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\(M=\dfrac{26}{7.11}+\dfrac{65}{11.21}+\dfrac{39}{21.27}+\dfrac{143}{27.49}\)
\(M=\dfrac{13.2}{7.11}+\dfrac{13.5}{11.21}+\dfrac{13.3}{21.27}+\dfrac{13.11}{27.49}\)
\(M=13.\dfrac{2}{7.11}+13.\dfrac{5}{11.21}+13.\dfrac{3}{21.27}+13.\dfrac{11}{27.49}\)
\(M=13.\left(\dfrac{2}{7.11}+\dfrac{5}{11.21}+\dfrac{3}{21.27}+\dfrac{11}{27.49}\right)\)
\(M=13.\dfrac{1}{2}\left(\dfrac{4}{7.11}+\dfrac{10}{11.21}+\dfrac{6}{21.27}+\dfrac{22}{27.49}\right)\)
\(M=\dfrac{13}{2}\left(\dfrac{1}{7}-\dfrac{1}{49}\right)\)
\(M=\dfrac{13}{2}.\dfrac{6}{49}=\dfrac{39}{49}\)
N sai đề rồi chuyển 35 thành 85 tính tương tự xong lấy \(P=\dfrac{M}{N}+\dfrac{34}{29}\)
P=2
M=26/7x11+65/11x21+39/21x27+143/37x49
M=26/77+65/231+13/189+143/1813
M=0,766705481
N=51/7x16+17/16x19+35/19x34+85/34x49
N=51/112+17/304+35/646+5/98
N=0,6164781702
P=0,766705481/0,6164781702+29/34
P=2,096627519
\(\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)
\(\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)
\(\frac{3}{x}=\frac{-1}{2}\)
\(\Rightarrow3.2=\left(-1\right).x\)
\(\Rightarrow6=\left(-1\right).x\)
\(\Rightarrow x=6:\left(-1\right)\)
\(\Rightarrow x=-6\)
\(\frac{x}{2}-\frac{2}{y}=\frac{1}{2}\)
\(\Rightarrow\frac{x}{2}-\frac{1}{2}=\frac{2}{y}\)
\(\Rightarrow\frac{x-1}{2}=\frac{2}{y}\)
\(\Rightarrow\hept{\begin{cases}x-1=2\\2=y\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}\)
\(b,\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)
\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)
\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{8}{6}\)
\(\Rightarrow\frac{3}{x}=\frac{-3}{6}\)
\(\Rightarrow x\cdot(-3)=18\Rightarrow x=-6\)
\(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{9.9}\)
\(N\)bé hơn \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{8.9}=N_1\)
\(N_1=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{8.9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\) \((1)\)
\(N\)lớn hơn \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}=N_2\)
\(\Rightarrow N_2=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{5}{10}-\frac{1}{10}=\frac{2}{5}\) \((2)\)
Từ \((1)\)và \((2)\)suy ra ; \(\frac{2}{5}\)bé hơn N bé hơn \(\frac{8}{9}\)
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