lm từng bước giúp mình nhé, cảm ơn
lm từng bước giúp mình nhé, cảm ơn
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
lm từng bước giúp mình nhé, cảm ơn
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
lm từng bước giúp mình nhé, cảm ơn
\(p,71\cdot70+35\cdot\left(-7\right)-13\cdot35\)
\(=4970+\left(-245\right)-455=4725-455=4270\)
\(q,18\cdot\left(23-17\right)-13\cdot\left(23+18\right)\)
\(=18\cdot6-13\cdot41=108-533=-425\)
lm từng bước giúp mình nhé, cảm ơn
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
lm từng bước giúp mình nhé, cảm ơn
\(1,-\dfrac{5}{12}\left(\dfrac{4}{9}+-\dfrac{5}{16}\right)\\ =-\dfrac{5}{12}\left(\dfrac{64}{144}+-\dfrac{45}{144}\right)\\ =-\dfrac{5}{12}.\dfrac{19}{144}\\ =-\dfrac{95}{1728}\\ 2,\left(-0,75+\dfrac{1}{2}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}\right)\\ =\left(-\dfrac{3}{4}+\dfrac{1}{2}\right)\left(\dfrac{4}{12}-\dfrac{3}{12}\right)\\ =\left(-\dfrac{3}{4}+\dfrac{2}{4}\right).\dfrac{1}{12}\\ =-\dfrac{1}{4}.\dfrac{1}{12}\\ =-\dfrac{1}{48}\\ 3,-\dfrac{5}{7}.\dfrac{4}{13}+-\dfrac{5}{7}.\dfrac{9}{13}+-\dfrac{5}{7}\\ =-\dfrac{5}{7}.\dfrac{4}{13}+-\dfrac{5}{7}.\dfrac{9}{13}+-\dfrac{5}{7}.1=-\dfrac{5}{7}\left(\dfrac{4}{13}+\dfrac{9}{13}+1\right)\\ =-\dfrac{5}{7}\left(\dfrac{4}{13}+\dfrac{9}{13}+\dfrac{13}{13}\right)\\ =-\dfrac{5}{7}.\dfrac{26}{13}\\ =-\dfrac{5}{7}.2\\ =-\dfrac{10}{7}\)
lm từng bước giúp mình nhé, cảm ơn
\(b,4\dfrac{1}{2}x-\dfrac{1}{2}x=\dfrac{2}{3}:\dfrac{4}{9}\\ \dfrac{9}{2}x-\dfrac{1}{2}x=\dfrac{3}{2}\\ \left(\dfrac{9}{2}-\dfrac{1}{2}\right)x=\dfrac{3}{2}\\ 4x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:4\\ x=\dfrac{3}{8}\)
lm từng bước giúp mk nhé, cảm ơn
a) \(-x+80=-220-5x\)
\(\Rightarrow-x+5x=-220-80\)
\(\Rightarrow4x=-300\)
\(\Rightarrow x=-\dfrac{300}{4}\)
\(\Rightarrow x=-75\)
b) \(98+\left(x-12\right)+68=-80-x\)
\(\Rightarrow166+\left(x-12\right)=-80-x\)
\(\Rightarrow166+x-12=-80-x\)
\(\Rightarrow154+x=-80-x\)
\(\Rightarrow154+80=-x-x\)
\(\Rightarrow-2x=234\)
\(\Rightarrow x=-\dfrac{234}{2}\)
\(\Rightarrow x=-117\)
c) \(122+x-78=-55-4x-1\)
\(\Rightarrow44+x=-56-4x\)
\(\Rightarrow x+4x=-56-44\)
\(\Rightarrow5x=-100\)
\(\Rightarrow x=-\dfrac{100}{5}\)
\(\Rightarrow x=-20\)
d) \(663+9x=-x-37\)
\(\Rightarrow9x+x=-37-663\)
\(\Rightarrow10x=-700\)
\(\Rightarrow x=-\dfrac{700}{10}\)
\(\Rightarrow x=-70\)
lm từng bước giúp mk nhé, cảm ơn
a) \(\left(2x-4\right)\left(x-2\right)^3=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\\left(x-2\right)^3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\x-2-0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)
\(\Rightarrow x=2\)
b) \(3^{x-1}:81=3^3\)
\(\Rightarrow3^{x-1}:3^4=3^3\)
\(\Rightarrow3^{x-1-4}=3^3\)
\(\Rightarrow3^{x-5}=3^3\)
\(\Rightarrow x-5=3\)
\(\Rightarrow x=8\)
c) \(x^{13}=x\)
\(\Rightarrow x^{13}-x=0\)
\(\Rightarrow x\left(x^{12}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{12}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{12}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d) \(\left(x-9\right)^4=\left(x-9\right)^2\)
\(\Rightarrow\left(x-9\right)^2=x-9\)
\(\Rightarrow\left(x-9\right)^2-\left(x-9\right)=0\)
\(\Rightarrow\left(x-9\right)\left(x-9-1\right)=0\)
\(\Rightarrow\left(x-9\right)\left(x-10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x-10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)
lm từng bước ra cho mình nhé, cảm ơn
`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`
\(a,\left(-2022\right).\left(x+8\right)=0\\ \Rightarrow\left(x+8\right)=0:\left(-2022\right)\\ \Rightarrow x+8=0\\ \Rightarrow x=-8\\ b,\left(7-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right. \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\\ c,2023x.\left(14-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2023x=0\\14-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\\ d,x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(\left(-2020\right).\left(x+8\right)=0\)
\(=>x+8=0=>x=-8\)
\(\left(7-x\right).\left(x+3\right)=0\)
\(=>\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
\(2023x.\left(14-x\right)=0\)
\(=>\left[{}\begin{matrix}x=0\\14-x=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\)
\(x^2-x=0\)
\(x\left(x-1\right)=0\)
\(=>\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)