Tính: \(A=2\frac{1}{113}.\frac{1}{115}-\frac{3}{113}.3\frac{114}{115}-\frac{4}{113.115}+\frac{12}{113}\)
Chứng minh rằng : \(\frac{1}{112^2}+\frac{1}{112^2}+\frac{1}{113^2}+\frac{1}{114^2}+\frac{1}{115^2}<\frac{3}{2.5.11.23}\)
Chứng minh rằng : \(\frac{1}{112^2}+\frac{1}{112^2}+\frac{1}{113^2}+\frac{1}{114^2}+\frac{1}{115^2}<\frac{1}{2.5.11.23}\)
Tính A=112×113×114....×112016×112017A=1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}....\times1\frac{1}{2016}\times1\frac{1}{2017}A=121×131×141....×120161×120171
Cho A=\(\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}\)
CMR 1<A<2
\(\frac{50}{111}>\frac{1}{4};\frac{50}{112}>\frac{1}{4};\frac{50}{113}>\frac{1}{4};\frac{50}{114}>\frac{1}{4}\)
\(A=\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}>\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)(1)
\(\frac{50}{111}< \frac{1}{2};\frac{50}{112}< \frac{1}{2};\frac{50}{113}< \frac{1}{2};\frac{50}{114}< \frac{1}{2}\)
\(\Rightarrow A=\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}< \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)(2)
từ (1) và (2) \(\Rightarrow1< A< 2\)
Tính giá trị biểu thức:
A = \(2\frac{1}{309}\times\frac{1}{785}-\frac{1}{103}\times3\frac{784}{785}-\frac{4}{309\times785}+\frac{4}{103}\)
B = \(4\frac{1}{113}\times\frac{1}{371}-\frac{2}{113}\times5\frac{370}{371}-\frac{3}{113\times371}-\frac{4}{371}\)
C = \(x^3-31x^2-32x+7\) tại x = 32
D = \(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)-x^2\)tại x = a + b + c
1) Cho n thuộc N; Chứng tỏ: A= \(\frac{14n+3}{21n+5}\) là phân số tối giản
2) Tính : B = \(\frac{1010+1007+\frac{2017}{113}+\frac{2017}{117}-\frac{1010}{119}-\frac{1007}{119}}{1010+1008+\frac{2018}{113}+\frac{2018}{117}-\frac{1010}{119}-\frac{1008}{110}}\)
Cho A=\(\frac{50}{111}+\frac{50}{112}+\frac{50}{113}+\frac{50}{114}\). Chứng tơ \(1< A< 2\)
Ta có :
\(\frac{50}{111}>\frac{50}{200}\)
\(\frac{50}{112}>\frac{50}{200}\)
\(\frac{50}{113}>\frac{50}{200}\)
\(\frac{50}{114}>\frac{50}{200}\)
\(\Rightarrow A>\frac{50}{200}+\frac{50}{200}+\frac{50}{200}+\frac{50}{200}\)hay \(A>\frac{50}{200}.4\left(1\right)\)
Mặt khác :
\(\frac{50}{111}< \frac{50}{100}\)
\(\frac{50}{112}< \frac{50}{100}\)
\(\frac{50}{113}< \frac{50}{100}\)
\(\frac{50}{114}< \frac{50}{100}\)
\(\Rightarrow A< \frac{50}{100}+\frac{50}{100}+\frac{50}{100}+\frac{50}{100}\)hay \(A< \frac{50}{100}.4\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\Rightarrow1< A< 2\left(đpcm\right)\)
Tính nhanh:
\(\frac{3-3^2+3^3-3^4+...+3^{99}}{\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}}.\left(11-\sqrt{91}\right)\left(11-\sqrt{95}\right)\left(11+\sqrt{99}\right)\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)...\left(11-\sqrt{113}\right)\left(11-\sqrt{104}\right)\)
Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)
\(=0\)
Do đó biểu thức trên đầu bài bằng 0
bạn ơi, trong dãy này không có số \(\sqrt{121}\)đâu
A = \(2\frac{1}{309}\times\frac{1}{785}-\frac{1}{103}\times3\frac{784}{785}-\frac{4}{309\times785}+\frac{4}{103}\)
B = \(4\frac{1}{113}\times\frac{1}{371}-\frac{2}{113}\times5\frac{370}{371}-\frac{3}{113\times371}-\frac{4}{371}\)
C = \(x^3-31x^2-32x+7\) tại \(x=32\)
D = \(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)-x^2\)tại \(x=a+b+c\)