Toán

h: \(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)

\(=6+\dfrac{3}{10}-3-\dfrac{4}{7}-2-\dfrac{3}{10}\)

\(=1-\dfrac{4}{7}=\dfrac{3}{7}\)

i: \(\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}\)

\(=7+\dfrac{4}{9}+4+\dfrac{7}{11}-3-\dfrac{4}{9}\)

\(=8+\dfrac{7}{11}=\dfrac{95}{11}\)

a: \(2x\left(3x^2-6x+4\right)=2x\cdot3x^2-2x\cdot6x+2x\cdot4\)

\(=6x^3-12x^2+8x\)

b: \(\left(4x-3\right)\left(2x+4\right)\)

\(=4x\cdot2x+4x\cdot4-3\cdot2x-3\cdot4\)

\(=8x^2+16x-6x-12=8x^2+10x-12\)

`a)2x(3x^2-6x-4)`

`=2x.3x^2-2x.6x-2x.4`

`=6x^3-12x^2-8x`

_

`b)(4x-3)(2x+4)`

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

`=8x^2+16x-6x-12`

`=8x^2+10x-12`

Tọa độ A là:

\(\left\{{}\begin{matrix}y=0\\\left(1-m\right)x+m+2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=0\\\left(1-m\right)x=-m-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=0\\x=\dfrac{m+2}{m-1}\end{matrix}\right.\)

=>\(A\left(\dfrac{m+2}{m-1};0\right)\)

\(OA=\sqrt{\left(\dfrac{m+2}{m-1}-0\right)^2+\left(0-0\right)^2}=\left|\dfrac{m+2}{m-1}\right|\)

Tọa độ B là:

\(\left\{{}\begin{matrix}x=0\\y=0\left(1-m\right)+m+2=m+2\end{matrix}\right.\)

=>B(0;m+2)

\(OB=\sqrt{\left(0-0\right)^2+\left(m+2-0\right)^2}=\left|m+2\right|\)

Để ΔAOB vuông cân thì OA=OB

=>\(\left|m+2\right|=\dfrac{\left|m+2\right|}{\left|m-1\right|}\)

=>\(\left|m+2\right|\left(\dfrac{1}{\left|m-1\right|}-1\right)=0\)

=>\(\left[{}\begin{matrix}m=-2\\\left|m-1\right|=1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}m=-2\\m-1=1\\m-1=-1\end{matrix}\right.\)

=>\(m\in\left\{-2;0;2\right\}\)