^{17}\left(3^4-1\right)}=\dfrac{2^2}{3^2}\cdot\dfrac{21-16}{80}=\dfrac{4}{9}\cdot\dfrac{5}{80}\)
\(=\dfrac{20}{720}=\dfrac{1}{36}\)
c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)
\(=\dfrac{255}{103}\)
a) \(12.79-64.37+12.21-36.37+12000\)
\(=\left(12.79+12.21\right)-\left(64.37+36.37\right)+12000\)
\(=\left[12\left(79+21\right)-37\left(64+36\right)+12000\right]\)
\(=12.100-37.100+12000\)
\(=1200-3700+12000\)
\(=9500\)
b) \(112.39+1237+112.61\)
\(=\left(112.39+112.61\right)+1237\)
\(=112\left(39+61\right)+1237\)
\(=112.100+1237\)
\(=11200+1237\)
\(=12437\)
c) \(35.762+351-35.662\)
\(=35\left(762-662\right)+351\)
\(=35.100+351\)
\(=3500+351\)
\(=3851\)
d) \(\frac{20}{4}-\frac{5^9}{5^8}=5-5=0\)
e) \(\left(\frac{5^{19}}{5^{17}}-4\right):7=\left(5^2-4\right):7=21:7=3\)
f) \(\frac{84}{4}+\frac{3^9}{3^7}+5^0=21+3^2+1=31\)