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a) \(3^4.\left(-4\right)^2=3^4.4^2=3^4.2^4=\left(3.2\right)^4=6^4\)
b) \(=1\frac{3}{5}+\left(6+\frac{5}{11}-4-\frac{5}{11}\right).\frac{1}{5}=1+\frac{3}{5}+2.\frac{1}{5}=2\)
c)\(=4.\left(0,75-0,25\right)=4.0,5=2\)
d) \(=\frac{\frac{1}{3}+\frac{3}{4}+\frac{3}{2}}{\frac{1.4}{3.4}+\frac{3.3}{4.3}+\frac{3.6}{2.6}}=\frac{\frac{1}{3}+\frac{3}{4}+\frac{3}{2}}{\frac{1}{3}+\frac{3}{4}+\frac{3}{2}}=1\)
a) \(\left(6\dfrac{1}{9}+3\dfrac{7}{11}\right)-1\dfrac{1}{9}\)
\(=\left(\left(6+3\right)+\left(\dfrac{1}{9}+\dfrac{7}{11}\right)\right)-\dfrac{10}{9}\)
\(=\left(9+\dfrac{74}{99}\right)-\dfrac{10}{9}\)
\(=9\dfrac{74}{99}-\dfrac{10}{9}\)
\(=\dfrac{965}{99}-\dfrac{10}{9}\)
\(=\dfrac{95}{11}\)
b) \(1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:5\%\)
\(=1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:\dfrac{1}{20}\)
\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+3\cdot20\)
\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+60\)
\(=\dfrac{1275}{22}\)
c) \(4\dfrac{3}{4}-0,37+\dfrac{1}{8}-1,28-2,5+3\cdot\dfrac{1}{2}\)
\(=\dfrac{19}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{3}{2}\)
\(=\dfrac{89}{40}\)
d) \(\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)
\(=\dfrac{5}{11}+\dfrac{10}{77}-5\cdot\dfrac{2}{11}\)
\(=\dfrac{5}{11}+\dfrac{10}{77}-\dfrac{10}{11}\)
\(=-\dfrac{25}{77}\)
e) \(\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
(câu này chờ mình một chút)
Câu e) anh rút thừa số chung là 2 cùng mũ số ra ngoài thì phân số thành số nguyên. Em nghĩ thế!
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
Bài 1:
a) Ta có: \(\frac{5}{6}-\frac{2}{3}+\frac{1}{4}\)
\(=\frac{10}{12}-\frac{8}{12}+\frac{3}{12}\)
\(=\frac{2+3}{12}=\frac{5}{12}\)
b) Ta có: \(1\frac{11}{12}-\frac{5}{12}\cdot\left(\frac{4}{5}-\frac{1}{10}\right):\frac{-5}{12}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\left(\frac{8}{10}-\frac{1}{10}\right)\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\frac{7}{10}\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{-7}{10}\)
\(=\frac{115}{60}+\frac{42}{60}=\frac{157}{60}\)
Bài 2:
a) Ta có: \(\frac{1}{2}\cdot x-\frac{2}{5}=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
b) Ta có: \(\left(1-2x\right)\cdot\frac{4}{3}=\left(-2\right)^3\)
\(\Leftrightarrow\left(1-2x\right)\cdot\frac{4}{3}=-8\)
\(\Leftrightarrow1-2x=-8:\frac{4}{3}=-8\cdot\frac{3}{4}=-6\)
\(\Leftrightarrow-2x=-6-1=-7\)
hay \(x=\frac{7}{2}\)
Vậy: \(x=\frac{7}{2}\)
\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{2}{4}\)
\(=\frac{3}{4}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)
\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)
\(=\frac{25}{20}-\frac{-1}{20}\)
\(=\frac{26}{20}\)
\(=\frac{13}{10}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
Vậy x=\(\frac{20}{27}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=1-\frac{4}{5}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=\frac{1}{5}\)
\(\frac{9}{11}-x=\frac{1}{5}\cdot\frac{-10}{11}\)
\(\frac{9}{11}-x=\frac{-2}{11}\)
\(x=\frac{9}{11}-\frac{-2}{11}\)
\(x=1\)
Vậy x=1
\(\frac{-11}{12}\cdot x+\frac{3}{4}=\frac{-1}{6}\)
\(\frac{-11}{12}\cdot x=\frac{-1}{6}-\frac{3}{4}\)
\(\frac{-11}{12}\cdot x=\frac{21}{12}\)
\(x=\frac{-21}{11}\)
Vậy x=\(\frac{-21}{11}\)
\(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)
\(\frac{3}{2}+x=\frac{-5}{4}-\frac{9}{2}\)
\(\frac{3}{2}+x=\frac{23}{4}\)
\(x=\frac{17}{4}\)
Vậy x=\(\frac{17}{4}\)
\(\left(\frac{3}{4}-x:\frac{2}{15}\right)\cdot\frac{1}{5}=-2,6\)
\(\frac{3}{4}-x:\frac{2}{15}=\frac{-13}{5}:\frac{1}{5}\)
\(\frac{3}{4}-x:\frac{2}{15}=-13\)
\(x:\frac{2}{15}=\frac{3}{4}-\left(-13\right)\)
\(x:\frac{2}{15}=\frac{45}{4}\)
\(x=\frac{3}{2}\)
Vậy x=\(\frac{3}{2}\)
\(3-\left(\frac{1}{6}-x\right)\cdot\frac{2}{3}=\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=\frac{2}{3}:\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=1\)
\(\frac{1}{6}-x=2\)
\(x=\frac{1}{6}-2\)
\(x=\frac{-11}{6}\)
Vậy x=\(\frac{-11}{6}\)
\(\left(1-2x\right)\cdot\frac{4}{5}=\left(-2\right)^3\)
\(1-2x=\frac{-1}{10}\)
\(2x=1-\frac{-1}{10}\)
\(2x=\frac{11}{10}\)
\(x=\frac{11}{20}\)
Vậy x=\(\frac{11}{20}\)
\(\frac{1}{6}-\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{8}\)
\(\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{7}{12}\) \(\frac{1}{2}x-\frac{1}{3}=\frac{-7}{12}\)
\(\frac{1}{2}x=\frac{11}{12}\) \(\frac{1}{2}x=\frac{-1}{4}\)
\(x=\frac{11}{6}\) \(x=\frac{-1}{2}\)
Vậy \(x\in\left\{\frac{11}{6};\frac{-1}{2}\right\}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{6}{9}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
tk mình đi mình làm nốt cho hjhj ^^
\(6\frac{5}{12}:2\frac{3}{4}+11\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{5}\right)=\frac{77}{12}:\frac{11}{4}+\frac{45}{4}.\frac{2}{15}=\frac{7}{3}+\frac{3}{2}=\frac{23}{6}\)
<=>77/12:11/3+45/4(2/15)
=7/4+3/2=13/4