Trắc nghiệm Hệ thức Vi – et và ứng dụng có đáp án (Thông hiểu) !!

A. $\frac{1}{6}$

B. 3

C. 6

D. 7

A. $\frac{-5}{6}$

B. $\frac{5}{6}$

C. $\frac{-5}{3}$

D. $\frac{5}{3}$

A. ${\mathrm{x}}_1=-1; {\mathrm{x}}_2=-\frac{5}{18};$$\mathrm{A}=18 (\mathrm{x} + 1)\left(\mathrm{x}+\frac{5}{18}\right)$

B. ${\mathrm{x}}_1=-1; {\mathrm{x}}_2=-\frac{5}{18};$$\mathrm{A}= (\mathrm{x} + 1)\left(\mathrm{x}+\frac{5}{18}\right)$

C. ${\mathrm{x}}_1=-1; {\mathrm{x}}_2=\frac{5}{18};$$\mathrm{A}=18 (\mathrm{x} + 1)\left(\mathrm{x}-\frac{5}{18}\right)$

D. ${\mathrm{x}}_1=1; {\mathrm{x}}_2=-\frac{5}{18};$$\mathrm{A}=18 (\mathrm{x} + 1)\left(\mathrm{x}+\frac{5}{18}\right)$

A. ${\mathrm{x}}_1=1; {\mathrm{x}}_2=-\frac{26}{5};$$\mathrm{B}=(\mathrm{x}-1)\left(\mathrm{x}+\frac{26}{5}\right)$

B. ${\mathrm{x}}_1=1; {\mathrm{x}}_2=-\frac{26}{5};$$\mathrm{B}=5(\mathrm{x}+1)\left(\mathrm{x}-\frac{26}{5}\right)$

C. ${\mathrm{x}}_1=1; {\mathrm{x}}_2=-\frac{26}{5};$$\mathrm{B}=5(\mathrm{x}-1)\left(\mathrm{x}+\frac{26}{5}\right)$

D. ${\mathrm{x}}_1=1; {\mathrm{x}}_2=\frac{26}{5};$$\mathrm{B}=5(\mathrm{x}-1)\left(\mathrm{x}-\frac{26}{5}\right)$

A. 8

B.12

C. 9

D. 10

A. −6

B. 16

C. −16

D. 6

A. ${\mathrm{x}}^2$ − 6x – 4 = 0

B. ${\mathrm{x}}^2$ − 6x + 4 = 0

C. ${\mathrm{x}}^2$ + 6x + 4 = 0

D. −${\mathrm{x}}^2$ − 6x + 4 = 0

A. ${\mathrm{x}}^2$ − 4x − 3 = 0

B. ${\mathrm{x}}^2$ + 3x − 4 = 0

C. ${\mathrm{x}}^2$ − 4x + 3 = 0

D. ${\mathrm{x}}^2$ + 4x + 3 = 0

A.${\mathrm{x}}_1=-1; {\mathrm{x}}_2=\frac{\mathrm{m}+7}{\mathrm{m}-2}$

B.${\mathrm{x}}_1=1; {\mathrm{x}}_2=-\frac{\mathrm{m}+7}{\mathrm{m}-2}$

C.${\mathrm{x}}_1=1; {\mathrm{x}}_2=\frac{\mathrm{m}+7}{\mathrm{m}-2}$

D.${\mathrm{x}}_1=-1; {\mathrm{x}}_2=-\frac{\mathrm{m}+7}{\mathrm{m}-2}$

A.${\mathrm{x}}_1=-1; {\mathrm{x}}_2=\frac{1-2\mathrm{m}}{\mathrm{m}}$

B.${\mathrm{x}}_1=1; {\mathrm{x}}_2=\frac{2\mathrm{m}-1}{\mathrm{m}}$

C.${\mathrm{x}}_1=1; {\mathrm{x}}_2=\frac{1-2\mathrm{m}}{\mathrm{m}}$

D.${\mathrm{x}}_1=-1; {\mathrm{x}}_2=\frac{2\mathrm{m}-1}{\mathrm{m}}$