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15 tháng 1 2022
Bài 2:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
P
1 tháng 5 2018
\(-\dfrac{1}{3}< \dfrac{A}{36}< \dfrac{B}{18}< -\dfrac{1}{4}\)
<=>\(-\dfrac{12}{36}< \dfrac{A}{36}< \dfrac{2B}{36}< -\dfrac{9}{36}\)
<=> -12 < x + 1 < 2(2 - y) < -9
<=> -12 < x + 1 < 4 - 2y < -9
=> x + 1 = -11 => x = -12
4 - 2y = -10 => y = 7
Vậy (x; y) = (-12; 7)
1 tháng 5 2018
−13<A36<B18<−14−13<A36<B18<−14
<=>−1236<A36<2B36<−936−1236<A36<2B36<−936
<=> -12 < x + 1 < 2(2 - y) < -9
<=> -12 < x + 1 < 4 - 2y < -9
=> x + 1 = -11 => x = -12
4 - 2y = -10 => y = 7
Vậy (x; y) = (-12; 7)
9 tháng 4 2018
4,
a,\(\dfrac{x-1}{9}\)=\(\dfrac{8}{3}\)
[x- 1].3=9.8
[x- 1].3=72
x-1=72:3
x-1=24
x=24+1
x=25
22 tháng 2 2022
a: =>2x-1=-2
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)
c: x/8=9/4
nên x/8=18/8
hay x=18
d: \(\Leftrightarrow\left(x-3\right)^2=36\)
=>x-3=6 hoặc x-3=-6
=>x=9 hoặc x=-3
e: =>-1,7x=6,12
hay x=-3,6
h: =>x-3,4=27,6
hay x=31
R
22 tháng 2 2022
a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)
\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)
\(\dfrac{1}{3}=-2x+1\div6\)
\(x=-\dfrac{1}{2}\)
b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(TH1:3x+2=0\)
\(3x=0-2\)
\(3x=-2\)
\(x=\dfrac{-2}{3}\)
\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)
\(\left(\dfrac{-2}{5}x-7\right)=0\)
\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)
\(\left(\dfrac{-2x-35}{5}\right)=0\)
\(-2x-35=0\)
\(-2x=0+35\)
\(x=-\dfrac{35}{2}\)
c) \(\dfrac{x}{8}=\dfrac{9}{4}\)
\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)
\(x=18\)
d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(x-3=18+2\)
\(x=20-3\)
\(x=17\)
e) \(4,5x-6,2x=6,12\)
\(\dfrac{9x}{2}-6,2.x=6,12\)
\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)
\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)
\(\dfrac{45x-62x}{10}=6.12\)
\(=-17x\div10=6.12\)
\(-17x=10.6.12\)
\(x=-3,6\)
h) \(11,4-\left(x-3,4\right)=-16,2\)
\(x-3,4=-16,2+11,4\)
\(x-3,4=-4,8\)
\(x=-1,4\)
26 tháng 3 2021
Bài 2:
a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)
\(=\dfrac{4+6-3}{n-1}\)
\(=\dfrac{7}{n-1}\)
Để A là số tự nhiên thì \(7⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(7\right)\)
\(\Leftrightarrow n-1\in\left\{1;7\right\}\)
hay \(n\in\left\{2;8\right\}\)
Vậy: \(n\in\left\{2;8\right\}\)
27 tháng 3 2021
ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2 Để B là STN thì 4n+10⋮n+2 4n+8+2⋮n+2 4n+8⋮n+2 ⇒2⋮n+2 n+2∈Ư(2) Ư(2)={1;2} Vậy n=0
6 tháng 3 2017
b, \(\dfrac{x-3}{4}=\dfrac{15}{20}\)
<=> \(\dfrac{x-3}{4}=\dfrac{3}{4}\)
=> x-3=3
<=> x=6
Vậy x=6
TT
9 tháng 6 2017
\(a,\dfrac{x}{15}=\dfrac{4}{y}=\dfrac{-2}{5}\)
* \(\dfrac{x}{15}=\dfrac{-2}{5}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{-6}{15}\)
\(\Rightarrow x=-6\)
*\(\dfrac{4}{y}=\dfrac{-2}{5}\)
\(\Rightarrow\dfrac{4}{y}=\dfrac{4}{-10}\)
\(\Rightarrow y=-10\)
Vậy x = - 6 ; y = - 10
\(b,\dfrac{x-3}{4}=\dfrac{15}{20}\)
=> ( x - 3 ) . 20 = 4. 15
=> 20x - 60 = 60
=> 20x = 60 + 60
=> 20x = 120
=> x = 120 : 20
=> x = 6
Vậy x = 6
\(c,\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{22}{-9}+\dfrac{-7}{15}< x\le\dfrac{-1}{3}+\dfrac{-1}{4}+\dfrac{-5}{12}\)
\(\Rightarrow\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{-22}{9}+\dfrac{-7}{15}< x\le\dfrac{-4}{12}+\dfrac{-3}{12}+\dfrac{-5}{12}\)
\(\Rightarrow\left(\dfrac{-5}{9}+\dfrac{-22}{9}\right)+\left(\dfrac{-8}{15}+\dfrac{-7}{15}\right)< x\le-1\)
\(\Rightarrow-3+\left(-1\right)< x\le-1\)
\(\Rightarrow-4< x\le-1\)
\(\Rightarrow x=-3;-2;-1\)
\(\dfrac{a}{9}-\dfrac{3}{b}=\dfrac{1}{18}\)
⇔ \(\dfrac{2a-1}{18}=\dfrac{3}{b}\)
⇒ \(\left(2a-1\right).b=18.3\)
⇔ \(\left(2a-1\right).b=54\)
Ta thấy \(2a-1\) là 1 số nguyên lẻ. Ta có các trường hợp sau:
TH1: \(\left\{{}\begin{matrix}2a-1=1\\b=54\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a=1\\b=54\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}2a-1=3\\b=18\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a=2\\b=18\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}2a-1=9\\b=6\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a=5\\b=6\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}2a-1=27\\b=2\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a=14\\b=2\end{matrix}\right.\)
TH5: \(\left\{{}\begin{matrix}2a-1=-1\\b=-54\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a=0\\b=-54\end{matrix}\right.\)
TH6: \(\left\{{}\begin{matrix}2a-1=-3\\b=-18\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a=-1\\b=-18\end{matrix}\right.\)
TH7: \(\left\{{}\begin{matrix}2a-1=-9\\b=-6\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a=-4\\b=-6\end{matrix}\right.\)
TH8: \(\left\{{}\begin{matrix}2a-1=-27\\b=-2\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}a=-13\\b=-2\end{matrix}\right.\)
Vậy \(\left(a,b\right)\in\left\{\left(1;54\right);\left(2;18\right);\left(5;6\right);\left(14;2\right);\left(0;-54\right);\left(-1;-18\right);\left(-4;-6\right);\left(-13;-2\right)\right\}\)
lâu ngày k lm dạng này, k bt có đúng k nx. Nếu có gì sai sót xin thứ lỗi