\(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x-5\right)\left(x+5\right)=264\)

\(\Leftrightarrow x^3+64-x^3+25x=264\)

\(\Leftrightarrow25x=200\)

hay x=8

\(=\left(x^3-2x^2+x+2x^2-4x+2-2x+7\right):\left(x^2-2x+1\right)\\ =\left[\left(x^2-2x+1\right)\left(x+2\right)-2x+7\right]:\left(x^2-2x+1\right)\\ =x+2\left(dư:-2x+7\right)\)

Trl:

\(\left(x+3\right).\left(y-6\right)=-4\)

\(\Rightarrow\orbr{\begin{cases}x+3=-4\\y-6=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-4-3\\y=-4+6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-7\\y=2\end{cases}}\)

Hc tốt

(x+3).(y-6)=-4

* x+3=-4           * y-6=-4

       x=-4-3             y=-4+6

       x=-7                y=2

vậy x=-7 hoặc y=2

Bài 2:

a: \(f\left(x\right)=x^3-9x^2+27x-27\)

\(=x^3-3x^2-6x^2+18x+9x-27\)

\(=x^2\left(x-3\right)-6x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3\)

b: \(g\left(x\right)=x^2-6x+9\)

\(=x^2-2\cdot x\cdot3+3^2\)

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

=>f(x)⋮g(x)

=>f(x) chia g(x) thì dư 0

\(\frac{f\left(x\right)}{k\left(x\right)}=\frac{x^3-9x^2+27x-27}{x^2-6x+10}\)

\(=\frac{x^3-6x^2+10x-3x^2+18x-30-x+3}{x^2-6x+10}=x-3+\frac{-x+3}{x^2-6x+10}\)

=>f(x) chia k(x) dư -x+3

\(\Leftrightarrow18x^2\left(x+4\right)-12x\left(x+4\right)=0\)

\(\Leftrightarrow6x\left(x+4\right)\left[3x-2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\\3x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)

chúc em thành công trong bài kiểm tra này và được 10 điểm để lên đỉnh olypia 

k cho anh nha

Chúc bn thi tốt , mà mai lớp 5 thi j

2 B

3 D

4 A

5 B

6 D

7 C

8 A

9 D

10 A

11 B

12 C

13 A

14 C

15 C

II 

1 majestic

a: Xét ΔABC có \(\hat{A}+\hat{C}+\hat{B}=180^0\)

=>\(\hat{C}-10^0+\hat{C}+\hat{C}+10^0=180^0\)

=>\(3\cdot\hat{C}=180^0\)

=>\(\hat{C}=60^0\)

=>\(\hat{A}=60^0-10^0=50^0\)

\(\hat{B}=60^0+10^0=70^0\)

b: Xét ΔABC có \(\hat{A}+\hat{C}+\hat{B}=180^0\)

=>\(\hat{C}+\frac76\cdot\hat{C}+\frac56\cdot\hat{C}=180^0\)

=>\(3\cdot\hat{C}=180^0\)

=>\(\hat{C}=60^0\)

\(\hat{B}=\frac76\cdot60^0=70^0;\hat{A}=\frac56\cdot60^0=50^0\)

c: Đặt \(a=\hat{A};b=\hat{B};c=\hat{C}\)

\(21\cdot\hat{A}=14\cdot\hat{B}=6\cdot\hat{C}\)

=>21a=14b=6c

=>\(\frac{21a}{42}=\frac{14b}{42}=\frac{6c}{42}\)

=>\(\frac{a}{2}=\frac{b}{3}=\frac{c}{7}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{7}=\frac{a+b+c}{2+3+7}=\frac{180^0}{12}=15^0\)

=>\(\begin{cases}a=15^0\cdot2=30^0\\ b=15^0\cdot3=45^0\\ c=15^0\cdot7=105^0\end{cases}\Rightarrow\hat{A}=30^0;\hat{B}=45^0;\hat{C}=105^0\)