\(\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{3}{7}\right)...\left(1-\frac{10}{7}\right)\)

\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)...\left(1-\frac{10}{7}\right)\)

\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\right)...\left(1-\frac{10}{7}\right)\)

\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...0...\left(1-\frac{10}{7}\right)\)

= 0

\(\frac{x+3}{-4}=-\frac{9}{x+3}\)

\(\Leftrightarrow\left(x+3\right)\left(x+3\right)=-4\cdot\left(-9\right)\)

\(\Leftrightarrow\left(x+3\right)^2=36\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=6^2\\\left(x+3\right)^2=\left(-6\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=6\\x+3=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-9\end{cases}}\)

Vậy ....

quy đồng

\(\left(x+3\right)^2=36\)

\(\left(x+3\right)^2-6^2=0\)

áp dụng định lí "  \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) ta được

\(\left(x+3-6\right)\left(x+3+6\right)=0\)

\(x=3,x=-9\)

b) \(\left(2,4.x-36\right)\div1\frac{5}{7}=-1\)

\(\left(2,4.x-36\right)=-1.\frac{12}{7}\)

\(2,4.x-36=-\frac{12}{7}\)

\(2,4.x=-\frac{12}{7}+36\)

\(2,4.x=\frac{240}{7}\)

\(x=\frac{240}{7}\div2,4\)

\(x=\frac{100}{7}\)

a) \(\frac{3}{x-5}=\frac{-4}{x+2}\)

\(\Rightarrow3.\left(x+2\right)=-4.\left(x-5\right)\)

\(\Rightarrow3x+6=-4x+20\)

\(\Rightarrow7x=14\)

\(\Rightarrow x=2\)

nhân chéo là đc:

3(x+2)=-4(x-5)

3x+6=-4x+20

3x+4x=20-6

7x     =14

 x      =2

Vậy x=2

(1 - 1/7) x (1 - 2/7) x ............(1 - 10/7)

= 6/7 x 5/7 x ..... x -3/7

=6/7 x 5/7x 4/7 x 3/7x 2/7 x 1/7 x 0 x -1/7x -2/7 x -3/7

=0

nhớ tk cho mình nha

thanks nha