= 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) \(\dfrac{7.25}{14.10}\)

= \(\dfrac{7.5.5}{7.2.5.2}\)

= \(\dfrac{5}{4}\)

c) \(\dfrac{1^3.3^2}{6^3.5}\)

= \(\dfrac{3^2}{\left(2.3\right)^3.5}\)

= \(\dfrac{1}{2^3.3.5}\)

= \(\dfrac{1}{120}\)

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: \(=\dfrac{x^2\left(x-3\right)}{x-3}=x^2\)

`@` `\text {Ans}`

`\downarrow`

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: \(=\dfrac{3}{2}\cdot\left(\dfrac{7}{3}-\dfrac{5}{3}\cdot4\right)\)

\(=\dfrac{7}{2}-\dfrac{5}{2}\cdot4=\dfrac{7}{2}-\dfrac{20}{2}=\dfrac{-13}{2}\)

phá ngoặc ra mà tính đi bn ơi

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

\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)

đùa thì kệ con mẹ mày xàm loz

Đáp án a là 0,75 

Đáp án b: là 0.86222222222

\(2\left(x-5\right)+3\left(2-3x\right)=5x+7\)

\(\Leftrightarrow2x-10+6-9x=5x+7\)

\(\Leftrightarrow\left(2x-9x\right)+\left(6-10\right)=5x+7\)

\(\Leftrightarrow-7x-4=5x+7\)

\(\Leftrightarrow-7x-5x=4+7\)

\(\Leftrightarrow-12x=11\)

\(\Leftrightarrow x=\frac{-11}{12}\)

\(3x-5\left(x-2\right)+7=4x-12\)

\(\Leftrightarrow3x-5x-10+7=4x-12\)

\(\Leftrightarrow\left(3x-5x\right)-\left(10-7\right)=4x-12\)

\(\Leftrightarrow-2x-3=4x-12\)

\(\Leftrightarrow-2x-4x=3-12\)

\(\Leftrightarrow-6x=-15\)

\(\Leftrightarrow x=\frac{-15}{-6}=\frac{5}{2}\)

\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}=\frac{101}{1540}.3\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Leftrightarrow\frac{5}{5\left(x+3\right)}=\frac{1}{5}-\frac{303}{1540}\)

\(\Leftrightarrow\frac{5}{5x+15}=\frac{308}{1540}-\frac{303}{1540}\)

\(\Leftrightarrow\frac{5}{5x+15}=\frac{5}{1540}\)

\(\Leftrightarrow5x+15=1540\)

\(\Leftrightarrow5x=1540-15\)

\(\Leftrightarrow5x=1525\)

\(\Leftrightarrow x=1525\div5\)

\(\Leftrightarrow x=305\)