Bn ghi rõ đề đc ko

\(\frac{2}{1\times4}+\frac{2}{4\times7}+\frac{2}{7\times10}+...+\frac{2}{37\times40}\)

\(=\frac{2}{3}\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{37\times40}\right)\)

\(=\frac{2}{3}\times\left(\frac{4-1}{1\times4}+\frac{7-4}{4\times7}+\frac{10-7}{7\times10}+...+\frac{40-37}{37\times40}\right)\)

\(=\frac{2}{3}\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

\(=\frac{2}{3}\times\left(1-\frac{1}{40}\right)=\frac{13}{20}\)

\(\dfrac{3}{4}-\left\{2-3.\left[\dfrac{7}{2}-\left(\dfrac{2}{3}\right)^2\right]\right\}=\dfrac{3}{4}-\left\{2-3.\left[\dfrac{7}{2}-\dfrac{4}{9}\right]\right\}=\dfrac{3}{4}-\left\{2-3.\dfrac{55}{18}\right\}=\dfrac{3}{4}-\left\{2-\dfrac{55}{6}\right\}=\dfrac{3}{4}-\left\{\dfrac{-43}{6}\right\}=\dfrac{95}{12}\)

\(\dfrac{3}{4}-\left\{2-3.\left[\dfrac{7}{2}-\left(\dfrac{2}{3}\right)^2\right]\right\}\\ =\dfrac{3}{4}-\left[2-3.\left(\dfrac{7}{2}-\dfrac{4}{9}\right)\right]\\ =\dfrac{3}{4}-\left(2-3\cdot\dfrac{55}{18}\right)\\ =\dfrac{3}{4}-\left(2-\dfrac{55}{6}\right)\\ =\dfrac{3}{4}-\left(-\dfrac{43}{6}\right)\\ =\dfrac{3}{4}+\dfrac{43}{6}\\ =\dfrac{95}{12}\)

`@` `\text {Ans}`

`\downarrow`

`b)`

`3/4 - {2 - 3*[7/2 - (2/3)^2]}`

`= 3/4 - [2 - 3*(7/2 - 4/9)]`

`= 3/4 - (2 - 3*55/18)`

`= 3/4 - (2 - 55/6)`

`= 3/4 - (-43/6)`

`= 95/12`

1. a) \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=14x^5+21x^7\)

b) \(\left(x^3-x^2+x-1\right):\left(x-1\right)=\dfrac{x^3-x^2+x-1}{x-1}\)

\(=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{x-1}=\dfrac{\left(x-1\right)\left(x^2+1\right)}{x-1}=x^2+1\)

2: \(x^2-8x+7=0\)

=>\(x^2-x-7x+7=0\)

=>\(x\left(x-1\right)-7\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x-7\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

1:

a: \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=21x^7+14x^5\)

b: \(\dfrac{x^3-x^2+x-1}{x-1}=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{\left(x-1\right)}\)

\(=x^2+1\)

1)

\(a,7x^2\cdot(2x^3+3x^5)\\=7x^2\cdot2x^3+7x^2\cdot3x^5\\=14x^5+21x^7\\---\\b,(x^3-x^2+x-1):(x-1)(dkxd:x\ne 1)\\=[x^2(x-1)+(x-1)]:(x-1)\\=(x-1)(x^2+1):(x-1)\\=x^2+1\)

2)

\(x^2-8x+7=0\\\Leftrightarrow x^2-x-7x+7=0\\\Leftrightarrow x(x-1)-7(x-1)=0\\\Leftrightarrow (x-1)(x-7)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

\(\text{#}Toru\)

2008x2002<2006x2004

điều khác biệt ở chỗ: 8x2<6x4.

2008*2002=2006*2004

2008 x 2002 < 2006 x 2004 

`8x^4y^5z^3 : 2x^3y^4z`

`= 4xyz^2`.

1, 14/9

2, 9/23 x 5/8 + 9/23 x 3/08 - 9/23

   = 9/23 x ( 5/8 + 3/8 -1 )

   = 9/23 x 0

   = 0