2.1 

\(\Leftrightarrow x^3+3x^2+2x-3x^2-9x-6=0\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

2.2

\(\Leftrightarrow x^3-2x^2-2x-x^2+2x+2=0\)

\(\Leftrightarrow x\left(x^2-2x-2\right)-\left(x^2-2x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-2x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\pm\sqrt{3}\end{matrix}\right.\)

2.3

\(\Leftrightarrow3x^3-3x^2+2x+3x^2-3x+2=0\)

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

\(\Leftrightarrow\left(x+1\right)\left(3x^2-3x+2\right)=0\)

\(\Leftrightarrow x=-1\)

2.5

\(\Leftrightarrow2x^3+x^2+3x-4x^2-2x-6=0\)

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

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

\(\Leftrightarrow x=2\)

2.9

\(\Leftrightarrow3x^3-3x^2-3x=1\)

\(\Leftrightarrow3x^3=3x^2+3x+1\)

\(\Leftrightarrow4x^3=x^3+3x^2+3x+1\)

\(\Leftrightarrow4x^3=\left(x+1\right)^3\)

\(\Leftrightarrow\left(x\sqrt[3]{4}\right)^3=\left(x+1\right)^3\)

\(\Leftrightarrow x\sqrt[3]{4}=x+1\)

\(\Leftrightarrow x\left(\sqrt[3]{4}-1\right)=1\)

\(\Leftrightarrow x=\dfrac{1}{\sqrt[3]{4}-1}\)

Bài 2

5 C

Bài 3

1 D

6 C

Còn lại ol r nhé

2) 5. C

3) 2. D

6. C

Còn lại ok nha

1: Thay m=2 vào phương trình, ta được:

\(x^2+4\left(2-1\right)x-12=0\)

=>\(x^2+4x-12=0\)

=>(x+6)(x-2)=0

=>\(\left[\begin{array}{l}x+6=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-6\\ x=2\end{array}\right.\)

2: \(x^2+4\left(m-1\right)x-12=0\)

a=1; b=4(m-1); c=-12

Vì \(a\cdot c=1\cdot\left(-12\right)=-12<0\)

nên phương trình luôn có hai nghiệm phân biệt trái dấu với mọi m

3: Theo Vi-et, ta có:

\(\begin{cases}x_1+x_2=-\frac{b}{a}=-4\left(m-1\right)\\ x_1x_2=\frac{c}{a}=-12\end{cases}\)

Vì x1 là nghiệm của phương trình nên ta có: \(x_1^2+4\left(m-1\right)\cdot x_1-12=0\)

=>\(x_1^2+4m\cdot x_1-4x_1-12=0\)

=>\(x_1^2-4x_1+4+4m\cdot x_1-16=0\)

=>\(\left(x_1-2\right)^2=-4mx_1+16=4\left(-mx_1+4\right)\)

=>\(\left|x_1-2\right|=\sqrt{4\left(-mx_1+4\right)}=2\cdot\sqrt{-mx_1+4}\)

\(4\left|x_1-2\right|\cdot\sqrt{4-mx_2}=\left(x_1+x_2-x_1x_2-8\right)^2\)

=>\(4\cdot2\cdot\sqrt{4-mx_1}\cdot\sqrt{4-mx_2}=\left(-4m+4+12-8\right)^2=\left(-4m+8\right)^2\)

=>\(8\cdot\sqrt{16-4mx_2-4mx_1+m^2\cdot x_1x_2}=16\cdot\left(m-2\right)^2\)

=>\(\sqrt{m^2\cdot\left(-12\right)-4m\left(x_1+x_2\right)+16}=2\cdot\left(m-2\right)^2\)

=>\(\sqrt{-12m^2-4m\cdot\left(-4m+4\right)+16}=2\left(m-2\right)^2\)

=>\(\sqrt{-12m^2+16m^2-16m+16}=2\left(m-2\right)^2\)

=>\(\sqrt{4m^2-16m+16}=2\left(m-2\right)^2\)

=>\(2\cdot\sqrt{m^2-4m+4}=2\left(m-2\right)^2\)

=>\(\left|m-2\right|=\left(m-2\right)^2\)

=>m-2=0

=>m=2

Bài 2: 

Ta có: \(3n^3+10n^2-5⋮3n+1\)

\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow3n\in\left\{0;-3;3\right\}\)

hay \(n\in\left\{0;-1;1\right\}\)

Bài 1: 

a: Xét ΔABI và ΔACI có

AB=AC

AI chung

BI=CI

Do đó: ΔABI=ΔACI

`@` `\text {Ans}`

`\downarrow`

3:

c: Xét ΔCAM có KI//AM

nên KI/AM=CI/CM

Xét ΔCMB có HI//MB

nên HI/MB=CI/CM

=>KI/AM=HI/MB

=>KI=HI

=>I là trung điểm của HK

III

1 It's colder today than yesterday

2 It takes 4 hours to travel by car and fives hours by train

3 We were busier at work today than everyday

4 Jane's sister cooks worse than her

5 Nobody in this team can play football as well as Tom

IV

1 D
2 A

\(2b,=\left(2x^3-4x^2-4x^2+8x-2x+4-9\right):\left(2x-4\right)\\ =\left[\left(2x-4\right)\left(x^2-2x-2\right)-9\right]:\left(2x-4\right)\\ =x^2-2x-2\left(\text{ dư -9}\right)\)