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![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) Ta có:
\(\left(9x+5y\right)⋮17\Rightarrow4\left(9x+5y\right)⋮17\)
Hay \(\left(36x+20y\right)⋮17.\) Mà \(34x⋮17;17y⋮17\)
\(\Rightarrow36x+20y-34x-17y=\left(2x+3y\right)⋮17\)
Vậy \(\left(2x+3y\right)⋮17\Leftrightarrow\left(9x+5y\right)⋮17\) (Đpcm)
b) Ta có:
\(\left(a+4b\right)⋮13\Rightarrow10\left(a+4b\right)⋮13\)
\(\Rightarrow10a+40b=\left(10a+b+39b\right)⋮13\)
Mà \(39b⋮13\Rightarrow\left(10a+b\right)⋮13\) (Đpcm)
c) Ta có:
\(\left(10a+b\right)⋮17\Rightarrow2\left(10a+b\right)⋮17\)
Hay \(\left(20a+2b\right)⋮17.\) Mà \(\left(3a+2b\right)⋮17\)
\(\Rightarrow\left(20a+2b\right)-\left(3a+2b\right)=17a⋮17\)
\(\Rightarrow\) Đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>3x-15-2x+6=51
=>x-9=51
=>x=60
b: =>432-2x+22+3x-15=200
=>x+439=200
=>x=-239
c: =>3x+17-4x+13=0
=>30-x=0
=>x=30
d: =>x-3-3x+6=-11-21=-32
=>-2x+3=-32
=>-2x=-35
=>x=35/2
\(3\cdot\left(x-5\right)-2\cdot\left(x-3\right)=51\)
\(\Rightarrow3x-15-2x+6=51\)
\(\Rightarrow x-9=51\)
\(\Rightarrow x=51+9\)
\(\Rightarrow x=60\)
b) \(432-2\cdot\left(x-11\right)+3\cdot\left(x-5\right)=200\)
\(\Rightarrow432-2x+22+3x-15=200\)
\(\Rightarrow x+439=200\)
\(\Rightarrow x=200-439\)
\(\Rightarrow x=-239\)
c) \(3x+17=4x-13\)
\(\Rightarrow4x-3x=17+13\)
\(\Rightarrow x=30\)
d) \(\left(x-3\right)+21=3\cdot\left(x-2\right)-11\)
\(\Rightarrow x-3+21=3x-6-11\)
\(\Rightarrow x+18=3x-17\)
\(\Rightarrow3x-x=18+17\)
\(\Rightarrow2x=35\)
\(\Rightarrow x=\dfrac{35}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.4./x-5/-54=-34
4./x-5/=-34+54
4./x-5/=20
/x-5/=20:4
/x-5/=5
=>x-5=5 va x-5=-5
TH1:x-5=-5
=>x=0
TH2:x-5=5
=>x=10
2. /x-1/+(-5)=2
/x-1/=2-(-5)
/x-1/=7
=>x-1=7 va x-1=-7
TH1: x-1=7
=>x=8
TH2:x-1=-7
=>x=-6
vay x=-6 hoac x=8
![](https://rs.olm.vn/images/avt/0.png?1311)
17⋮x; 21⋮x và 51⋮x \(\Rightarrow\) x \(\in\) ƯC(17,21,51)
Vì 17 là SNtố \(\Rightarrow\) ƯCLN(17,21,51) = 1
\(\Rightarrow\) x \(\in\) Ư(1) = {1; -1}
Giải:
\(\left[{}\begin{matrix}17⋮x\\21⋮x\\51⋮x\end{matrix}\right.\) \(\Rightarrow x\inƯC\left(17;21;51\right)\)
\(17=17\)
\(21=3.7\)
\(51=3.17\)
\(\RightarrowƯCLN\left(17;21;51\right)=1\)
\(\RightarrowƯC\left(17;21;51\right)=Ư\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow x\in\left\{\pm1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Thực hiện phép tính bằng cách hợp lí :
a) (-46) + (-125) + 46 + 25 = [(-46)+46] + [(-125)+25]
= 0+(-100) = -100
b) 25.(-15) + 25.(-5) + (-20).75 = 25.[(-15)+(-5)] + (-20).75
= 25.(-20) + (-20).75 = (-20).(25+75) = (-20).100 = -2000
c) (-151)+(-37)+(-42)+(-63)+142 =(-151)+[(-37)+(-63)]+[(-42)+142]
= (-151) + [(-100) + 100] = -151
d)32+(-149)+(-311)+(-89)+(-51) = 32+[(-149)+(-51)] + [(-311)+(-89)]
= 32+[(-200)+(-400)] = 32+(-600) = -568
e)-65.(87-17)-87.(17-65) = (-65).87 - (-65).17 - 87.17 + 87.65
= (-65).87 + 65.17 - 87.17 + 87.65 = [(-65).87+87.65] + 65.(17-87)
= 65.(-70) = -4550
g) -43.(53-16) - 53.(16-43) = (-43).53 - (-43).16 - 53.16 + 53.43
= (-43).53 + 43.16 - 53.16 + 53.43 = [(-43).53+53.43] + 16.(43-53)
= 16.(-10) = -160
236 - x = 3.2 - 2.3 + 51 : 17
236 - x = 6 - 6 + 3
236 - x = 3
x = 236 - 3
x = 233
~ HT ~
Tìm số tự nhiên xx thỏa mãn: 236 - x = 3.2^4- 2.3^2 + 51 : 17236−x=3.24−2.32+51:17.