Đáp án đúng là: B

Ta có

$30{\left(4-2\text{x}\right)}^2+3\text{x}-6=30{\left(2x-4\right)}^2+3\left(x-2\right)$

$={30.2}^2\left(x-2\right)+3\left(x-2\right)$

$=120{\left(x-2\right)}^2+3\left(x-2\right)$

$=3\left(x-2\right)\left(40\left(x-2\right)+1\right)$

$=3\left(x-2\right)\left(40x-79\right)$

Nhân tử chung có thể là 3(x - 2)

Đáp án đúng là: B

$x^2-2xy+y^2-81=\left(x^2-2xy+y^2\right)-81$ (nhóm 3 hạng tử đầu để xuất hiện bình phương một hiệu)

$={\left(x-y\right)}^2-9^2$ (áp dụng hằng đẳng thức ${\text{A}}^{\text{2}}-{\text{B}}^{\text{2}}\text{=}\left(\text{A}-\text{B}\right)\left(\text{A + B}\right)$

$=\left(\text{x}-\text{y}-9\right)\left(\text{x}-\text{y + 9}\right)$

Đáp án đúng là: A

Vì $x^5+x^3+x^2+1$

$=x^3\left(x^2+1\right)+x^2+1$

$=\left(x^2+1\right)\left(x^3+1\right)$

Nên $\left(x^5+x^3+x^2+1\right):\left(x^3+1\right)$

$=\left(x^2+1\right)\left(x^3+1\right):\left(x^3+1\right)$

$=\left(x^2+1\right)$

Ta có:

$4\left(x-5\right)-2x\left(5-x\right)=0$

$\iff 4\left(x-5\right)+2x\left(x-5\right)=0$

$\iff \left(x-5\right)\left(4+2x\right)=0$

$\iff \left[\begin{array}{c}x-5=0\\ 4+2x=0\end{array}\right.$

$\iff \left[\begin{array}{c}x=5\\ x=-2\end{array}\right.$

$\Rightarrow x_1+x_2=5-2=3$

Đáp án đúng là: C

Đáp án đúng là: D

Ta có:

$x^3+2x^2-9x-18=0$

$\iff \left(x^3+2x^2\right)-\left(9x-18\right)=0$

$\iff x^2\left(x+2\right)-9\left(x-2\right)=0$

$\iff \left(x^2-9\right)\left(x+2\right)=0$

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

$\iff \left[\begin{array}{c}x-3=0\\ x+3=0\\ x-2=0\end{array}\right.$

$\iff \left[\begin{array}{c}x=3\\ x=-3\\ x=-2\end{array}\right.$

Câu 6. Chọn câu trả lời đúng nhất:

$x^2{y}^{2}z+x{y}^2{z}^2+x^{2}y{z}^2=$

A. $\text{x}\left({\text{xy}}^{\text{2}}{\text{z + y}}^{\text{2}}{\text{z}}^{\text{2}}{\text{+ xyz}}^{\text{2}}\right)$

B. $\text{y}\left({\text{x}}^{\text{2}}{\text{yz + xyz}}^{\text{2}}{\text{+ x}}^{\text{2}}{\text{z}}^{\text{2}}\right)$

C. $\text{z}\left({\text{x}}^{\text{2}}{\text{y}}^{\text{2}}{\text{+ xy}}^{\text{2}}{\text{z + x}}^{\text{2}}\text{yz}\right)$

D. $\text{xyz}\left(\text{xy + yz + xz}\right)$

Câu 7. Cho ${\text{4a}}^{\text{2}}\left(\text{x + 1}\right)-\text{7bx}-\text{7b =}\left(\text{x + 1}\right)\left(\text{…}\right)\text{.}$ Biểu thức thích hợp vào dấu … là

A. $4{\text{a}}^2-\text{b}$

B. $4{\text{a}}^2+7\text{b}$

C. $4{\text{a}}^2-7\text{b}$

D. $4{\text{a}}^2\text{+ b}$

Câu 8. Phân tích đa thức thành nhân tử: $x^2+6x+9$

A. $\left(x+3\right)\left(x-3\right)$

B. $\left(x-1\right)\left(x+9\right)$

C. ${\left(x+3\right)}^2$

D. $\left(x+6\right)\left(x-3\right)$

Câu 9. Kết quả phân tích đa thức ${\text{x}}^2-\text{xy}+\text{x}-\text{y}$ thành nhân tử là

A. $\left(\text{x}+1\right)\left(\text{x}-\text{y}\right)$

B. $\left(\text{x}-\text{y}\right)\left(\text{x}-1\right)$

C. $\left(\text{x}-\text{y}\right)\left(\text{x}+y\right)$

D. $x\left(x-y\right)$

Câu 10. Chọn câu sai.

A. ${\left(\text{x}-1\right)}^3+2{\left(\text{x}-1\right)}^2={\left(\text{x}-1\right)}^2\left(\text{x}+1\right)$

B. ${\left(\text{x}-1\right)}^3+2\left(\text{x}-1\right)=\left(\text{x}-1\right)\left[{\left(\text{x}-1\right)}^2+2\right]$

C. ${\left(\text{x}-1\right)}^3+2{\left(\text{x}-1\right)}^2=\left(\text{x}-1\right)\left[{\left(\text{x}-1\right)}^2+2x-2\right]$

D. ${\left(\text{x}-1\right)}^3+2{\left(\text{x}-1\right)}^2=\left(\text{x}-1\right)\left(\text{x}+3\right)$

Đáp án đúng là: A

Ta có:

$5{\text{x}}^2-10\text{x}+5=0$

$\iff 5\left({\text{x}}^2-2\text{x}+1\right)=0$

$\iff {\left(\text{x}-1\right)}^2=0$

$\iff \text{x}-1=0$

$\iff \text{x}=1$

Đáp án đúng là: A

Theo đề ra ta có:

$3x^3-8x^2-41x+30$

$=3x^3-2x^2-6x^2+4x-45x+30$

$=\left(3x^3-2x^2\right)-\left(6x^2-4x\right)-\left(45x-30\right)$

$=x^2\left(3x-2\right)-2x\left(3x-2\right)-15\left(3x-2\right)$

$=\left(x^2-2x-15\right)\left(3x-2\right)$

$=\left(x^2+3x-5x-15\right)\left(3x-2\right)$

$=\left[\left(x^2+3x\right)-\left(5x+15\right)\right]\left(3x-2\right)$

$=\left[x\left(x+3\right)-5\left(x+3\right)\right]\left(3x-2\right)$

$=\left(3x-2\right)\left(x-5\right)\left(x+3\right)$

Đáp án đúng là: C

Ta có:

$A=x^4+3x^3-27x-81$

$=\left(x^4-81\right)+\left(3x^3-27x\right)$

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

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

Ta có: $x^2+3x+9=x^2+2.\frac{3}{2}x+\frac{9}{4}+\frac{27}{4}\ge \frac{27}{4}>0,\forall x$

Mà $\left|\mathrm{x}\right|<3\iff {\mathrm{x}}^2<9\iff {\mathrm{x}}^2-9<0$

$\Rightarrow \mathrm{A}=\left({\mathrm{x}}^2-9\right)\left({\mathrm{x}}^2+3\mathrm{x}+9\right)<0$ khi $\left|\mathrm{x}\right|<3$

Đáp án đúng là: C

Ta có:

${\left(3{\text{x}}^2+3\text{x}-5\right)}^2-{\left(3{\text{x}}^2+3\text{x}+5\right)}^2$

$=\left(3x^2+3x-5-3x^2-3x-5\right)\left(3x^2+3x-5+3x^2+3x+5\right)$

$=-10\left(6x^2+6x\right)$

$=-10.6x\left(x+1\right)$

$=-60x\left(x+1\right)$

$=mx\left(x+1\right)$

$\Rightarrow m=-60<0$

Đáp án đúng là: D

Ta có:

${\text{B = x}}^{\text{3}}{\text{+ 3x}}^{\text{2}}{\text{y + 3xy}}^{\text{2}}{\text{+ y}}^{\text{3}}{\text{+ x}}^{\text{2}}{\text{+ 2xy + y}}^{\text{2}}$

$=\left(x^3+3x^{2}y+3x{y}^2+y^3\right)+\left({\text{x}}^{\text{2}}{\text{+ 2xy + y}}^{\text{2}}\right)$

$={\left(x+y\right)}^3+{\left(x+y\right)}^2$

$={\left(x+y\right)}^2\left(x+y+1\right)$

Vì x = 20 – y nên x + y = 20. Thay x + y = 20 vào $B={\left(x+y\right)}^2\left(x+y+1\right)$ ta được $B={20}^2\cdot \left(20+1\right)=400.21=8400$

Vậy B > 8 300 khi x = 20 – y.