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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: \(x=\frac{7}{2}\)
a) \(\left(\frac{11}{12}:\frac{44}{16}\right).\left(\frac{-1}{3}+\frac{1}{2}\right)\) \(=\left(\frac{11}{12}.\frac{16}{44}\right).\left(\frac{-2}{6}+\frac{3}{6}\right)\) \(=\frac{1}{3}.\frac{1}{6}\) \(=\frac{1}{18}\)
b) \(\frac{\left(-5\right)^2.\left(-5\right)^3.16}{5^4.\left(-2\right)^4}\) \(=\frac{\left(-5\right)^5.2^4}{5^4.\left(-2\right)^4}\) \(=5\) (Có sửa đề lại, nếu có sai thì ib mình sửa lại nhé!)
c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\) \(=\frac{15}{2}.\left(\frac{-3}{5}\right)+\frac{5}{2}.\left(\frac{-3}{5}\right)\) \(=\frac{-3}{5}.\left(\frac{15}{2}+\frac{5}{2}\right)\)
\(=\frac{-3}{5}.10\) \(=-6\)
d) \(\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{5}{4}\) \(=\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right).\frac{4}{5}\)
\(=\frac{4}{5}.\left(\frac{-1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\) \(=\frac{4}{5}.\left(\frac{0}{2}+1\right)\) \(=\frac{4}{5}.1=\frac{4}{5}\)
a) (1112:4416).(−13+12)(1112:4416).(−13+12) =(1112.1644).(−26+36)=(1112.1644).(−26+36) =13.16=13.16 =118=118
b) (−5)2.(−5)3.1654.(−2)4(−5)2.(−5)3.1654.(−2)4 =(−5)5.2454.(−2)4=(−5)5.2454.(−2)4 =5=
c) 7,5:(−53)+212:(−53)7,5:(−53)+212:(−53) =152.(−35)+52.(−35)=152.(−35)+52.(−35) =−35.(152+52)=−35.(152+52)
=−35.10=−35.10 =−6=−6
d) (−12+13).45+(23+12):54(−12+13).45+(23+12):54 =(−12+13).45+(23+12).45=(−12+13).45+(23+12).45
=45.(−12+13+23+12)=45.(−12+13+23+12) =45.(02+1)=45.(02+1) =45.1=45
Bài 1 :
\(\left(-2\right)\left(x+1\right)-3\left(1-x\right)=4\)
\(\Leftrightarrow-2x-2-3+3x=4\)
\(\Leftrightarrow x=4+2+3=9\)
Bài 2 :
Cho \(S=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\)
\(\Leftrightarrow S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)
\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\Rightarrow S< \left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)
\(+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)
\(\Leftrightarrow S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)(1)
Lại có :
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)
\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\Leftrightarrow S>\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)
\(+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)
\(\Leftrightarrow S>\frac{10}{40}+\frac{10}{50}+\frac{10}{60}=\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\)(2)
Từ (1) và (2) , ta có :
\(\frac{3}{5}< S< \frac{4}{5}hay\frac{3}{5}< \frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}< \frac{4}{5}\)
a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)
Do đó \(x\in\left\{0;1;2\right\}\)
b)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
a) \(\frac{2}{3}+\frac{1}{3}\cdot\left(-\frac{2}{5}\right)\\ =\frac{2}{3}+\frac{-2}{15}\\ =\frac{10}{15}+\frac{-2}{15}\\ =\frac{8}{15}\)
b) \(0,75\cdot1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\\ =\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}\cdot\frac{-20}{21}\\ =\frac{4}{3}-\frac{-4}{3}\\ =\frac{4}{3}+\frac{4}{3}\\ =\frac{4}{3}\cdot2\\ =\frac{8}{3}\)
c) \(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}-\frac{-4}{19}+\frac{8}{23}\\ =\frac{-2}{17}+\frac{15}{23}+\frac{-15}{17}+\frac{4}{19}+\frac{8}{23}\\ =\left(\frac{-2}{17}+\frac{-15}{17}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\\ =\left(-1\right)+1+\frac{4}{19}\\ =0+\frac{4}{19}\\ =\frac{4}{19}\)
d) \(2019^0\cdot\left(6-2\frac{4}{5}\right)\cdot3\frac{1}{8}-1\frac{3}{5}:25\%\\ =1\cdot\left(\frac{30}{5}-\frac{14}{5}\right)\cdot\frac{25}{8}-\frac{8}{5}:\frac{1}{4}\\ =1\cdot\frac{16}{5}\cdot\frac{25}{8}-\frac{8}{5}\cdot4\\ =\frac{16}{5}\cdot\frac{25}{8}-\frac{32}{5}\\ =\frac{50}{5}-\frac{32}{5}\\ =\frac{18}{5}\)
e) \(\left(\frac{7}{8}-\frac{1}{2}\right)\cdot2\frac{2}{3}-\frac{3}{7}\cdot\left(2,5^2\right)\\ =\left(\frac{7}{8}-\frac{4}{8}\right)\cdot\frac{8}{3}-\frac{3}{7}\cdot6,25\\ =\frac{3}{8}\cdot\frac{8}{3}-\frac{3}{7}\cdot\frac{25}{4}\\ =1-\frac{75}{28}\\ =\frac{28}{28}-\frac{75}{28}\\ =\frac{-47}{28}\)
a, \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-2}{5}\right)\)
= \(\frac{2}{3}+\frac{-2}{15}=\frac{8}{15}\)
b, \(0,75.1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\)
= \(\frac{3}{4}.\frac{16}{9}-\frac{7}{5}.\frac{-20}{21}\)
= \(\frac{4}{3}-\left(\frac{-4}{3}\right)=\frac{8}{3}\)
c, \(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}+\frac{4}{19}+\frac{8}{23}\)
= \(\left(\frac{-2}{17}+\frac{-15}{17}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\)
= \(\left(-1\right)+1+\frac{4}{19}=0+\frac{4}{19}=\frac{4}{19}\)
d, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}-1\frac{3}{5}:25\%\)
=> \(\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}:25\%\)
= \(\frac{16}{5}.\frac{25}{8}-\frac{8}{5}.25:100\)
= 10 - 0,4 = 9,6
e, \(\left(\frac{7}{8}-\frac{1}{2}\right).2\frac{2}{3}-\frac{3}{7}.\left(2,5^2\right)\)
=> \(\frac{3}{8}.\frac{8}{3}-\frac{3}{7}.6,25\)
= \(1-\frac{75}{28}=\frac{-47}{28}\)
4a) \(\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\)
\(\Leftrightarrow x=\frac{1}{10}:\frac{-2}{3}=\frac{1}{10}.\frac{3}{-2}=\frac{3}{-20}\)
Vậy x=\(\frac{3}{-20}\)
b) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\Leftrightarrow\left(\frac{2}{3}-\frac{3}{2}\right)x=\frac{5}{12}\)
\(\Leftrightarrow\frac{-5}{6}x=\frac{5}{12}\)
\(\Leftrightarrow x=\frac{5}{12}:\frac{-5}{6}=\frac{5}{12}.\frac{6}{-5}=\frac{1}{-2}\)
Vậy x=\(\frac{1}{-2}\)
g)Sửa đề: \(\left|4x-1\right|=\left(-3\right)^2\)
\(\Leftrightarrow\left|4x-1\right|=9\)
\(\Rightarrow\left[{}\begin{matrix}4x-1=9\\4x-1=\left(-9\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{2};-2\right\}\)
i) \(\left(x-1^3\right)=125\)
\(\Leftrightarrow x-1=125\)
\(\Leftrightarrow x=125+1=126\)
Vậy x=126
k) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
xin lỗi câu h tui xin chữa lại là:\(|x+70\%|=2\frac{1}{5}\)