(1/2.x+3/4)^2:-3/14=-7/24
giúp mik nhá
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= 51/20
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= 3/14
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= 4/7 - 2/7
= 2/7
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= 17/45
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= 2/3
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= 2 x 1/4
= 1/2
a) Ta có: \(\dfrac{7\cdot25}{14\cdot10}\)
\(=\dfrac{7\cdot5\cdot5}{7\cdot2\cdot2\cdot5}\)
\(=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{24\cdot15-14\cdot9}{36\cdot12}\)
\(=\dfrac{9\cdot8\cdot5-14\cdot9}{36\cdot12}\)
\(=\dfrac{9\cdot\left(8\cdot5-14\right)}{9\cdot4\cdot12}\)
\(=\dfrac{40-14}{4\cdot12}\)
\(=\dfrac{13}{24}\)
a) \(\frac{3}{4}\)+ \(\frac{4}{3}\):\(\frac{2}{7}\)- \(\frac{4}{3}\)
=\(\frac{3}{4}\) +\(\frac{4}{3}\) .\(\frac{7}{2}\) -\(\frac{4}{3}\)
=\(\frac{3}{4}\) + \(\frac{14}{3}\) -\(\frac{4}{3}\)
=\(\frac{9}{12}\) +\(\frac{56}{12}\)
=\(\frac{56}{12}\) - \(\frac{4}{3}\)
=\(\frac{56}{12}\)- \(\frac{16}{12}\)
=\(\frac{10}{3}\)
b) \(\frac{5}{2}\) - [\(\frac{1}{2}\) + (\(\frac{-2}{3}\)) .\(\frac{6}{5}\)].\(\frac{2}{5}\)
= \(\frac{5}{2}\) - [\(\frac{1}{2}\) + (\(\frac{-4}{5}\))].\(\frac{2}{5}\)
= \(\frac{5}{2}\) - (\(\frac{5}{10}\) +\(\frac{-8}{10}\)).\(\frac{2}{5}\)
=\(\frac{5}{2}\)-\(\frac{3}{10}\).\(\frac{2}{5}\)
=\(\frac{5}{2}\)-\(\frac{3}{25}\)
= \(\frac{125}{50}\)-\(\frac{6}{50}\)
= \(\frac{119}{50}\)
(Mình không chắc chắn lắm!^^)
a: \(=\dfrac{2\left(x+2\right)\left(x-1\right)}{x+2}=2x-2\)
b: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=x^2-3x+1\)
c: \(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}=x^2-2x+2\)
d: \(=\dfrac{x^2\left(x-3\right)}{x-3}=x^2\)
a: =2/3+1/5*10/7
=2/3+2/7
=14/21+6/21=20/21
b: \(=\dfrac{1}{2}\cdot\dfrac{-3+2}{4}=\dfrac{1}{2}\cdot\dfrac{-1}{4}=\dfrac{-1}{8}\)
c: \(=\dfrac{3}{4}+\dfrac{9}{5}:\dfrac{3}{2}-1\)
=-1/4+9/5*2/3
=-1/4+18/15
=-1/4+6/5
=-5/20+24/20=19/20
d: \(=\dfrac{3}{2}\cdot\left(\dfrac{7}{3}-\dfrac{5}{3}\cdot4\right)\)
\(=\dfrac{7}{2}-\dfrac{5}{2}\cdot4=\dfrac{7}{2}-\dfrac{20}{2}=\dfrac{-13}{2}\)
a. \(\left(x^2-2x+1\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow x^2-2x+1-3x^2+3x=0\)
\(\Leftrightarrow-2x^2+x+1=0\)
\(\Leftrightarrow-2x^2+2x-x+1=0\)
\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow-\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)
Vậy \(x\in\left\{-\frac{1}{2};1\right\}\)
b. \(4\left(7x-3\right)-\left(7x^2-3x\right)=0\)
\(\Leftrightarrow4\left(7x-3\right)-x\left(7x-3\right)=0\)
\(\Leftrightarrow\left(4-x\right)\left(7x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\7x-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{3}{7}\end{cases}}\)
Vậy \(x\in\left\{4;\frac{3}{7}\right\}\)
c.\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
\(\Leftrightarrow\left(5-x\right)\left(2+3x\right)=\left(2-3x\right)\left(2+3x\right)\)
\(\Leftrightarrow\left(2+3x\right)\left(5-x-2+3x\right)=0\)
\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=0\\2x+3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)
Vậy \(x\in\left\{-\frac{2}{3};-\frac{3}{2}\right\}\)
d. \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow7-4+4=-x+2x\)
\(\Leftrightarrow7=x\)
Vậy x = 7
e. \(\left(x-1\right)-\left(2x-1\right)=9\)
\(\Leftrightarrow x-1-2x+1=9\)
\(\Leftrightarrow-x=9\)
\(\Leftrightarrow x=-9\)
g. \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}}\)Mà : \(x^2+1\ge1>0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy x = -1
\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)
\(2\left(x-5\right)+3\left(2-3x\right)=5x+7\)
\(\Leftrightarrow2x-10+6-9x=5x+7\)
\(\Leftrightarrow\left(2x-9x\right)+\left(6-10\right)=5x+7\)
\(\Leftrightarrow-7x-4=5x+7\)
\(\Leftrightarrow-7x-5x=4+7\)
\(\Leftrightarrow-12x=11\)
\(\Leftrightarrow x=\frac{-11}{12}\)
\(3x-5\left(x-2\right)+7=4x-12\)
\(\Leftrightarrow3x-5x-10+7=4x-12\)
\(\Leftrightarrow\left(3x-5x\right)-\left(10-7\right)=4x-12\)
\(\Leftrightarrow-2x-3=4x-12\)
\(\Leftrightarrow-2x-4x=3-12\)
\(\Leftrightarrow-6x=-15\)
\(\Leftrightarrow x=\frac{-15}{-6}=\frac{5}{2}\)
Bài làm:
Ta có: \(\left(\frac{1}{2}x+\frac{3}{4}\right)^2\div\left(-\frac{3}{14}\right)=-\frac{7}{24}\)
\(\Leftrightarrow\left(\frac{1}{2}x+\frac{3}{4}\right)^2=\frac{1}{16}=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x+\frac{3}{4}=\frac{1}{4}\\\frac{1}{2}x+\frac{3}{4}=-\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=-\frac{1}{2}\\\frac{1}{2}x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
(\(\frac{1}{2}\).x+\(\frac{3}{4}\))2: \(\frac{-3}{14}\)=\(\frac{-7}{24}\)
(\(\frac{1}{2}\).x+\(\frac{3}{4}\))2 =\(\frac{1}{16}\)
(\(\frac{1}{2}\).x+\(\frac{3}{4}\))2=(\(\frac{1}{4}\))2
\(\hept{\begin{cases}\frac{1}{2}.x+\frac{3}{4}=\frac{1}{4}\\\frac{1}{2}.x+\frac{3}{4}=\frac{-1}{4}\end{cases}}\)<=>\(\hept{\begin{cases}\frac{1}{2}.x=\frac{1}{4}-\frac{3}{4}\\\frac{1}{2}.x=\frac{-1}{4}-\frac{3}{4}\end{cases}}\)<=>\(\hept{\begin{cases}\frac{1}{2}.x=-\frac{1}{2}\\\frac{1}{2}.x=-1\end{cases}}\)
<=>\(\hept{\begin{cases}x=\frac{-1}{2}:\frac{1}{2}=-1\\x=-1:\frac{1}{2}=-2\end{cases}}\)