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làm mẫu một bài thôi nha
3n+2=3.(n-1)+5
hay 3(n-1)+5 phải chia hết cho n-1, mà 3(n-1) chia hết cho n-1, vậy 5 phải chia hết cho n-1, U(5)=1;5 =>n=2 hoặc n=6
Ta có:\(\dfrac{x}{-12}=\dfrac{-3}{x}\)
\(\Rightarrow x.x=-3.\left(-12\right)\)
\(x^2=36\)
Vì \(x\in Z\)\(\Rightarrow x=\pm6\)
bài 4 dễ mà , bạn làm xong rồi gửi cho mik , đễ mik xem có đúng k nhé
Bài 1 :
a) \(x+2\dfrac{3}{4}=5\dfrac{2}{3}\)
\(x+\dfrac{11}{4}=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}-\dfrac{11}{4}\)
\(x=\dfrac{35}{12}\)
Vậy .........................
b) \(x.3\dfrac{1}{2}=4\dfrac{3}{4}\)
\(x.\dfrac{7}{2}=\dfrac{19}{4}\)
\(x=\dfrac{19}{4}:\dfrac{7}{2}\)
\(x=\dfrac{19}{14}\)
Vậy .................
c) \(x:3\dfrac{1}{2}=4\dfrac{3}{4}\)
\(x:\dfrac{7}{2}=\dfrac{19}{4}\)
\(x=\dfrac{19}{4}.\dfrac{7}{2}\)
\(x=\dfrac{133}{8}\)
Vậy ...................
e) \(x-\dfrac{3}{4}=6.\dfrac{3}{8}\)
\(x-\dfrac{3}{4}=\dfrac{9}{4}\)
\(x=\dfrac{9}{4}+\dfrac{3}{4}\)
\(x=3\)
Vậy .............
f) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)
\(\dfrac{7}{8}:x=\dfrac{5}{2}\)
\(x=\dfrac{7}{8}:\dfrac{5}{2}\)
\(x=\dfrac{7}{20}\)
Vậy ................
g) \(x+\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{3}{4}\)
\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{1}{6}\)
\(x=\dfrac{7}{12}\)
Vậy .................
h) \(x+17,67=100-63,2\)
\(x+17,67=36,8\)
\(x=36,8-17,67\)
\(x=19,13\)
Vậy ................
i) \(x:0,01=10\)
\(x=10.0,01\)
\(x=0,1\)
Vậy ...............
k) \(8,01-x=1,99\)
\(x=8,01-1,99\)
\(x=6,02\)
Vậy ............
l) \(x.0,5=2,2\)
\(x=2,2:0,5\)
\(x=4,4\)
Vậy ............
m) \(x:7,5=3,7+4,1\)
\(x:7,5=7,8\)
\(x=7,8.7,5\)
\(x=58,5\)
Vậy ............
Giải:
4.Theo đề bài ta có:
\(A=7.a+4 \)
\(=17.b+3 \)
\(=23.c+11 (a,b,c ∈ N)\)
Nếu ta thêm 150 vào số đã cho thì ta lần lượt có:
\(A+150=7.a+4+150=7.a+7.22=7.(a+22)\)
\(=17.b+3+150=17.b+17.9=17.(b+9)\)
\(=23.c+11+150=23.c+23.7=23.(c+7) \)
\(\Rightarrow A+150⋮7;17;23\).Nhưng 7, 17 và 23 là ba số đôi một nguyên tố cùng nhau, suy ra \(A+150⋮7.17.13=2737\)
Vậy \(A+150=2737k\left(k=1;2;3;4;...\right)\)
Suy ra: \(A=2737k-150=2737k-2737+2587=2737(k-1)+2587=2737k+2587\)
Do \(2587<2737\)
\(\Rightarrow A\div2737\) dư \(2587\)
a)\(n+7⋮n+2\)
\(\Rightarrow\left(n+2\right)+5⋮n+2\)
\(\Rightarrow5⋮n+2\)
\(\Rightarrow n+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
| n+2 | 1 | -1 | 5 | -5 |
| n | -1 | -3 | 3 | -7 |
Vậy \(n\in\left\{-1;-3;3;-7\right\}\)
b)\(9-n⋮n-3\)
\(\Rightarrow6-\left(n-3\right)\)
\(\Rightarrow6⋮n-3\)
\(\Rightarrow n-3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
nếu n-3=1 thì n=4
nếu n-3=-1 thì n=2
nếu n-3=2 thì n=5
nếu n-3=-2 thì n=1
nếu n-3=3 thì n=6
nếu n-3=-3 thì n=0
nếu n-3=6 thì n=9
nếu n-3=-6 thì n=-3
Vậy \(n\in\left\{4;2;5;1;6;0;9;-3\right\}\)
c)\(n^2+n+17⋮n+1\)
\(\Rightarrow n\left(n+1\right)+17⋮n+1\)
\(\Rightarrow17⋮n+1\)
\(\Rightarrow n+1\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
nếu n+1=1 thì n=0
nếu n+1=-1 thì n=-2
nếu n+1=17 thì n=16
nếu n+1=-17 thì n=-18
Vậy \(n\in\left\{0;-2;16;-18\right\}\)
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
A=4+22+23+....+220
2A=8+23+24+...+221
=> A+2A-A = (8+23+24+...+221) - (4+22+23+....+220)
=>A=221+8 - (22+4)=221
=>A là 1 lũy thừa của 2
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