\(\left[\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8+1,75\right]:0,05=235\)

\(\Leftrightarrow\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8+1,75=11,75\)

\(\Leftrightarrow\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8=10\)

\(\Leftrightarrow\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right)\cdot2,8=10\)

\(\Leftrightarrow6\frac{3}{7}-\frac{0,75x-2}{0,35}=\frac{25}{7}\)

\(\Leftrightarrow\frac{0,75x-2}{0,35}=\frac{20}{7}\)

\(\Leftrightarrow0,75x-2=1\)

\(\Leftrightarrow0,75x=3\)

\(\Leftrightarrow x=4\)

Vậy \(x=4\)

C=-66.(1/2-1/3+1/11)+124.(-37)+63.(-124)

C=-66.17/66-124.(37+63)

C=-17-124.100

C=-12417

a/ \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\\3+2^{x+1}=24-\left[16-\left(4-1\right)\right]\)

\(3+2^{x+1}=24-\left(16-3\right)\\ 3+2^{x-1}=24-13\\ 3+2^{x-1}=11\\ 2^{x+1}=11-3\\ 2^{x-1}=8\)

\(2^{x-1}=2^3\\ \Rightarrow x-1=3\\x=3+1\\ x=4\)

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=205550\)

\(\left(x.100\right)+\left(1+2+3+....+100\right)=205550\)

Ta tính tổng \(1+2+3+...+100\\ \) trước

Số các số hạng: \(\left[\left(100-1\right):1+1\right]=100\)

Tổng :\(\left[\left(100+1\right).100:2\right]=5050\)

Thay số vào ta có được:

\(\left(x.100\right)+5050=205550\\ \\ x.100=205550-5050\\ \\x.100=20500\\ \\x=20500:100\\ \\\Rightarrow x=2005\)

\(\left|x-5\right|=18+2.\left(-8\right)\\\left|x-5\right|=18+\left(-16\right)\\\left|x-5\right|=2\: \)

\(\Rightarrow\left[\begin{array}{nghiempt}x-5=2\\\\x-5=\left(-2\right)\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=2+5\\\\x=\left(-2\right)+5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=7\\\\x=3\end{array}\right.}\)

=> x ϵ {7;3}

\(A=-66.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)

\(A=-66.\left(\frac{33}{66}-\frac{22}{66}+\frac{6}{66}\right)+\left(-124\right).\left(37+63\right)\)

\(A=-66.\frac{17}{66}+\left(-124\right).100\)

\(A=-17+\left(-12400\right)\)

A = -12417

\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)

\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)

\(=\left(\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\times\frac{2}{x}\right)\div\frac{x-1}{x^3}\)

\(=\left(\frac{1}{\left(x+1\right)^2}\times\left(\frac{x^2+1}{x^2}+\frac{2}{x}\right)\right)\div\frac{x-1}{x^3}\)

\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x^3+2x^2+x}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x^2+2x+1\right)}{x^3}\right)\div\frac{x-1}{x^3}\)

\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x+1\right)^2}{x^3}\right)\div\frac{x-1}{x^3}\)

\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)

\(=\frac{x}{x-1}\)

=\(2\left(\frac{1}{2}-\frac{1}{2.3}\right).2\left(\frac{1}{2}-\frac{1}{3.4}\right)....2\left(\frac{1}{2}-\frac{1}{99.100}\right)\)

=\(2^{89}\left(\frac{1}{2}.98-\frac{1}{2}+\frac{1}{100}\right)\)

\(=2^{98}.\left(49-\frac{49}{100}\right)=\frac{2^{98}.4851}{100}\)

a) \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right].\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)

\(=\left(-7\right)-\left(-40\right).\left(-3\right)-25\)

\(=\left(-7\right)-120-25\)

\(=-152\)

b) \(\left(-2\right)^3.3-\left(1^{10}+8\right):\left(-3\right)^2\)

\(=\left(-8\right).3-\left(1+8\right):9\)

\(=\left(-24\right)-9:9\)

\(=\left(-24\right)-1\)

\(=-25\)

                                                     Bài giải

a, \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right]\cdot\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)

\(=\left(-7\right)-\left(-40\right)\cdot\left(-3\right)-25\)

\(=-7-120-25\)

\(=-127-25\)

\(=-152\)

b, \(\left(-2\right)^3\cdot3-\left(1^{10}+8\right)\text{ : }\left(-3\right)^2\)

\(=-8\cdot3-\left(1+8\right)\text{ : }9\)

\(=-24-9\text{ : }9\)

\(=-24-1\)

\(=-25\)

a, \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right]\cdot\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)

\(=\left(-7\right)-\left(-40\right)\cdot\left(-3\right)-25\)

\(=-7-120-25\)

\(=-127-25\)

\(=-152\)

b, \(\left(-2\right)^3\cdot3-\left(1^{10}+8\right)\text{ : }\left(-3\right)^2\)

\(=-8\cdot3-\left(1+8\right)\text{ : }9\)

\(=-24-9\text{ : }9\)

\(=-24-1\)

\(=-25\)