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(23,4 + 19,5) x 7 + (23,4 + 19,5) x 3 + 11
= (23,4 + 19,5) x ( 7+3 ) + 11
= 42,9 x 10 + 11
= 429 + 11
= 440
\(\frac{\left(23,4+15,5\right)\cdot7+\left(23,4+19,5\right)\cdot3+11}{0,55\cdot2\cdot30+5\cdot11+2,75\cdot8}\)
\(=\frac{\left(23,4+15,5\right)\cdot\left(7+3\right)+12+11}{1,1\cdot30+5\cdot11+11\cdot2}\)
\(=\frac{38,9\cdot10+23}{11\cdot\left(3+5+2\right)}=\frac{412}{110}=\frac{206}{55}\)
Chúc em học tốt!
(Mong em đừng đăng câu hỏi linh tinh lên nữa)
a)
\(\frac{0,24x450+0.8x15x3+3x3x8}{65-60+55-50+...+25-2+5}\)
\(=\frac{0,24x45x10+0,8x45+3x3x8}{\left(65-60\right)+\left(55-50\right)+...+\left(25-20\right)+5}\)
\(=\frac{45\left(0,24x10+0,8\right)+3x3x8}{5+5+...+5}\)
\(=\frac{45x2,4+45x0,8+3x3x4x2}{5x6}\)
\(=\frac{108+36+36x2}{5x6}=\frac{36x3+36+36x2}{5x6}\)
\(=\frac{36\left(3+1+2\right)}{5x6}=\frac{36x6}{5x6}=\frac{36}{5}\)
b)
\(\frac{\left(23,4+19,5\right)\times7+\left(23,4+19,5\right)\times3+11}{0,55\times2\times30+5\times11+2,75\times8}\)
\(=\frac{\left(23,4+19,5\right)\times\left(7+3\right)+11}{1,1\times30+5\times11+2\times11}\)
\(=\frac{42,9\times10+11}{33+11\left(5+2\right)}=\frac{440}{33+77}=\frac{440}{110}=4\)
Cách này không hay lắm. Mong bạn thông cảm!
23,4x 12,5 +12,5 x 75,6 + 12,5
23,4x 12,5 +12,5 x 75,6 + 12,5 x 1
12,5 x(23,4+75,6+1)
12,5 x 100
1250
23,4x12,5+12,5x75,6+12,5=12,5x(23,4+75,6+1)=12,5x100=1250
\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)
\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)
\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)
\(\Leftrightarrow X=\frac{109}{6075}\)
Vậy X=109/6075
Chắc Sai kết quả chứ công thức đúng nha!!!...
Fighting!!!...
Đặt:
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)
Giải phương trình:
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)
\(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)
\(12x+\frac{12}{25}=11x+\frac{121}{243}\)
\(12x-11x=\frac{121}{243}-\frac{12}{25}\)
\(x=\frac{109}{6075}\)
