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4: =4+5+{12+(-8)-(-8)-(-5)+(-12)+16-5+3}
=9+[12+5-12+16-5+3}
=9+19=28
5: =15-[13+8-10-16]-{5-7+9-3}
=15-(-5)-4
=15+5-4
=20-4
=16
m) 63.83-9.7.(-17)
= 5229-63.(-17)
= 5229 - (-1071)
= 5229 + 1071
= 6300
n) 42.3 - 7[(-34)+18]
= 126 - 7. (-16)
= 126 - (-112)
=126+112
= 238
\(p,71\cdot70+35\cdot\left(-7\right)-13\cdot35\)
\(=4970+\left(-245\right)-455=4725-455=4270\)
\(q,18\cdot\left(23-17\right)-13\cdot\left(23+18\right)\)
\(=18\cdot6-13\cdot41=108-533=-425\)
g: =(32-34)+(36-38)+(40-42)
=(-2)+(-2)+(-2)
=-6
l: =-5-{-[7-10]-[5+12]}-[-3+9-4-5]
=-5+(-3)+17-[-3]
=-8+17+3
=20-8=12
9: =>x-2/3=1/5*-7/15=-7/75
=>x=-7/75+50/75=43/75
11: =>x-4/5=3/10*(-7)/3=-7/10
=>x=-7/10+4/5=8/10-7/10=1/10
12:=>7/5*9+x=4/9
=>x=4/9-63/5=-547/45
10: =>x+5/6=-8/147
=>x=-87/98
\(1,-\dfrac{5}{12}\left(\dfrac{4}{9}+-\dfrac{5}{16}\right)\\ =-\dfrac{5}{12}\left(\dfrac{64}{144}+-\dfrac{45}{144}\right)\\ =-\dfrac{5}{12}.\dfrac{19}{144}\\ =-\dfrac{95}{1728}\\ 2,\left(-0,75+\dfrac{1}{2}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}\right)\\ =\left(-\dfrac{3}{4}+\dfrac{1}{2}\right)\left(\dfrac{4}{12}-\dfrac{3}{12}\right)\\ =\left(-\dfrac{3}{4}+\dfrac{2}{4}\right).\dfrac{1}{12}\\ =-\dfrac{1}{4}.\dfrac{1}{12}\\ =-\dfrac{1}{48}\\ 3,-\dfrac{5}{7}.\dfrac{4}{13}+-\dfrac{5}{7}.\dfrac{9}{13}+-\dfrac{5}{7}\\ =-\dfrac{5}{7}.\dfrac{4}{13}+-\dfrac{5}{7}.\dfrac{9}{13}+-\dfrac{5}{7}.1=-\dfrac{5}{7}\left(\dfrac{4}{13}+\dfrac{9}{13}+1\right)\\ =-\dfrac{5}{7}\left(\dfrac{4}{13}+\dfrac{9}{13}+\dfrac{13}{13}\right)\\ =-\dfrac{5}{7}.\dfrac{26}{13}\\ =-\dfrac{5}{7}.2\\ =-\dfrac{10}{7}\)
`a,-2022.(x+8)=0`
`=>x+8=0:(-2022)`
`=>x+8=0`
`=>x=0-8`
`=>x=-8`
`b,(7-x)(x+3)=0`
`@ TH1`
`7-x=0`
`=>x=7-0`
`=>x=7`
`@ TH2`
`x+3=0`
`=>x=0-3`
`=>x=-3`
`c,2023x .(14-x)=0`
`@ TH1`
`2023x=0`
`=>x=0:2023`
`=>x=0`
`@ TH2`
`14-x=0`
`=>x=14-0`
`=>x=14`
`d,x^2-x=0`
`=>x(x-1)=0`
`@ TH1`
`x=0`
`@ TH2`
`x-1=0`
`=>x=0+1`
`=>x=1`
\(a,\left(-2022\right).\left(x+8\right)=0\\ \Rightarrow\left(x+8\right)=0:\left(-2022\right)\\ \Rightarrow x+8=0\\ \Rightarrow x=-8\\ b,\left(7-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right. \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\\ c,2023x.\left(14-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2023x=0\\14-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\\ d,x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Ta có \(x\inƯ\left(30\right)\)\(\left(ĐKXĐ:x\le8\right)\)
\(< =>x\in\left\{1;2;3;5;6;10;15;30\right\}\)
Do \(x\le8\)suy ra ta có bộ số x thỏa mãn sau :
\(x\in\left\{1;2;3;5;6\right\}\)
\(b,4\dfrac{1}{2}x-\dfrac{1}{2}x=\dfrac{2}{3}:\dfrac{4}{9}\\ \dfrac{9}{2}x-\dfrac{1}{2}x=\dfrac{3}{2}\\ \left(\dfrac{9}{2}-\dfrac{1}{2}\right)x=\dfrac{3}{2}\\ 4x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:4\\ x=\dfrac{3}{8}\)