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a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
a: \(\Leftrightarrow x^2=\dfrac{-5}{2}\cdot\dfrac{-10}{9}=\dfrac{50}{18}=\dfrac{25}{9}\)
=>x=5/3hoặc x=-5/3
c: \(\Leftrightarrow4\left(x-\dfrac{5}{8}\right)=\dfrac{1}{4}+\dfrac{3}{4}=1\)
=>x-5/8=1/4
hay x=2/8+5/8=7/8
d: \(\Leftrightarrow\left|x-3\right|=\dfrac{2}{5}+\dfrac{3}{5}=1\)
=>x-3=1 hoặc x-3=-1
=>x=4 hoặc x=2
e: =>1-1/2x=-3
=>1/2x=4
hay x=8
9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
a) \(\dfrac{3}{7}x-2\dfrac{1}{3}=0,5\)
\(\Leftrightarrow\dfrac{3}{7}x-\dfrac{7}{3}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3}{7}x=\dfrac{1}{2}+\dfrac{7}{3}\)
\(\Leftrightarrow\dfrac{3}{7}x=\dfrac{17}{6}\)
\(\Leftrightarrow x=\dfrac{17}{6}:\dfrac{3}{7}\)
\(\Leftrightarrow x=\dfrac{119}{18}\)
b) \(\dfrac{4}{7}-\dfrac{2}{3}:x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4}{7}-\dfrac{2}{3}:x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{4}{7}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{-19}{28}\)
\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{-19}{28}\)
\(\Leftrightarrow x=\dfrac{56}{-57}\)
c) \(\left(\dfrac{2}{3}x+2\dfrac{1}{4}\right):3\dfrac{1}{5}=0,75\)
\(\Leftrightarrow\left(\dfrac{2}{3}x+\dfrac{9}{4}\right):\dfrac{16}{6}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{9}{4}=\dfrac{3}{4}.\dfrac{16}{6}\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{9}{4}=2\)
\(\Leftrightarrow\dfrac{2}{3}x=2-\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{-1}{4}\)
\(\Leftrightarrow x=\dfrac{-1}{4}:\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{-3}{8}\)
d) \(\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}-\dfrac{2}{3}=1\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}=1+\dfrac{2}{3}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|:\dfrac{1}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|=\dfrac{5}{3}.\dfrac{1}{4}\)
\(\Leftrightarrow\left|\dfrac{4}{5}-\dfrac{2}{3}x\right|=\dfrac{5}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{4}{5}-\dfrac{2}{3}x=\dfrac{5}{12}\\\dfrac{4}{5}-\dfrac{2}{3}x=-\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{5}-\dfrac{5}{12}\\\dfrac{2}{3}x=\dfrac{4}{5}-\left(-\dfrac{5}{12}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{23}{60}\\\dfrac{2}{3}x=\dfrac{73}{60}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{60}:\dfrac{2}{3}\\x=\dfrac{73}{60}:\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{20}\\x=\dfrac{73}{20}\end{matrix}\right.\)
a)\(\dfrac{3}{7}x-2\dfrac{1}{3}=0,5\)
\(\dfrac{3}{7}x=0,5+2\dfrac{1}{3}\)
\(\dfrac{3}{7}x=\dfrac{17}{6}\)
\(x=\dfrac{17}{6}:\dfrac{3}{7}\)
\(x=\dfrac{119}{18}\)
b)\(\dfrac{4}{7}-\dfrac{2}{3}:x=1\dfrac{1}{4}\)
\(\dfrac{2}{3}:x=\dfrac{4}{7}-1\dfrac{1}{4}\)
\(x=\dfrac{2}{3}:\left(-\dfrac{19}{28}\right)\)
\(x=-\dfrac{56}{57}\)
c)\(\left(\dfrac{2}{3}x+2\dfrac{1}{4}\right):3\dfrac{1}{5}=0,75\)
\(\dfrac{2}{3}x+2\dfrac{1}{4}=0,75:3\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{15}{64}-2\dfrac{1}{4}\)
\(x=-\dfrac{129}{64}:\dfrac{2}{3}\)
\(x=-\dfrac{387}{128}\)
b) \(\dfrac{4}{5}-\dfrac{3}{4}:x=0,3\)
\(\Rightarrow0,8-0,75:x=0,3\)
\(\Rightarrow0,75:x=0,5\)
\(\Rightarrow x=1,5\)
c) \(\dfrac{-3}{2}-\dfrac{1}{4}x=1\dfrac{1}{3}-0,2x\)
\(\Rightarrow\dfrac{-3}{2}-\dfrac{4}{3}=\dfrac{1}{4}x-\dfrac{1}{5}x\)
\(\Rightarrow x=\dfrac{-17}{6}\cdot20\)
\(\Rightarrow x=\dfrac{-170}{3}\)
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!
6. \(\dfrac{x}{4}=\dfrac{9}{x}\)
=>x2=4.9=36
=>x\(\in\)\(\left\{-6;6\right\}\)
\((\dfrac{2x}{5}+2):\left(-4\right)=-1\dfrac{1}{2}\)
(\(\dfrac{2x}{5}+2):\left(-4\right)=-\dfrac{3}{2}\)
\(\dfrac{2x}{5}=-\dfrac{3}{2}.\left(-4\right)\)
\(\dfrac{2x}{5}=6\)
\(\dfrac{2x}{5}=\dfrac{30}{5}\)
2x = 30
x = 30 : 2 = 15
tớ sẽ ko làm cách chuyển vế đâu nha!
mong thông cảm
câu 1:
x = -3/4 - 25/100
x = -1
câu 2:
x = -25/10 . 13/3
x = -65/6