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PK
1
2 tháng 11 2021
Bài 4:
\(a,\Rightarrow5⋮x\Rightarrow x\inƯ\left(5\right)=\left\{1;5\right\}\\ b,\Rightarrow x-2+7⋮x-2\\ \Rightarrow x-2\inƯ\left(7\right)=\left\{1;7\right\}\\ \Rightarrow x\in\left\{3;9\right\}\\ c,\Rightarrow3\left(x+1\right)+4⋮x+1\\ \Rightarrow x+1\inƯ\left(4\right)=\left\{1;2;4\right\}\\ \Rightarrow x\in\left\{0;1;3\right\}\\ d,\Rightarrow10x+6⋮2x-1\\ \Rightarrow5\left(2x-1\right)+11⋮2x-1\\ \Rightarrow2x-1\inƯ\left(11\right)=\left\{1;11\right\}\\ \Rightarrow x\in\left\{1;6\right\}\\ e,\Rightarrow x\left(x+3\right)+11⋮x+3\\ \Rightarrow x+3\inƯ\left(11\right)=\left\{1;11\right\}\\ \Rightarrow x=8\left(x\in N\right)\\ f,\Rightarrow x\left(x+3\right)+2\left(x+3\right)+5⋮x+3\\ \Rightarrow x+3\inƯ\left(5\right)=\left\{1;5\right\}\\ \Rightarrow x=2\left(x\in N\right)\)
6 tháng 1 2022
Câu 2:
1: \(\Leftrightarrow x\cdot\dfrac{7}{2}=\dfrac{9}{2}+3=\dfrac{15}{2}\)
hay x=15/7
2: \(\Leftrightarrow x=\dfrac{5}{2}\cdot\dfrac{8}{5}=4\)
3: \(\Leftrightarrow x=\dfrac{-11\cdot10}{5}=-11\cdot2=-22\)
4: =>2x=90
hay x=45
giúp mình bài 4 và bài 5 với ạ
B
11 tháng 1 2023
`a, 18(x-16)=-36`
`=>x-16=-36:18`
`=>x-16=-2`
`=>x=-2+16`
`=>x= 14`
`b,(-12-x):(-5)=3`
`=>-12-x=3 . (-5)`
`=>-12-x=-15`
`=>x= -12-(-15)`
`=>x=-12+15`
`=>x= 3`
`c,11-(-53+x)=97`
`=>-53+x=11-97`
`=>-53+x=-86`
`=>x=-86-(-53)`
`=>x=-86+53`
`=>x= -33`
`d,-9.x+(-7).x=-48`
`=> [-9+(-7) ].x=-48`
`=>-16.x=-48`
`=>x=-48:(-16)`
`=>x= 3`
`e,(2x-1)^3=-27`
`=> (2x-1)^3=-3^3`
`=> 2x-1=-3`
`=>2x=-3+1`
`=>2x=-2`
`=>x=-2:2`
`=>x=-1`
bn tách ra nhé!
Y
11 tháng 1 2023
Bài 5 :
\(a,5x-16=40+x\)
\(\Leftrightarrow5x-x=40+16\)
\(\Leftrightarrow4x=56\)
\(\Leftrightarrow x=14\)
\(b,-41+2x-\left(-24\right)=4x-23-x\)
\(\Leftrightarrow-41+2x+24=4x-23-x\)
\(\Leftrightarrow2x-4x+x=41-24-23\)
\(\Leftrightarrow-x=-6\)
\(\Leftrightarrow x=6\)
\(c,2\left(x-5\right)-\left(x+6\right)=\left(-7\right)^2\)
\(\Leftrightarrow2x-10-x-6=49\)
\(\Leftrightarrow x=65\)
\(d,120-4\left(1-x\right)=106-3x\)
\(\Leftrightarrow120-4+4x=106-3x\)
\(\Leftrightarrow4x+3x=106-120+4\)
\(\Leftrightarrow7x=-10\)
\(\Leftrightarrow x=-\dfrac{10}{7}\)
giúp em bài 5 ạ
10 tháng 3 2022
Bài 2:
a: =>10(3x-1)=110
=>3x-1=11
hay x=4
b: =>720:(21-2x+5)=(3+5)x5=40
=>-2x+26=18
=>-2x=-8
hay x=4
19 tháng 4 2021
(1-1/2)×(1-1/3)×(1-1/4)×....×(1-1/100)
=(2/2-1/2)x(3/3-1/3)x...x(100/100-1/100)
=1/2x2/3x...x99/100
=1/100
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
5 tháng 5 2022
Giả sử
\(n^2+13n+44⋮49\Rightarrow n^2+13n+44⋮7\)
Ta có
\(n^2+13n+30+14⋮7\Rightarrow n^2+13n+30⋮7\)
\(n^2+13n+30=\left(n+3\right)\left(n+10\right)⋮7\)
\(\Rightarrow n+3⋮7\) hoặc \(n+10⋮7\) Giả sử \(n+10⋮7\)
\(\Rightarrow\left(n+10\right)-\left(n+3\right)=7⋮7\Rightarrow n+3⋮7\)
\(\Rightarrow\left(n+3\right)\left(n+10\right)⋮49\Rightarrow n^2+13n+30⋮49\)
Ta đã giả sử
\(n^2+13n+30+14⋮49\Rightarrow14⋮49\) => vô lý vậy \(n^2+13n+44\) không chia hết cho 49
27 tháng 2 2022
\(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{19.20}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{3}-\dfrac{1}{20}< \dfrac{1}{3}\)
\(B=\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{19.20}=2\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{19.20}\right)=2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)=2\left(\dfrac{1}{3}-\dfrac{1}{20}\right)=\dfrac{2}{3}-\dfrac{2}{20}< \dfrac{2}{3}\)
\(C=\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{18.20}=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{18.20}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{18}-\dfrac{1}{20}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{4}-\dfrac{1}{40}< \dfrac{1}{4}\)
\(D=\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{18.20}=\dfrac{1}{2}\left(\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{18.20}\right)=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{18}-\dfrac{1}{20}\right)=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{20}\right)=\dfrac{1}{8}-\dfrac{1}{40}< \dfrac{1}{8}\)
Bài 5:
AB<AC
nên \(\widehat{C}< \widehat{B}\)
=>\(\widehat{ADB}< \widehat{ADC}\)
AB<AC
nên ˆC<ˆBC^<B^
=>ˆADB<ˆADC