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\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
Bài 1:
a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)
\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)
\(=\frac{29}{35}\)
Bài 2:
a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)
Câu 2 :
a) \(\frac{1}{4}\) + \(\frac{1}{3}\): (3x) = -5
⇒ \(\frac{1}{3}\): (3x) = -5 - \(\frac{1}{4}\)
⇒ \(\frac{1}{3}\): (3x) = \(\frac{-21}{4}\)
⇒ 3x = \(\frac{1}{3}\): \(\frac{-21}{4}\)
⇒ 3x = \(\frac{-4}{63}\)
⇒ x = \(\frac{-4}{63}\):3
⇒ x = \(\frac{-4}{189}\)
Vậy x = \(\frac{-4}{189}\)
b) (3x-4) . (5x+15)=0
⇒ xảy ra 2 trường hợp 3x-4=0 ; 5x+15=0
* 3x-4 =0
⇒ 3x =0+4
⇒ 3x =4
⇒ x =4:3
⇒ Vô lý không tính được bạn nhé
* 5x+15 = 0
⇒ 5x = 0-15
⇒ 5x = -15
⇒ x = -15:5
⇒ x = -3
Vậy x ∈ ∅ và x ∈ 3
c) |2x-1| = \(\frac{11}{2}\)
⇒ xảy ra 2 trường hợp 2x-1 = \(\frac{11}{2}\); 2x-1 = \(\frac{-11}{2}\)
* 2x-1=\(\frac{11}{2}\)
⇒ 2x = \(\frac{11}{2}\)+1
⇒ 2x = \(\frac{9}{2}\)
⇒ x = \(\frac{9}{2}\):2
⇒ x = \(\frac{9}{4}\)
* 2x-1 = \(\frac{-11}{2}\)
⇒ 2x = 1 + \(\frac{-11}{2}\)
⇒ 2x = \(\frac{-9}{2}\)
⇒ x = \(\frac{-9}{2}\):2
⇒ x = \(\frac{-9}{4}\)
Vậy x ∈ { \(\frac{9}{4}\); \(\frac{-9}{4}\)}
Câu 2:
a) \(\frac{1}{4}+\frac{1}{3}:\left(3x\right)=-5\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{1}{3x}=-5-\frac{1}{4}=-\frac{21}{4}\)
\(\Leftrightarrow\frac{1}{9x}=\frac{-21}{4}\)
\(\Leftrightarrow9x=\frac{4\cdot1}{-21}=-\frac{4}{21}\)
hay \(x=-\frac{4}{21}:9=-\frac{4}{189}\)
Vậy: \(x=-\frac{4}{189}\)
b) Ta có: \(\left(3x-4\right)\left(5x+15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-4=0\\5x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=4\\5x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{4}{3};-3\right\}\)
c) Ta có: \(\left|2x-1\right|=\frac{11}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{11}{2}\\2x-1=-\frac{11}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{11}{2}+1=\frac{13}{2}\\2x=-\frac{11}{2}+1=-\frac{9}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{2}:2=\frac{13}{4}\\x=-\frac{9}{2}:2=-\frac{9}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{13}{4};\frac{-9}{4}\right\}\)
Câu 3:
a) Ta có: \(1-3\cdot\left[4-30:\left(-18+3\right)\right]\)
\(=1-3\cdot\left[4-30:\left(-15\right)\right]\)
\(=1-3\cdot\left[4-\left(-2\right)\right]\)
\(=1-3\cdot6=1-18=-17\)
b) Ta có: \(\frac{5\cdot7+5\cdot\left(-4\right)}{21\cdot5}\)
\(=\frac{5\cdot\left(7-4\right)}{5\cdot21}=\frac{3}{21}=\frac{1}{7}\)
c) Ta có: \(\frac{-2}{9}+\frac{5}{4}+\left(-\frac{1}{6}\right):\frac{3}{5}+\frac{1}{18}\)
\(=-\frac{2}{9}+\frac{5}{4}-\frac{5}{18}+\frac{1}{18}\)
\(=-\frac{2}{9}+\frac{5}{4}-\frac{4}{18}\)
\(=\frac{-8}{36}+\frac{45}{36}-\frac{8}{36}=\frac{29}{36}\)
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)
hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)
Vậy: \(x=\frac{9}{10}\)
b) Ta có: \(5\frac{4}{7}:x=13\)
\(\Leftrightarrow\frac{39}{7}:x=13\)
\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)
Vậy: \(x=\frac{3}{7}\)
c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)
\(\Leftrightarrow x\cdot\frac{14}{5}=84\)
\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)
Vậy: x=30
d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\)
hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)
Vậy: x=-5
e) Ta có: \(8\frac{2}{3}:x-10=-8\)
\(\Leftrightarrow\frac{26}{3}:x=2\)
hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)
Vậy: \(x=\frac{13}{3}\)
g) Ta có: \(x+30\%=-1.3\)
\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)
hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)
Vậy: \(x=\frac{-8}{5}\)
i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)
\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)
\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)
\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)
Vậy: x=-9
k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)
\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)
hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)
Vậy: x=30
m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)
\(\Leftrightarrow\left|2x-1\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)
