Tính:\(A=3\dfrac{1}{417}.\dfrac{1}{762}-\dfrac{1}{139}+\dfrac{761}{762}-\dfrac{4}{417.762}+\dfrac{5}{139}\)
LƯU Ý:\(3\dfrac{1}{417}\)là hỗn số
Những câu hỏi liên quan
Tính tổng M=\(3\dfrac{1}{417}\times\dfrac{1}{762}-\dfrac{1}{139}\times4\dfrac{761}{762}-\dfrac{4}{417\times762}+\dfrac{5}{139}\)
\(M=\left(\dfrac{1252}{417.762}-\dfrac{4}{417.762}\right)-\left(\dfrac{1}{139}+\dfrac{5}{139}\right).\dfrac{3809}{762}\)
\(M=\left(\dfrac{1252-4}{317754}\right)-\dfrac{6}{139}.\dfrac{3809}{762}\)
\(M=\left(\dfrac{1248}{317754}\right)-\dfrac{22854}{105918}\)
\(M=\dfrac{208}{52959}-\dfrac{3809}{17653}\)
\(M=\dfrac{3671824}{934885227}-\dfrac{201720831}{934885227}\)
\(M=\dfrac{-198049007}{934885227}\)
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Tính tổng :\(3\dfrac{1}{417}.\dfrac{1}{762}-\dfrac{1}{139}.4\dfrac{761}{762}-\dfrac{4}{761.762}+\dfrac{5}{139}\)
Tìm các số a1 , a2 , a3, ...., a9
\(\dfrac{a1-1}{9}=\dfrac{a2-2}{8}=...=\dfrac{a9-9}{1}\)và a1 +a2+a3+...+a9=90
Bài 2:
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=...=\dfrac{a_9-9}{1}=\dfrac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}=\dfrac{\left(a_1+a_2+...+a_9\right)-\left(1+2+...+9\right)}{9+8+...+1}\)
\(=\dfrac{90-45}{45}=1\)
+) \(\dfrac{a_1-1}{9}=1\Rightarrow a_1=10\)
+) \(\dfrac{a_2-2}{8}=1\Rightarrow a_2=10\)
...
+) \(\dfrac{a_9-9}{1}=1\Rightarrow a_9=10\)
Vậy \(a_1=a_2=...=a_9=10\)
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sai đề tí nha tú chỗ 4/761.762 chuyển thành 4/417.762
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Câu 1:
a, Tính M 3dfrac{1}{417}cdotdfrac{1}{762}-dfrac{1}{139}cdot4dfrac{761}{762}-dfrac{4}{417cdot762}+dfrac{5}{139}
b, Tính left(dfrac{3}{4}-81right)left(dfrac{3^2}{5}-81right)left(dfrac{3^3}{6}-81right)...left(dfrac{3^{2000}}{2003}-81right)
Câu 2: Cho left(a+3right)left(b-4right)-left(a-3right)left(b+4right)0 . Chứng minh dfrac{a}{3}dfrac{b}{4}.
Đọc tiếp
Câu 1:
a, Tính M =\(3\dfrac{1}{417}\cdot\dfrac{1}{762}-\dfrac{1}{139}\cdot4\dfrac{761}{762}-\dfrac{4}{417\cdot762}+\dfrac{5}{139}\)
b, Tính \(\left(\dfrac{3}{4}-81\right)\left(\dfrac{3^2}{5}-81\right)\left(\dfrac{3^3}{6}-81\right)...\left(\dfrac{3^{2000}}{2003}-81\right)\)
Câu 2: Cho \(\left(a+3\right)\left(b-4\right)-\left(a-3\right)\left(b+4\right)=0\) . Chứng minh \(\dfrac{a}{3}=\dfrac{b}{4}\).
Câu 2
(a+3)(b-4)-(a-3)(b+4)=0
=>ab-4a+3b-12-ab-4a+3b+12=0
=>-8a=-6b
=>a/b=3/4
=>a/3=b/4
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Tính M = 3\(\dfrac{1}{437}\) * \(\dfrac{1}{762}\) - \(\dfrac{1}{139}\) * 4\(\dfrac{761}{762}\) - \(\dfrac{4}{417\cdot762}\) + \(\dfrac{5}{139}\) .
Giúp mình giải chi tiết bài này nha, các bạn.
Đặt 761=a; 139=b
\(M=\left(3+\dfrac{1}{3b}\right)\cdot\dfrac{1}{a+1}-\dfrac{1}{b}\cdot\left(4+\dfrac{a}{a+1}\right)-\dfrac{4}{\left(a+1\right)\cdot3b}+\dfrac{5}{b}\)
\(=\dfrac{9b+1}{3b}\cdot\dfrac{1}{a+1}-\dfrac{1}{b}\cdot\dfrac{4a+4}{a+1}-\dfrac{4}{3b\left(a+1\right)}+\dfrac{5}{b}\)
\(=\dfrac{9b+1-3\left(4a+4\right)-4+15\left(a+1\right)}{3b\left(a+1\right)}\)
\(=\dfrac{9b+1-12a-12-4+15a+15}{3b\left(a+1\right)}\)
\(=\dfrac{3a+9b}{3b\left(a+1\right)}=\dfrac{3\left(a+3b\right)}{3b\left(a+1\right)}=\dfrac{a+3b}{b\left(a+1\right)}\)
\(=\dfrac{761+3\cdot139}{139\left(761+1\right)}=\dfrac{589}{52959}\)
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(3+1/417).1/762-1/139.(4+761/762)-4/417.762+5/139
Tính :\(3\frac{1}{417}.\frac{1}{762}-\frac{1}{139}.4\frac{761}{762}-\frac{4}{417.762}+\frac{5}{139}\)
Tính D=\(3\frac{1}{417}.\frac{1}{762}-\frac{1}{139}.4\frac{761}{762}-\frac{4}{417.762}+\frac{5}{139}\)
\(D=\left(3+\frac{1}{417}\right).\frac{1}{762}-\frac{1}{139}\left(4+\frac{761}{762}\right)-\frac{4}{417.762}+\frac{5}{139}\)
=\(\frac{3}{762}+\frac{1}{417.762}-\frac{4}{139}-\frac{761}{139.762}-\frac{4}{417.762}+\frac{5}{139}\)
=\(\frac{3}{762}-\frac{1}{139.762}+\frac{1}{139}-\frac{761}{139.762}=\frac{3}{762}+\frac{1}{139}\left(-\frac{1}{762}+1\right)-\frac{761}{139.762}=\)
\(\frac{3}{762}+\frac{761}{139.762}-\frac{761}{139.762}=\frac{3}{762}\)
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Tính nhanh:
\(3\frac{1}{417}.\frac{1}{762}-\frac{1}{139}.4\frac{761}{762}-\frac{4}{417.762}+\frac{5}{139}\)
Tính nhanh \(P=3\frac{1}{417}.\frac{1}{762}-\frac{1}{139}.4\frac{761}{762}-\frac{4}{417.762}+\frac{5}{139}\)
