\(\dfrac{10}{n-1}\in Z\Rightarrow n-1=Ư\left(10\right)\)

\(\Rightarrow n-1=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)

\(\Rightarrow n=\left\{-9;-4;-1;0;2;3;6;11\right\}\)

Để \(\frac{10}{n-1}\)nguyên thì \(n-1\inƯ\left(10\right)\)

\(\Rightarrow n-1\in\left\{-1,1,10,-10\right\}\)

\(\Rightarrow n-1\in\left\{0,2,-9,11\right\}\)

Vậy \(n\in\left\{0;2;-9;11\right\}\)thì \(\frac{10}{n-1}\)nguyên

\(f\left(x\right)=ax^2+bx+c\)

\(f\left(4\right)=a.4^2+b.4+c=16a+4b+c\)

\(f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c=4a-2b+c\)

\(f\left(4\right)-f\left(-2\right)=\left(16a+4b+c\right)-\left(4a-2b+c\right)=12a+6b\)

\(=6\left(2a+b\right)=0\)

\(\Leftrightarrow f\left(4\right)=f\left(-2\right)\)

\(f\left(4\right)+2f\left(-2\right)=\left(16a+4b+c\right)+2\left(4a-2b+c\right)=24a+3c=3\left(8a+c\right)\ne0\)

Suy ra \(f\left(4\right)=f\left(-2\right)\ne0\)suy ra đpcm. 

`Answer:`

a. `2+3x=20`

`<=>3x=20-2`

`<=>3x=18`

`<=>x=18:3`

`<=>x=6`

b. `3x+7=3-x`

`<=>3+x+7=3`

`<=>4x+7=3`

`<=>4x=3-7`

`<=>4x=-4`

`<=>x=-1`

c. `2x^3-8x=0`

`<=>2x.(x^2-4)=0`

`<=>2x.(x-2).(x+2)=0`

`<=>2x=0` hoặc `x-2=0` hoặc `x+2=0`

`<=>x=0` hoặc `x=2` hoặc `x=-2`

`Answer:`

a. \(A=\left(\frac{1}{2}xy^2\right)\left(4xy^3\right)\)

\(=\left(\frac{1}{2}.4\right).\left(x.x\right).\left(y^2.y^3\right)\)

\(=2x^2y^5\)

b. Thay `x=-1` và `y=2` vào đơn thức `A`, ta được

`A=2.(-1)^2 .2^5=2.1.32=2.32=64`

1.A

2.B

3.C

4.B

`Answer:`

Tam giác \(MNP\)có \(MN=MP\)nên tam giác \(MNP\)cân tại \(M\).

\(I\)là giao điểm hai đường cao suy ra \(MI\)vuông góc với \(NP\). 

Do đó \(MI\)đồng thời là đường phân giác của tam giác \(MNP\)

suy ra \(\widehat{IMN}=\frac{1}{2}\widehat{NMP}=\frac{1}{2}\left(180^o-67^o-67^o\right)=23^o\)

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