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A
1
26 tháng 8 2021
h: Ta có: \(\dfrac{5}{x+3}=\dfrac{x+3}{5}\)
\(\Leftrightarrow\left(x+3\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=-5\\x+3=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=2\end{matrix}\right.\)
TL
1
10 tháng 1 2022
e: Để 4n+1/3n-1 là số nguyên thì \(12n+3⋮3n-1\)
\(\Leftrightarrow3n-1\in\left\{1;-1;7;-7\right\}\)
hay \(n\in\left\{0;-2\right\}\)
3 tháng 1 2022
b: \(\Leftrightarrow n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;-2;1;-3;3;-5\right\}\)
c: \(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-1;-3;3;-7\right\}\)
d: \(\Leftrightarrow n+2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{-1;-3;0;-4;2;-6\right\}\)
a: \(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{2;0;6;-4\right\}\)
TL
1
3 tháng 1 2022
a: =100x54-100x(-6)
=100x60
=6000
b: =99(123-56+66-123)=990
c: =547x(1+103-4)=54700
d: =-76x10=-760
TL
1
2 tháng 1 2022
Bài 4:
a: \(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{2;0;6;-4\right\}\)
TL
1
10 tháng 1 2022
a: \(-\dfrac{6}{13}=-\dfrac{12}{26}=\dfrac{-18}{39}=-\dfrac{24}{52}=\dfrac{-30}{65}=\dfrac{-36}{78}=\dfrac{-42}{91}\)
b: \(\dfrac{15}{-7}=\dfrac{-15}{7}=\dfrac{-30}{14}=\dfrac{-45}{21}=\dfrac{-60}{28}=\dfrac{-75}{35}=-\dfrac{90}{42}\)
TL
1
24 tháng 8 2021
a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\Rightarrow x=-\dfrac{5}{8}+\dfrac{3}{4}\Rightarrow x=\dfrac{1}{8}\)
b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\Rightarrow x=-\dfrac{1}{4}-\dfrac{5}{8}\Rightarrow x=-\dfrac{7}{8}\)
c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\Rightarrow x=\left(\dfrac{5}{24}-\dfrac{5}{6}\right):\dfrac{3}{4}\Rightarrow x=-\dfrac{5}{6}\)
d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{3}{8}+\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\Rightarrow x=\dfrac{2}{3}:\dfrac{19}{24}=\dfrac{16}{19}\)
24 tháng 8 2021
a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\\ \Rightarrow x=\dfrac{1}{8}\)
b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{7}{8}\)
c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{5}{8}\\ \Rightarrow x=-\dfrac{5}{6}\)
d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\\ \Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\\ \Rightarrow x=\dfrac{16}{19}\)
e) \(\left(6,5-2x\right):\dfrac{5}{13}=\dfrac{13}{10}\\ \Rightarrow6,5-2x=\dfrac{1}{2}\\ \Rightarrow2x=6\\ \Rightarrow x=3\)
f) \(\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|-\dfrac{3}{4}=-\dfrac{1}{6}\\ \Rightarrow\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|=\dfrac{7}{12}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{2}=\dfrac{7}{12}\\\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{7}{12}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{13}{4}\end{matrix}\right.\)
g) \(\dfrac{x-3}{3}=\dfrac{2x+3}{5}\\ \Rightarrow5x-15=6x+9\\ \Rightarrow-x=24\\ \Rightarrow x=-24\)
h) \(\dfrac{x-5}{6}=\dfrac{6}{x-5}\\ \Rightarrow\left(x-5\right)^2=6^2\\ \Rightarrow\left[{}\begin{matrix}x-5=-6\\x-5=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=11\end{matrix}\right.\)
10 tháng 1 2022
Bài 4:
a) Áp dụng t/c dtsbn:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x+y}{4+3}=\dfrac{14}{7}=2\Rightarrow\left\{{}\begin{matrix}x=2.4=8\\y=2.3=6\end{matrix}\right.\)
Vậy....
b) Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{2x}{16}=\dfrac{3y}{36}=\dfrac{2x+3y}{16+36}=\dfrac{13}{52}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}.8=2\\y=\dfrac{1}{4}.12=3\end{matrix}\right.\)(nhận)
Vậy...
Bài 4:
a) Ta có: \(\widehat{yOz}+\widehat{xOy}=180^0\)(2 góc kề bù)
\(\Rightarrow\widehat{yOz}=180^0-\widehat{xOy}=180^0-50^0=130^0\)
b) Ta có: \(\widehat{zOt}=\widehat{yOt}=\dfrac{1}{2}\widehat{yOz}=\dfrac{1}{2}.130^0=65^0\)(do Ot là tia phân giác \(\widehat{yOz}\))
c) Ta có: \(\widehat{xOt}=\widehat{yOt}+\widehat{xOy}=65^0+50^0=115^0\)
Bài 5:
a) Ta có: \(\widehat{xOz}+\widehat{xOy}=180^0\)(2 góc kề bù)
\(\Rightarrow\widehat{xOz}=180^0-\widehat{xOy}=180^0-110^0=70^0\)
b) Ta có: \(\widehat{zOt}=\dfrac{1}{2}\widehat{xOz}=\dfrac{1}{2}.70^0=35^0\)( Ot là tia phân giác \(\widehat{xOz}\))
c) Ta có: \(\widehat{xOt}=\widehat{zOt}=35^0\)( Ot là tia phân giác \(\widehat{xOz}\))
Bài 4:
a: Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)
\(\Leftrightarrow\widehat{yOz}=180^0-50^0\)
\(\Leftrightarrow\widehat{yOz}=130^0\)
b: \(\widehat{zOt}=\dfrac{\widehat{yOz}}{2}=65^0\)