Tính giá trị biểu thức:
B=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).......\left(\frac{1}{98^2}-1\right).\left(\frac{1}{99^2}-1\right)\)
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Tính giá trị biểu thức sau
\(\left(100+\frac{99}{2}+\frac{98}{3}+.....+\frac{1}{100}\right)\div\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{101}\right)-2\)
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
\(=\frac{\left[\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{1}{100}+1\right)+\frac{101}{101}\right]}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}+\frac{101}{101}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{101.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=101-2\)( vì \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\ne0\))
\(=99\)
Tham khảo nhé~
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Tính giá trị biểu thức:
A=\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right).....\left(\frac{1}{99}+1\right)\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{100}{2}=50\)
Vậy \(A=50\).
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}=\frac{3.4.5.....100}{2.3.4.....99}\)
\(\Leftrightarrow A=\frac{100}{2}=50\)
mk ko viết lại đề bài đâu nhá:
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}........\frac{99}{98}.\frac{100}{99}\)
\(A=\frac{100}{2}=50\)
Xem thêm câu trả lời
Tính \(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+......+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\) ta được B=
Tính giá trị biểu thức :
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{99}\right)\cdot\left(1-\frac{1}{100}\right)\)
Ta có:
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right).\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\) \(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)
nha
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Tính \(T=\left(\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right)X\left(\frac{1}{99}+\frac{2}{98}+...+\frac{98}{2}\right)-\left(\frac{1}{99}+\frac{2}{98}+..+\frac{99}{1}\right)X\left(\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}\right)\)
tính giá trị :
A=\(\frac{2^{19}\cdot27^3-15\cdot\left(-4\right)^9\cdot9^4}{^{6^9\cdot2^{10}+\left(-12\right)}^{^{10}}}\)
B= \(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{98^2}-1\right)\cdot\left(\frac{1}{99^2}-1\right)\)
Tính B = \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+....\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)
Còn thiếu mũ 99 ở cuối cùng nha
Tính \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)
bn chắc đề đúng chứ?chổ (1/2)^99 đó,2 cái liền hả?
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\(\frac{1}{2}\)+\(\frac{1^2}{2^2}\)+\(\frac{1^3}{2^3}\)+...+\(\frac{1^{98}}{2^{98}}\)+\(\frac{1^{99}}{2^{99}}\)
=\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+...+\(\frac{1}{2^{99}}\)
=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{2^2}\)+...+\(\frac{1}{2^{98}}\)-\(\frac{1}{2^{99}}\) Còn lại tự làm nhá kết quả cuối cùng là 299-1/299
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Tính giá trị biểu thức : A = \(\left(\frac{1}{2^2}-1\right)\) . \(\left(\frac{1}{3^2}-1\right)\) . \(\left(\frac{1}{4^2}-1\right)\) .........\(\left(\frac{1}{98^2}-1\right)\) \(\left(\frac{1}{99^2}-1\right)\)