23/31 - ( -8/31 - 8/23 )
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\(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)\) =\(\frac{31}{23}-\frac{417}{736}=\frac{25}{32}\)
31/23-(7/32+8/23)
=31/23-7/32-8/23
=(31/23-8/23)-7/32
=23/23-7/32
=1-7/32=32/32-7/32=25/32
\(\frac{19}{15}+\frac{31}{23}-\frac{4}{15}-\frac{8}{23}=\frac{19-4}{15}+\frac{31-8}{23}=\frac{15}{15}+\frac{23}{23}=1+1=2\)
\(\frac{19}{15}+\frac{31}{23}-\frac{4}{15}-\frac{8}{23}\)
\(=\frac{19-4}{15}+\frac{31-8}{23}\)
\(=\frac{15}{15}+\frac{23}{23}\)
\(=1+1=2\)
a: =-12/20+5/20-6/20
=-13/20
b: =10/50-45/50-14/50
=-49/50
c: =31/23-7/32-8/23=1-7/32=25/32
\(a\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)\)
\(=\frac{31}{23}-\frac{7}{23}-\frac{8}{23}\)
\(=\frac{16}{23}\)
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tốt
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a)\(\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)=\frac{31}{23}-\frac{8}{23}+\frac{7}{23}=1+\frac{7}{23}=1\frac{7}{23}\)
b)\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)=\left(\frac{13}{41}+\frac{28}{41}\right)+\left(\frac{79}{67}-\frac{12}{67}\right)-\frac{1}{3}=1+1-\frac{1}{3}\)
\(=2-\frac{1}{3}=\frac{5}{3}\)
Bài 1:
1) Ta có: \(\left(-12\right)+6\cdot\left(-3\right)\)
\(=-12-18\)
=-30
2) Ta có: \(\left(36-2020\right)+\left(2019-136\right)-27\)
\(=36-2020+2019-136-27\)
\(=1-100-27\)
\(=-126\)
3) Ta có: \(\left(144-97\right)-\left(244-197\right)\)
\(=144-97-244+197\)
\(=-100+100=0\)
4) Ta có: \(\left(-24\right)\cdot13-24\cdot\left(-3\right)\)
\(=-24\cdot13+24\cdot3\)
\(=24\cdot\left(-13+3\right)\)
\(=24\cdot\left(-10\right)=-240\)
5) Ta có: \(54+55+56+57+58-\left(64+65+66+67+68\right)\)
\(=54+55+56+57+58-64-65-66-67-68\)
\(=\left(54-64\right)+\left(55-65\right)+\left(56-66\right)+\left(57-67\right)+\left(58-68\right)\)
\(=\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)\)
=-50
6) Ta có: \(24\cdot\left(16-5\right)-16\cdot\left(24-5\right)\)
\(=24\cdot16-24\cdot5-16\cdot24+16\cdot5\)
\(=-24\cdot5+16\cdot5\)
\(=5\cdot\left(-24+16\right)\)
\(=-5\cdot8=-40\)
7) Ta có: \(47\cdot\left(23+50\right)-23\cdot\left(47+50\right)\)
\(=47\cdot23+47\cdot50-23\cdot47-23\cdot50\)
\(=47\cdot50-23\cdot50\)
\(=50\cdot\left(47-23\right)\)
\(=50\cdot24=1200\)
8) Ta có: \(\left(-31\right)\cdot47+\left(-31\right)\cdot52+\left(-31\right)\)
\(=-31\cdot\left(47+52+1\right)\)
\(=-31\cdot100=-3100\)
Bài 2:
1) Ta có: \(-17-\left(2x-5\right)=-6\)
\(\Leftrightarrow-17-2x+5+6=0\)
\(\Leftrightarrow-2x-6=0\)
\(\Leftrightarrow-2x=6\)
hay x=-3
Vậy: x=-3
2) Ta có: \(10-2\left(4-3x\right)=-4\)
\(\Leftrightarrow10-8+6x+4=0\)
\(\Leftrightarrow6x+6=0\)
\(\Leftrightarrow6x=-6\)
hay x=-1
Vậy: x=-1
3) Ta có: \(-12+3\left(-x+7\right)=-18\)
\(\Leftrightarrow-12-3x+21+18=0\)
\(\Leftrightarrow-3x+27=0\)
\(\Leftrightarrow-3x=-27\)
hay x=9
Vậy: x=9
4) Ta có: \(-45:\left[5\cdot\left(-3-2x\right)\right]=3\)
\(\Leftrightarrow5\cdot\left(-3-2x\right)=-15\)
\(\Leftrightarrow-2x-3=-3\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: x=0
5) Ta có: x(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3\right\}\)
6) Ta có: (x-2)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-4\right\}\)
7) Ta có: \(x\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-1;3\right\}\)
Bài 1:
1) Ta có: (−12)+6⋅(−3)(−12)+6⋅(−3)
=−12−18=−12−18
=-30
2) Ta có: (36−2020)+(2019−136)−27(36−2020)+(2019−136)−27
=36−2020+2019−136−27=36−2020+2019−136−27
=1−100−27=1−100−27
=−126
Tớ chcs cậu học thật giỏi nha !
a) 2016 - 100.(x+31)=27:23
<=> 2016 - 100.(x+31) = 24 = 16
<=> 100.(x+31) = 2016 - 16 = 2000
<=> x+31 = 2000:100 = 20
<=> x = 20 - 31 = -11
b) (x+8)3 = 125
<=> (x+8)3 = 53
<=> x+8 = 5
<=> x = 5 - 8
<=> x = -3
= \(\dfrac{15}{23}\)
15/23