tinh gia tri bieu thuc: B=1*2*3+2*3*4+3*4*5+...+17*18*19
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A
\(\frac{3}{5}\div\frac{4}{5}+\frac{1}{2}\times\frac{2}{3}\)
\(=\frac{3}{4}+\frac{1}{3}\)
\(=\frac{13}{12}\)
B
\(\frac{5}{4}\times x=\frac{3}{8}+\frac{1}{4}\)
\(\frac{5}{4}\times x =\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{5}{4}\)
\(x=\frac{4}{8}=\frac{1}{2}\)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
B = 1.2.3 + 2.3.4 + 3.4.5 + .... + 17.18.19
<=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 17.18.19.4
<=> 4B = 1.2.3.(4 - 0) + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + .... + 17.18.19.(20 - 16)
<=> 4B = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .... + 17.18.19.20 - 16.17.18.19
<=> 4B = 17.18.19.20 = 116280
<=> B = 116280 : 4 = 29070
B = 1.2.3 + 2.3.4 + 3.4.5 + .... + 17.18.19
<=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 17.18.19.4
<=> 4B = 1.2.3.(4 - 0) + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + .... + 17.18.19.(20 - 16)
<=> 4B = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .... + 17.18.19.20 - 16.17.18.19
<=> 4B = 17.18.19.20 = 116280
<=> B = 116280 : 4 = 29070