a) C1 : 2/5 x 3/7 + 2/7 x 4/7

         = 2/5 x (3/7 + 4/7)

        = 2/5 x 1

        = 2/5

C2 : 2/5 x 3/7 + 2/5 x 4/7

= 6/35 + 8/35

= 2/5 

b) (2/3 - 4/7) x 5/5

= 2/21 x 5/5

= 2/21

C2 : (2/3 - 4/7) x 5/5

= 2/3 x 1 - 4/7 x 1

= 2/3 - 4/7

= 2/21

c) 3/4 x 2/5 - 3/4 x 2/7

= 3/4 x (2/5 - 2/7)

= 3/4 x 4/35

= 3/35

C2 : 3/4 x 2/5 - 3/4 x 2/7

= 3/10 - 3/14

= 3/35

a, 

Cách 1

\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)

\(=\frac{6}{35}+\frac{8}{35}\)

\(=\frac{2}{5}\)

Cách 2 :

\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)

\(=\frac{2}{5}\times\left(\frac{3}{7}+\frac{4}{7}\right)\)

\(=\frac{2}{5}\times1\)

\(=\frac{2}{5}\)

b,

Cách 1 :

\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)

\(=\frac{2}{21}\times1\)

\(=\frac{2}{21}\)

Cách 2 :

\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)

\(=\frac{2}{3}\times\frac{5}{5}-\frac{4}{7}\times\frac{5}{5}\)

\(=\frac{2}{3}-\frac{4}{7}\)

\(=\frac{2}{21}\)

c, 

Cách 1 :

\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)

\(=\frac{3}{10}-\frac{3}{14}\)

\(=\frac{3}{35}\)

Cách 2 :

\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)

\(=\frac{3}{4}\times\left(\frac{2}{5}-\frac{2}{7}\right)\)

\(=\frac{3}{4}\times\frac{4}{35}\)

\(=\frac{3}{35}\)

Mk nhanh nhất đó

Đúng 100%

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Cảm ơn bạn nhiều

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Tớ chưa học đến

ai lm nhanh tui tick cho

chỉ 2 bn đầu tiên lm xong

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

à

kh hiểu bn ơi

`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`

a) Cách 1 : (3/4 + 1/2) x 5/7

               = 5/4 x 5/7

                = 25/28

Cách 2 : (3/4 + 1/2) x 5/7

           = 3/4 x 5/7 + 1/2 x 5/7

            = 15/28 + 5/14

            = 25/28

b) Cách 1 : 5/7 x 13/21 + 2/7 x 13/21

= 65/147 + 26/147

= 13/21

Cách 2 : 5/7 x 13/21 + 2/7 x 13/21

= (5/7 + 2/7) x 13/21

= 1 x 13/21

= 13/21

câu c tự làm

8 /15 :2/3+7/15:2/3 =                                              tính bằng 2 cách nhé các bạn

Cách 1:6/8+4/8+5/7=10/8*5/7=50/56=25/28

Cách 2 :3/4*5/7+1/2*
5/7=15/28+10/28=25/28

Cái này thầy tôi đã chữa yên tâm

\(a,\dfrac{4}{5}+\dfrac{1}{4}\times\dfrac{2}{5}\)

\(=\dfrac{4}{5}+\dfrac{1}{10}\)

\(=\dfrac{8}{10}+\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

\(---\)

\(b,\dfrac{7}{12}+\dfrac{3}{44}:\dfrac{3}{8}-\dfrac{1}{4}\)

\(=\left(\dfrac{7}{12}-\dfrac{1}{4}\right)+\dfrac{3}{44}\cdot\dfrac{8}{3}\)

\(=\left(\dfrac{7}{12}-\dfrac{3}{12}\right)+\dfrac{8}{44}\)

\(=\dfrac{4}{12}+\dfrac{8}{44}\)

\(=\dfrac{44}{132}+\dfrac{24}{132}\)

\(=\dfrac{68}{132}\)

\(=\dfrac{17}{33}\)

#\(Toru\)