Tính A=1*99 + 2*98 + 3*97 + .......+ 92*2 +99*1
B=\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+......+\frac{1}{101\cdot400}\)
làm ơn giúp mình với nha mình đang vội
Tính A=\(\frac{\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}}{\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}}\)
Gọi tử số là \(C\)và mẫu số là \(D\)
Ta có:
\(A=\frac{C}{D}\)
\(C=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.102}+...+\frac{1}{101.400}\)
\(C=\frac{1}{299}\left[\left(1-\frac{1}{300}\right)\right]+\left(\frac{1}{2}-\frac{1}{301}\right)+\left(\frac{1}{3}-\frac{1}{302}\right)+...+\left(\frac{1}{101}-\frac{1}{400}\right)\)
\(C=\frac{1}{299}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)\)
\(D=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(D=\frac{1}{101}\left[\left(1-\frac{1}{102}\right)+\left(\frac{1}{2}-\frac{1}{103}\right)+\left(\frac{1}{3}-\frac{1}{104}\right)+...+\left(\frac{1}{299}-\frac{1}{400}\right)\right]\)
\(D=\frac{1}{101}\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-\frac{1}{104}-...-\frac{1}{400}\right)\)
\(D=\frac{1}{101}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)\)
\(\Rightarrow A=\frac{C}{D}=\frac{\frac{1}{299}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)}{\frac{1}{101}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\frac{1}{302}-...-\frac{1}{400}\right)}\)
\(=\frac{\frac{1}{299}}{\frac{1}{101}}=\frac{101}{299}.\)
Vậy \(A=\frac{101}{299}.\)
Tính \(\frac{A}{B}\), biết rằng :
\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}\)
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}\)
1> Tính A/B
A= \(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+.....+\frac{1}{101\cdot400}\)
B = \(\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+.....+\frac{1}{299\cdot400}\)
2> Cho a,b thộc N*
CMR :\(\frac{a}{b}+\frac{b}{a}\)lớn hơn hoặc bằng 2
nhanh nha đang cần gấp
thanks so much
mk chỉ nghĩ đc bài 2 thui mong bn thông cảm.
2) Ta có\(\frac{a}{b}+\frac{b}{a}=\frac{a.a+b.b}{b.a}=\frac{a+b}{1}\)
Mà theo đề bài a,b\(\inℕ^∗\)
=> \(a,b\ge1\Rightarrow a+b\ge2\Rightarrow\frac{a+b}{1}\ge2\)
Thấy đúng thì tk nha
\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...\frac{1}{101\cdot400}\)=
Tính \(\frac{A}{B}\) biết :
\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}\)
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}\)
Ta có :
\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+...+\frac{1}{101\cdot400}\)
\(\Rightarrow299A=\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+...+\frac{299}{101\cdot400}\)
\(\Rightarrow299A=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}=C\)
\(\Rightarrow A=\frac{C}{299}\)
Lại có :
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}\)
\(\Rightarrow101B=\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+...+\frac{101}{299\cdot400}\)
\(\Rightarrow101B=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\)
\(\Rightarrow101B=\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+...+\frac{1}{400}\right)\)
\(\Rightarrow B=\frac{C}{101}\)
\(\Rightarrow\frac{A}{B}=\frac{101}{299}\)
Tính \(\frac{A}{B}\), biết rằng :
\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}\)
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}\)
299A=\(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+...+\frac{299}{101\cdot400}\)
299A=\(1-\frac{1}{300}+\frac{1}{300}-\frac{1}{301}-...-\frac{1}{101}+\frac{1}{101}-\frac{1}{400}\)
299A=\(1-\frac{1}{400}\)
299A=\(\frac{399}{400}\)
A=\(\frac{399}{400}:299\)
A=\(\frac{119310}{400}\)
tương tự tính câu B
Ta có: \(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
\(\Rightarrow A=\frac{1}{399}.\left(\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101.400}\right)\)
\(\Rightarrow A=\frac{1}{299.}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(\Rightarrow A=\frac{1}{299}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+..+\frac{1}{401}\right)\right]\)
Mặt khác \(B=\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)
\(\Rightarrow B=\frac{1}{101}.\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)
\(\Rightarrow B=\frac{1}{101}.\left[\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{203}+...+\frac{1}{400}\right)\right]\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}.\left[\left(1+\frac{1}{2}+..+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}{\frac{1}{101}.\left[\left(1+\frac{1}{2}+....+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}\)
\(=\frac{1}{299}:\frac{1}{101}=\frac{101}{299}\)
\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)chứng minh rằng \(\frac{1}{299}\left(\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right)\)=\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)
A=\(\frac{1}{1\cdot300}\)+\(\frac{1}{2\cdot301}\)+...+\(\frac{1}{101\cdot400}\)tinh \(\frac{A}{B}\)
B=\(\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}\)
các bạn ơi giúp mình với nha |
A=\(\frac{101+100+99+98+97+.....+3+2+1}{101-100+99-98+....+3-2+1}\)
Bài làm:
\(A=\frac{101+100+99+...+3+2+1}{101-100+99-98+...+3-2+1}=\frac{\left(101+1\right).101\div2}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}\)
\(A=\frac{5151}{1+1+...+1+1}=\frac{5151}{51}=101\)(51 số hạng 1)