Tìm x là số nguyên
\(\dfrac{x+46}{20}=x\dfrac{2}{5}\)(\(x\dfrac{2}{5}\)là hỗn số)
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30 tháng 6 2021
Giải:
a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;2\right\}\)
b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow-12.\left(x-6\right)=4.18\)
\(\Rightarrow-12x+72=72\)
\(\Rightarrow-12x=72-72\)
\(\Rightarrow-12x=0\)
\(\Rightarrow x=0:-12\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\)
\(\dfrac{x+46}{20}=\dfrac{2x}{5}\)
\(\Rightarrow5.\left(x+46\right)=2x.20\)
\(\Rightarrow5x+230=40x\)
\(\Rightarrow5x-40x=-230\)
\(\Rightarrow-35x=-230\)
\(\Rightarrow x=-230:-35\)
\(\Rightarrow x=\dfrac{46}{7}\)
Chúc bạn học tốt!
23 tháng 3 2021
Bài 1:
b) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=100\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{13;-7\right\}\)
21 tháng 3 2021
\(x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3-4}{6}=-\dfrac{1}{6}\) là phương án c
29 tháng 7 2021
1) Ta có: \(P=\dfrac{1}{\sqrt{x}-1}-\dfrac{x\sqrt{x}-\sqrt{x}}{x+1}\left(\dfrac{1}{x-2\sqrt{x}+1}+\dfrac{1}{1-x}\right)\)
\(=\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(x-1\right)}{x+1}\cdot\left(\dfrac{\sqrt{x}+1-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x+1}\cdot\dfrac{2}{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{x+1}\)
Để \(P=-\dfrac{2}{5}\) thì \(\dfrac{\sqrt{x}-1}{x+1}=\dfrac{-2}{5}\)
\(\Leftrightarrow-2x-2=5\sqrt{x}-5\)
\(\Leftrightarrow-2x-2-5\sqrt{x}+5=0\)
\(\Leftrightarrow-2x-5\sqrt{x}+3=0\)
\(\Leftrightarrow-2x-6\sqrt{x}+\sqrt{x}+3=0\)
\(\Leftrightarrow-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}+3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(-2\sqrt{x}+1\right)=0\)
\(\Leftrightarrow-2\sqrt{x}+1=0\)
\(\Leftrightarrow-2\sqrt{x}=-1\)
\(\Leftrightarrow x=\dfrac{1}{4}\)(thỏa ĐK)
9 tháng 6 2017
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
=> 2(2x+1) = 6.7
4x+2=42
4x=40
x=10
Vậy x=10
9 tháng 6 2017
a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\\ =>6.7=2.\left(2x+1\right)\\ =>2x+1=\dfrac{6.7}{2}=\dfrac{42}{2}=21\\ =>2x=21-1=20\\ =>x=\dfrac{20}{2}=10\)
b) \(\dfrac{24}{7x-3}=-\dfrac{4}{25}\\ =>24.25=-4.\left(7x-3\right)\\ =>7x-3=\dfrac{24.25}{-4}=-150\\ =>7x=-150+3=-147\\ =>x=\dfrac{-147}{7}=-21\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=-\dfrac{12}{18}\\ =>x-6=\dfrac{4.18}{-12}=-6\\ =>x=-6+6=0\\ y=\dfrac{-12.24}{18}=-16\)
d) \(-\dfrac{1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\\ < =>-\dfrac{8}{40}\le-\dfrac{5x}{40}\le\dfrac{10}{40}\\ =>-8\le-5x\le10\\ Mà:-8< -5.1< -5.0< -5.\left(-1\right)< -5.\left(-2\right)=10\\ =>x\in\left\{-2;-1;0;1\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\\ < =>\dfrac{x+46}{20}=\dfrac{5x+2}{5}\\ =>5\left(x+46\right)=20\left(5x+2\right)\\ < =>5x+230=100x+40\\ < =>230-40=100x-5x\\ < =>190=95x\\ =>x=\dfrac{190}{95}=2\)
f) \(y\dfrac{5}{y}=\dfrac{56}{y}\\ < =>\dfrac{y^2+5}{y}=\dfrac{56}{y}\\ =>y\left(y^2+5\right)=56y\\ =>y^2+5=\dfrac{56y}{y}=56\\ =>y^2=56-5=51\\ =>y=\sqrt{51}\)
MH
26 tháng 1 2022
a) \(\dfrac{x}{2}=\dfrac{2}{x}\)
⇔ \(x^2=4\)
⇒ \(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
b) \(\dfrac{x}{-5}=\dfrac{-5}{x}\)
⇔ \(x^2=25\)
⇒ \(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
NA
26 tháng 1 2022
\(a,\Rightarrow x^2=2^2\\ \Rightarrow x=2\\ b,x^2=\left(-5\right)^2\\ \Rightarrow x=-5\)
NV
1
H9
HT.Phong (9A5)
CTVHS
9 tháng 1
Ta có:
\(\dfrac{x+5}{x-2}=\dfrac{x-2+7}{x-2}=\dfrac{x-2}{x-2}+\dfrac{7}{x-2}=1+\dfrac{7}{x-2}\)
Để \(\dfrac{x+5}{x-2}\) là một số nguyên thì \(\dfrac{7}{x-2}\) phải nguyên
\(\Rightarrow7\) ⋮ \(x-2\)
\(\Rightarrow x-2\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)
\(\Rightarrow x\in\left\{3;1;9;-5\right\}\)
NH
15 tháng 3 2023
\(\dfrac{x-5}{x-2}=\dfrac{x-2-3}{x-2}=1-\dfrac{-3}{x-2}\)
để `(x-5)/(x-2)` là số nguyên thì -3 phải chia hết cho x-2
=> x-2 thuộc ước của -3
ta có bảng sau
| x-2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
vậy \(x\in\left\{3;1;5;-1\right\}\)
BF
15 tháng 3 2023
Ta có: `(x-5)/(x-2) = (x-2-3)/(x-2) = 1 - 3/(x-2)`
Để `(x-5)/(x-2)` là số nguyên thì `3/(x-2) ∈ Z`
`=> x - 2 ∈ Ư(3) = {-3;-1;1;3}`
`=> x∈ {-1;1;3;5}`
Vậy `(x-5)/(x-2)` là số nguyên khi `x ∈ {-1;1;3;5}`
\(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Leftrightarrow5\cdot\left(x+46\right)=20\cdot\left(5x+2\right)\)
\(\Leftrightarrow100x+40-5x-230=0\)
\(\Leftrightarrow95x-190=0\)
\(\Leftrightarrow x=2\)