10 Bài tập Bài tập liên quan đến tính chất của cấp số nhân (có lời giải)

A. (b – c)² + (c – a)² + (d – b)² = (a – d)²;

B. (b – c)² + (c – a)² + (d – b)² = (a + d)²;

C. (b – c)² + (c – a)² + (d – b)² = (a – b)²;

D. (b – c)² + (c – a)² + (d – b)² = (a + b)².

A. $a^2\cdot b^2\cdot c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=2(a^3+b^3+c^3)$

B. $a^2\cdot b^2\cdot c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=\frac{1}{2}\left(a^3+b^3+c^3\right)$

C. $a^2\cdot b^2\cdot c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=a^3+b^3+c^3$

D. $a^2\cdot b^2\cdot c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=\frac{1}{3}\left(a^3+b^3+c^3\right)$

A. $\left(a+b+c\right);\sqrt{\frac{1}{3}\left(ab+bc+ca\right)};\sqrt[3]{abc};$

B. $\frac{1}{3}\left(a+b+c\right);\sqrt{\frac{1}{3}\left(ab+bc+ca\right)};\sqrt[3]{abc};$

C. $\frac{1}{3}\left(a+b+c\right);\sqrt{\frac{1}{3}(ab+bc+ca)};\sqrt[3]{3abc};$

D. $\left(a+b+c\right);\sqrt{ab+bc+ca};\sqrt[3]{abc}.$

A. u_{k – m} . u_{k + m} = u_k²;

B. u_{k – m} . u_k = (u_{k + m})²;

C. u_k . u_{k + m} = (u_{k – m})²;

D. u_{k – m} . u_{k + m} = (u_{k + 1})².

A. (a² + b²)(b² + c²) = (ab + bc)²;

B. (a² – b²)(b² + c²) = (ab + bc)²;

C. (a² – b²)(b² – c²) = (ab + bc)²;

D. (a² + b²)(b² + c²) = (ab – bc)².

A. tan a; tan b; tan c;

B. cos a; cos b; cos c;

C. sin a; sin b; sin c;

D, sin 2a; sin 2b; sin 2c.

A. $\frac{b^2}{a^2{b}^2}-\frac{b}{ab}=\frac{a^2{c}^2}{a^2{b}^2}-\frac{ac}{ab}$

B. $\frac{b^4}{a^2{b}^2}-\frac{b^2}{ab}-\frac{ac}{ab}=\frac{a^2{c}^2}{a^2{b}^2}$

C. $\frac{b^4}{a^2{b}^2}-\frac{b^2}{ab}=\frac{a^2{c}^2}{a^2{b}^2}-\frac{ac}{ab}$

D. $\frac{b^4}{a^2{b}^2}+\frac{b^2}{ab}=\frac{a^2{c}^2}{a^4{b}^2}-\frac{ac}{ab}$

A. ac + 2ad + bd = (b + c)²;

B. ac + 2ad + bd = (a + c)²;

C. ac + 2ad + bd = (a – c)²;

D. ac + 2ad + bd = (b – c)².

A. $\frac{c}{d\left(a+c\right)}=\frac{2}{b+d};$

B. $\frac{c}{d\left(a+c\right)}=\frac{1}{b+d};$

C. $\frac{c}{d\left(a+c\right)}=\frac{1}{2}\cdot \frac{1}{b+d};$

D. $\frac{c}{d\left(a+c\right)}=\frac{1}{3}\cdot \frac{1}{b+d}.$