a) \(4.\left(-\dfrac{1}{2}\right)^3-2.\left(-\dfrac{1}{2}\right)^2+3.\left(-\dfrac{1}{2}\right)+1\)

\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}+3.\left(-\dfrac{1}{2}\right)+1\)

\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1\)

\(=-\dfrac{3}{2}\)

b) \(8.\sqrt{9}-\sqrt{64}\)

\(=8.3-8\)

\(=24-8\)

\(=16\)

c) \(\sqrt{\dfrac{9}{16}}+\dfrac{25}{46}:\dfrac{5}{23}-\dfrac{7}{4}\)

\(=\dfrac{3}{4}+\dfrac{5}{2}-\dfrac{7}{4}\)

\(=-1+\dfrac{5}{2}\)

\(=\dfrac{3}{2}\)

56:54=

a) 2/3 + 3/4 . (-4/9)

= 2/3 - 1/3

= 1/3

b) -5/7 . 31/33 + (-5/7) : 33/2

= -5/7 . 31/33 - 5/7 . 2/33

= -5/7 . (31/33 + 2/33)

= -5/7 . 1

= -5/7

c) -3/5 . 13/11 - (-3/5) . 2/11

= -3/5 . (13/11 - 2/11)

= -3/5 . 1

= -3/5

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)

\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1+1+1}{a+b+c}=\dfrac{3}{a+b+c}=\dfrac{3}{1}=3\)

\(\Rightarrow a=b=c=\dfrac{1}{3}\)

\(\Rightarrow A=\dfrac{a^3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=a^3=\left(\dfrac{1}{3}\right)^3=\dfrac{1}{27}\)

\(A=1^3+2^3+3^3+...+n^3\)

Ta chứng minh

\(A=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\) (1)

+ Với \(n=3\)

\(1^3+2^3+3^3=36\)

\(\left(1+2+3\right)^2=36\)

=> (1) đúng

+ Giả sử (1) đúng với \(n=k\)

\(\Rightarrow1^3+2^3+3^3+...+k^3=\left(1+2+3+...+k\right)^2\)

+ Ta cần chứng minh (1) đúng với \(n=k+1\) Khi đó

\(VT=1^3+2^3+3^3+...+k^3+\left(k+1\right)^3=\)

\(=\left(1+2+3+...+k\right)^2+\left(k+1\right)^3=\)

\(=\left[\dfrac{k\left(k+1\right)}{2}\right]^2+\left(k+1\right)^3=\)

\(=\dfrac{k^2\left(k+1\right)^2+4\left(k+1\right)^3}{4}=\dfrac{\left(k+1\right)^2\left(k^2+4k+4\right)}{4}\)

\(VP=\left[1+2+3+...+k+\left(k+1\right)\right]^2=\)

\(=\left[\dfrac{\left(k+1\right)\left(k+1+1\right)}{2}\right]^2=\)

\(\dfrac{\left(k+1\right)^2\left(k+2\right)^2}{4}=\dfrac{\left(k+1\right)^2\left(k^2+4k+4\right)}{4}\)

Như vậy VT=VP nên (1) đúng  với \(n=k+1\)

Theo nguyên tắc của phương pháp quy nạp => (1) đúng

Câu C đúng