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a; \(\dfrac{x-1}{12}\) = \(\dfrac{5}{3}\)
\(x-1\) = \(\dfrac{5}{3}\) \(\times\) 12
\(x\) - 1 = 20
\(x\) = 20 + 1
\(x\) = 21
b; \(\dfrac{-x}{8}\) = \(\dfrac{-50}{x}\)
-\(x\).\(x\) = -50.8
-\(x^2\) = -400
\(x^2\) = 400
\(\left[{}\begin{matrix}x=-20\\x=20\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-20; 20}
c; \(\dfrac{x}{3}\) = \(\dfrac{14}{x+1}\)
\(x\).(\(x\)+1) = 14.3
\(x^2\) + \(x\) = 42
\(x^2\) + \(x\) - 42 = 0
\(x^2\) - 6\(x\) + 7\(x\) - 42 = 0
\(x\).(\(x\) - 6) + 7.(\(x\) - 6) = 0
(\(x\) - 6).(\(x\) + 7) = 0
\(\left[{}\begin{matrix}x-6=0\\x+7=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6\\x=-7\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-7; 6}
d; \(x-\dfrac{2}{9}\) = \(\dfrac{1}{6}\)
\(x\) = \(\dfrac{1}{6}\) + \(\dfrac{2}{9}\)
\(x\) = \(\dfrac{7}{18}\)
Vậy \(x\) = \(\dfrac{7}{18}\)
Bài 4:
a; \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) = \(\dfrac{5}{20}\) - \(\dfrac{4}{20}\) = \(\dfrac{1}{20}\)
b; \(\dfrac{3}{5}\) - \(\dfrac{-1}{2}\) = \(\dfrac{6}{10}\) + \(\dfrac{5}{10}\) = \(\dfrac{11}{10}\)
c; \(\dfrac{3}{5}\) - \(\dfrac{-1}{3}\) = \(\dfrac{9}{15}\) + \(\dfrac{5}{15}\) = \(\dfrac{14}{15}\)
d; \(\dfrac{-5}{7}\) - \(\dfrac{1}{3}\)= \(\dfrac{-15}{21}\) - \(\dfrac{7}{21}\)= \(\dfrac{-22}{21}\)
Bài 5
a; 1 + \(\dfrac{3}{4}\) = \(\dfrac{4}{4}\) + \(\dfrac{3}{4}\) = \(\dfrac{7}{4}\) b; 1 - \(\dfrac{1}{2}\) = \(\dfrac{2}{2}\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\)
c; \(\dfrac{1}{5}\) - 2 = \(\dfrac{1}{5}\) - \(\dfrac{10}{5}\) = \(\dfrac{-9}{5}\) d; -5 - \(\dfrac{1}{6}\) = \(\dfrac{-30}{6}\) - \(\dfrac{1}{6}\) = \(\dfrac{-31}{6}\)
e; - 3 - \(\dfrac{2}{7}\)= \(\dfrac{-21}{7}\) - \(\dfrac{2}{7}\)= \(\dfrac{-23}{7}\) f; - 3 + \(\dfrac{2}{5}\) = \(\dfrac{-15}{5}\) + \(\dfrac{2}{5}\)= - \(\dfrac{13}{5}\)
g; - 3 - \(\dfrac{2}{3}\) = \(\dfrac{-9}{3}\) - \(\dfrac{2}{3}\) = \(\dfrac{-11}{3}\) h; - 4 - \(\dfrac{-5}{7}\) = \(\dfrac{-28}{7}\)+ \(\dfrac{5}{7}\) = - \(\dfrac{23}{7}\)
Có quá nhiều bài, thứ nhất em đăng tách ra, thứ hai chụp gần cận cho rõ, thứ ba em chỉ đăng bài cần giúp
\(a,MSC:180\\ Có:-5=\dfrac{-5.180}{180}=\dfrac{-900}{180};\dfrac{17}{-20}=\dfrac{17.\left(-9\right)}{\left(-9\right).\left(-20\right)}=\dfrac{-153}{180};\dfrac{-16}{9}=\dfrac{-16.20}{9.20}=\dfrac{-320}{180}\\ ---\\ b.MSC:75\\ Có:\dfrac{13}{-15}=\dfrac{13.\left(-5\right)}{\left(-15\right).\left(-5\right)}=\dfrac{-65}{75};\dfrac{-18}{25}=\dfrac{-18.3}{25.3}=\dfrac{-54}{75};-3=\dfrac{-3.75}{75}=\dfrac{-225}{75}\)
Bài 5:
a. Gọi $d=ƯCLN(n-2, n+1)$
$\Rightarrow n-2\vdots d; n+1\vdots d$
$\Rightarrow (n+1)-(n-2)\vdots d$
$\Rightarrow 3\vdots d\Rightarrow d\in \left\{1; 3\right\}$
Để ps tối giản thì $n-2\not\vdots 3$
$\Leftrightarrow n\neq 3k+2$ với $k$ là số tự nhiên bất kỳ.
b.
Gọi $d=ƯCLN(n+5, n-2)$
$\Rightarrow n+5\vdots d; n-2\vdots d$
$\Rightarrow (n+5)-(n-2)\vdots d$
$\Rightarrow 7\vdots d$
$\Rightarrow d\in \left\{1; 7\right\}$
Để ps tối giản thì $n-2\not\vdots 7$
$\Rightarrow n\neq 7k+2$ với $k$ là số tự nhiên bất kỳ.
Phân số | Đọc | Tử Số | Mẫu số |
\(\dfrac{5}{7}\) | Năm phần bẩy | 5 | 7 |
\(\dfrac{-6}{11}\) | âm sáu phần mười một | -6 | 11 |
\(\dfrac{-2}{13}\) | âm hai phần ba | -2 | 13 |
\(\dfrac{9}{-11}\) | chín phần âm mười một | 9 | -11 |
a: \(0,25:\left(\dfrac{1}{2}x\right)-\dfrac{2}{5}=\dfrac{3}{4}\)
=>\(0,25:\left(\dfrac{1}{2}x\right)=\dfrac{3}{4}+\dfrac{2}{5}=\dfrac{23}{20}\)
=>\(\dfrac{1}{2}x=\dfrac{1}{4}:\dfrac{23}{20}=\dfrac{1}{4}\cdot\dfrac{20}{23}=\dfrac{5}{23}\)
=>\(x=\dfrac{5}{23}:\dfrac{1}{2}=\dfrac{10}{23}\)
b: \(\dfrac{-3}{4}-\dfrac{3}{8}x=\dfrac{4}{5}+\dfrac{-2}{3}\)
=>\(-\dfrac{3}{4}-\dfrac{3}{8}x=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\)
=>\(\dfrac{3}{8}x=-\dfrac{3}{4}-\dfrac{2}{15}=\dfrac{-45-8}{60}=\dfrac{-53}{60}\)
=>\(x=-\dfrac{53}{60}:\dfrac{3}{8}=\dfrac{-53}{60}\cdot\dfrac{8}{3}=\dfrac{-106}{45}\)
c: \(\dfrac{2}{3}x-\dfrac{5}{7}=\dfrac{2}{7}\cdot\dfrac{11}{5}\)
=>\(\dfrac{2}{3}x-\dfrac{5}{7}=\dfrac{22}{35}\)
=>\(\dfrac{2}{3}x=\dfrac{22}{35}+\dfrac{5}{7}=\dfrac{47}{35}\)
=>\(x=\dfrac{47}{35}:\dfrac{2}{3}=\dfrac{47}{35}\cdot\dfrac{3}{2}=\dfrac{141}{70}\)
d:
ĐKXĐ: x<>-1/2
\(\dfrac{2x}{3}=\dfrac{10}{2x+1}\)
=>\(2x\left(2x+1\right)=10\cdot3=30\)
=>\(x\left(2x+1\right)=15\)
=>\(2x^2+x-15=0\)
=>\(2x^2+6x-5x-15=0\)
=>(x+3)(2x-5)=0
=>\(\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=\dfrac{5}{2}\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: x<>0
\(\dfrac{11}{8}+\dfrac{13}{6}=\dfrac{85}{2x}\)
=>\(\dfrac{85}{2x}=\dfrac{11\cdot3+13\cdot4}{24}\)
=>\(\dfrac{85}{2x}=\dfrac{85}{24}\)
=>2x=24
=>x=12(nhận)
f: \(1,4x-\dfrac{2}{11}=1\dfrac{1}{5}\)
=>\(1,4x-\dfrac{2}{11}=\dfrac{6}{5}\)
=>\(1,4x=\dfrac{6}{5}+\dfrac{2}{11}=\dfrac{66+10}{55}=\dfrac{76}{55}\)
=>\(x=\dfrac{76}{55}:1,4=\dfrac{76}{77}\)
g: \(\dfrac{2}{15}:\left(1,24x\right)=\dfrac{3}{5}+\dfrac{-2}{3}\)
=>\(\dfrac{2}{15}:\left(1,24x\right)=\dfrac{9-10}{15}\)
=>\(\dfrac{2}{15}:\left(1,24x\right)=\dfrac{-1}{15}\)
=>\(1,24x=-2\)
=>\(x=-\dfrac{2}{1,24}=\dfrac{-200}{124}=\dfrac{-50}{31}\)
h: \(\dfrac{3x}{91}=\dfrac{-3}{14}\cdot\dfrac{10}{13}\)
=>\(\dfrac{3x}{91}=\dfrac{-3\cdot5}{7\cdot13}=\dfrac{-15}{91}\)
=>3x=-15
=>x=-5