Đặt A = 1 + 2 + ... + 31 + 32

Số số hạng trong A là: (32 - 1) : 1 + 1 = 32 (số)

Tổng A bằng : (32 + 1).32 : 2 = 528

Thay A = 528 vào biểu thức (1 + 2 + 3 +...+ 31 + 32) : 11 + 2, ta được:

528 : 11 + 2

= 48 + 2 = 50

Có: \(\dfrac{7}{6}\times\dfrac{1}{2}+\dfrac{7}{6}\times\dfrac{3}{4}+\dfrac{7}{6}+\dfrac{15}{4}\)

\(=\dfrac{7}{6}\times\left(\dfrac{1}{2}+\dfrac{3}{4}+1\right)+\dfrac{15}{4}\)

\(=\dfrac{7}{6}\times\dfrac{9}{4}+\dfrac{15}{4}\)

\(=\dfrac{63}{24}+\dfrac{90}{24}=\dfrac{51}{8}\)

*Phân tích thêm:

\(=\left(x+1\right)\left[\left(x^3+1\right)-\left(x^2-1\right)\right]\)

\(=\left(x+1\right)\left[\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x+1\right)\right]\)

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

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

\(x^4-x^2+2x+2\)

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

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

\(=\left(x+1\right)\left[x^2\left(x-1\right)+2\right]\)

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

a) \(\dfrac{1}{27}.81^n=3^n\)

\(\Rightarrow\dfrac{3^{4n}}{3^3}=3^n\)

\(\Rightarrow3^{4n-3}=3^n\)

\(\Rightarrow4n-3=n\)

\(\Rightarrow4n-n=3\Leftrightarrow n=1\)

b) \(C=3+3^2+3^3+...+3^{100}\)

\(3.C=3^2+3^3+3^4+...+3^{100}+3^{101}\)

\(3C-C=3^2+3^3+3^4+...+3^{100}+3^{101}-\left(3+3^2+3^3+...+3^{100}\right)\)

\(2C=3^{101}-3\)

\(\Rightarrow2C+3=3^{101}\)

Mặt khác: \(2C+3=3^n\) \(\Rightarrow n=101\)

c) \(2.16\ge2^n>4\)

\(\Rightarrow2.2^4\ge2^n>2^2\)

\(\Rightarrow2^5\ge2^n>2^2\Leftrightarrow n\in\left\{3;4;5\right\}\) (do n là số tự nhiên)

\(B=\left(x+1\right)^2-2\left(2x-1\right)\left(1+x\right)+4x^2-4x+1\)

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

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

10.

a) \(\left(-3+\dfrac{3}{x}-\dfrac{1}{3}\right):\left(1+\dfrac{2}{5}+\dfrac{2}{3}\right)=-\dfrac{5}{4}\)

\(\Rightarrow\left(\dfrac{3}{x}-\dfrac{10}{3}\right):\dfrac{31}{15}=-\dfrac{5}{4}\)

\(\Rightarrow\dfrac{3}{x}-\dfrac{10}{3}=-\dfrac{5}{4}.\dfrac{31}{15}=-\dfrac{31}{12}\)

\(\Rightarrow\dfrac{3}{x}=-\dfrac{31}{12}+\dfrac{10}{3}=\dfrac{3}{4}\)

\(\Leftrightarrow x=4\)

b) \(\left(\dfrac{2}{5}-x\right):1\dfrac{1}{3}+\dfrac{1}{2}=-4\)

\(\Rightarrow\left(\dfrac{2}{5}-x\right):\dfrac{4}{3}=-4-\dfrac{1}{2}=-\dfrac{9}{2}\)

\(\Rightarrow\dfrac{2}{5}-x=-\dfrac{9}{2}.\dfrac{4}{3}=-6\)

\(\Rightarrow x=\dfrac{2}{5}-\left(-6\right)=\dfrac{32}{5}\)

c) \(-\dfrac{3x}{4}.\left(\dfrac{1}{x}+\dfrac{2}{7}\right)=0\)  

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{3x}{4}=0\\\dfrac{1}{x}+\dfrac{2}{7}=0\end{matrix}\right.\)                 \(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{x}=-\dfrac{2}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{7}{2}\end{matrix}\right.\)

d) \(2,5x+\dfrac{-4}{7}=\dfrac{1}{2}x\)

\(\Rightarrow2,5x-\dfrac{1}{2}x=\dfrac{4}{7}\)

\(\Rightarrow2x=\dfrac{4}{7}\Leftrightarrow x=\dfrac{2}{7}\)

9. a) \(\dfrac{4}{3}+\dfrac{2}{3}:x=-\dfrac{2}{5}\)

\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{26}{15}\)\(\Rightarrow x=\dfrac{2}{3}:\left(-\dfrac{26}{15}\right)=-\dfrac{5}{13}\)

b) \(\dfrac{7}{3}-1\dfrac{3}{5}.x=-1\dfrac{2}{3}\)

\(\Rightarrow\dfrac{8}{5}x=\dfrac{7}{3}-\left(-\dfrac{5}{3}\right)=\dfrac{7+5}{3}\)

\(\Rightarrow\dfrac{8}{5}x=4\Leftrightarrow x=\dfrac{5}{2}\)

c) \(\left(-2\dfrac{1}{4}:x+1,5\right)\left(\dfrac{-5}{3}.x-\dfrac{5}{12}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-2\dfrac{1}{4}:x=-1.5\\\dfrac{-5}{3}.x=\dfrac{5}{12}\end{matrix}\right.\)         \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

d) \(-\dfrac{3}{5}.x-\left(x+2,5\right)=-\dfrac{13}{15}\)

\(\Rightarrow-\dfrac{8}{5}.x=-\dfrac{13}{15}+2,5\)

\(\Rightarrow-\dfrac{8}{5}.x=\dfrac{49}{30}\)

\(\Rightarrow x=\dfrac{49}{30}:\left(\dfrac{-8}{5}\right)=-\dfrac{49}{48}\)