B= \(\dfrac{2x-11}{x-5}\)
Tìm x để B là số nguyên
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\(\dfrac{47}{3}=15\dfrac{2}{3}\)
\(\dfrac{105}{100}=1\dfrac{5}{100}\)
\(\dfrac{47}{3}=\dfrac{45+2}{3}=15+\dfrac{2}{3}=15\dfrac{2}{3}\)
\(\dfrac{105}{100}=1+\dfrac{5}{100}=1\dfrac{5}{100}\)
xy+3x-7y=21
=>x(y+3)-7y-21=0
=>(y+3)(x-7)=0
=>\(\left\{{}\begin{matrix}x-7=0\\y+3=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=7\\y=-3\end{matrix}\right.\)
đặt 1+3+...+3^2024 là A
3A=3+3^2+...+3^2025
3A-A=(3+3^2+...+3^2025)-(1+3+...+3^2024)
2A=3^2025-1
A=3^2025-1/2
7: \(\dfrac{x}{7}=\dfrac{-6}{21}\)
=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)
=>x=-2
8: \(\dfrac{x}{15}=\dfrac{3}{5}+\dfrac{-2}{3}\)
=>\(\dfrac{x}{15}=\dfrac{9}{15}-\dfrac{10}{15}=-\dfrac{1}{15}\)
=>x=-1
9: \(-\dfrac{2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
=>\(-\dfrac{2}{3}-\dfrac{2}{3}x+\dfrac{5}{3}=\dfrac{3}{2}\)
=>\(-\dfrac{2}{3}x+1=\dfrac{3}{2}\)
=>\(-\dfrac{2}{3}x=\dfrac{3}{2}-1=\dfrac{1}{2}\)
=>\(x=\dfrac{-1}{2}:\dfrac{2}{3}=\dfrac{-3}{4}\)
10: \(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
=>\(\left(2x+\dfrac{3}{5}+\dfrac{3}{5}\right)\left(2x+\dfrac{3}{5}-\dfrac{3}{5}\right)=0\)
=>\(2x\left(2x+\dfrac{6}{5}\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{6}{5}:2=-\dfrac{3}{5}\end{matrix}\right.\)
\(75\%\cdot x-2\dfrac{1}{2}=1\dfrac{2}{5}\)
=>\(x\cdot0,75=1,4+2,5=3,9\)
=>\(x=3,9:0,75=5,2\)
\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{7}{180}\)
=>\(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{7}{60}\)
=>\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{7}{60}\)
=>\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{7}{60}\)
=>\(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{7}{60}=\dfrac{12}{60}-\dfrac{7}{60}=\dfrac{5}{60}=\dfrac{1}{12}\)
=>x+3=12
=>x=9
1\(\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(\dfrac{28}{25}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)
= \(\dfrac{118}{100}.\dfrac{25}{100}.3+\left(-\dfrac{47}{60}\right):\dfrac{47}{24}\)
= \(\dfrac{177}{200}+-\dfrac{2}{5}\)
= \(\dfrac{177}{200}+-\dfrac{80}{200}=\dfrac{97}{200}\)
ĐKXĐ: x<>5
Để B là số nguyên thì \(2x-11⋮x-5\)
=>\(2x-10-1⋮x-5\)
=>\(-1⋮x-5\)
=>\(x-5\in\left\{1;-1\right\}\)
=>\(x\in\left\{6;4\right\}\)