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Câu 1 dễ thôi. Bạn tính tử, rồi tính mẫu sao cho khi phân phối ở cả tử và mẫu đều có phần thừa số có thể rút gọn cho nhau. Giờ mik bận quá nên ko thể giải dầy đủ. Thông cảm nha...
Câu 2: Cũng ko khó lắm đâu:
S=\(\frac{1}{1}\) - \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{7}\)+...+\(\frac{1}{n}\)-\(\frac{1}{n+3}\)
=1-\(\frac{1}{n+3}\)<1.
Vậy: S<1
Để làm dc bài sau, bạn nhớ giùm mik công thức: \(\frac{a}{b.\left(b+a\right)}\)=\(\frac{1}{b}\)-\(\frac{1}{b+a}\)
Câu 3: Đặt \(A=\frac{2003.2004-1}{2003.2004}\), \(B=\frac{2004.2005-1}{2004.2005}\)ta có:
\(A=\frac{2003.2004}{2003.2004}\)-\(\frac{1}{2003.2004}\)=1-\(\frac{1}{2003.2004}\)
\(B=\frac{2004.2005}{2004.2005}\)-\(\frac{1}{2004.2005}\)=1-\(\frac{1}{2004.2005}\)
Vì 2003.2004<2004.2005=>\(\frac{1}{2003.2004}\)>\(\frac{1}{2004.2005}\)
=>1-\(\frac{1}{2003.2004}\)<1-\(\frac{1}{2004.2005}\)
Vậy: \(\frac{2003.2004-1}{2003.2004}\)< \(\frac{2004.2005-1}{2004.2005}\)
Nhớ cho mik nha. Thanks
1, mình không ghi đề nha
A= \(\frac{1.1+1.1+1.1}{3+3.3+3.3+3}\)
A=\(\frac{1.3}{9.3}\)
A=\(\frac{1}{9}\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n\left(n+3\right)}\)
\(\Rightarrow S=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{\left(n+3\right)-n}{n\left(n+3\right)}\)
\(\Rightarrow S=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{n+3}{n\left(n+3\right)}-\dfrac{n}{n\left(n+3\right)}\)
\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{n}-\dfrac{1}{n+3}\)
\(\Rightarrow S=1-\dfrac{1}{n+3}< 1\Rightarrow S< 1\)
Vậy S < 1
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...................+\dfrac{3}{n\left(n+1\right)}\)
\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.............+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow S=1-\dfrac{1}{n+1}< 1\)
\(\Rightarrow S< 1\rightarrowđpcm\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n.\left(n+1\right)}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}\)\(< 1\)
\(\Leftrightarrow S< 1\)
tik cho mik nhé
a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
A = \(\dfrac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
= \(\dfrac{7.9+7.2.9.3+7.3.9.4}{21.27+21.2.27.3+21.3.27.4}\)
= \(\dfrac{\left(7.9\right)\left(2.3+3.4\right)}{\left(21.27\right)\left(2.3+.3.4\right)}\)
= \(\dfrac{1}{9}\)
@Nguyễn Ngọc Gia Hân
a) A=\(\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2004}=1-\dfrac{1}{2003.2004}\)
B = \(\dfrac{2004.2005-1}{2004.2005}=\dfrac{2004.2005}{2004.2005}-\dfrac{1}{2004.2005}=1-\dfrac{1}{2004.2005}\)
Vì \(\dfrac{1}{2003.2004}>\dfrac{1}{2004.2005}\)
\(\Rightarrow1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\)
Vậy A < B
b) \(\left(3X-2^4\right).7^5=2.7^6.\dfrac{1}{2009^0}\)
\(\left(3X-2^4\right).7^5=2.7^6.1\)
\(\left(3X-2^4\right).7^5=2.7^6\)
\(\left(3X-2^4\right).=2.7^6:7^5\)
\(3X-2^4=2.7\)
\(3X-16=14\)
\(3X=16+14=30\)
\(X=30:3=10\)
Vậy X = 10
1/ \(A=\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2003.2004}=1-\dfrac{1}{2003.2004}\)
\(B=\dfrac{2004.2005-1}{2004.2005}=\dfrac{2004.2005}{2004.2005}-\dfrac{1}{2004.2005}=1-\dfrac{1}{2004.2005}\)
Vì \(1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\Leftrightarrow A< B\)
2/ \(\left(3x-2^4\right).7^5=2.7^6.\dfrac{1}{2009^0}\)
\(\Leftrightarrow\left(3x-2^4\right).7^5=2.7^6.1\)
\(\Leftrightarrow3x-2^4=2.7^6:7^5\)
\(\Leftrightarrow3x-2^4=2.7\)
\(\Leftrightarrow3x-16=14\)
\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=10\left(tm\right)\)
Vậy ..