Bài làm

       \(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)

\(=\frac{84}{12}+\left(\frac{7}{12}-\frac{6}{12}+\frac{36}{12}\right)-\left(\frac{1}{12}+\frac{60}{12}\right)\)

\(=\frac{84}{12}+\frac{37}{12}-\frac{61}{12}\)

\(=\frac{60}{12}\)

\(=5\)

# Chúc bạn học tốt #

a: \(x^2\cdot x-2x^3=x^3-2x^3=-x^3\)

b: \(6x^2y\cdot3xy-2y^2\left(x+y\right)=18x^3y^2-2xy^2-2y^3\)

c: \(\left(4x^2+5x-1\right)\left(2x^3-3x\right)\)

\(=8x^5-12x^3+10x^4-15x^2-2x^3+3x\)

\(=8x^5+10x^4-14x^3-15x^2+3x\)

hihi cho mình đi

Ta có: \(M=1\cdot2014+2\cdot2013+\cdots+2014\cdot1\)

\(=2\left(1\times2014+2\times2013+\cdots+1007\times1008\right)\)

\(=2\left\lbrack1\times\left(2015-1\right)+2\times\left(2015-2\right)+\cdots+1007\times\left(2015-1007\right)\right\rbrack\)

\(=2\cdot\left\lbrack2015\times\left(1+2+\cdots+1007\right)-\left(1^2+2^2+\cdots+1007^2\right)\right\rbrack\)

\(=2\cdot\left\lbrack2015\times1007\times\frac{1008}{2}-\frac{1007\times\left(1007+1\right)\times\left(2\times1007+1\right)}{6}\right\rbrack\)

\(=2\cdot\left\lbrack2015\times1007\times504-1007\times168\times2015\right\rbrack=2\times2015\times1007\times168\left(3-1\right)=4\times168\times2015\times1007\)

\(N=1+\left(1+2\right)+\cdots+\left(1+2+\cdots+2014\right)\)

\(\) \(=\frac{1\times2}{2}+\frac{2\times3}{2}+\cdots+\frac{2014\times2015}{2}\)

\(=\frac12\times\left(1\times2+2\times3+\cdots+2014\times2015\right)\)

\(=\frac12\times\left\lbrack1\times\left(1+1\right)+2\times\left(2+1\right)+\cdots+2014\times\left(2014+1\right)\right\rbrack\)

\(=\frac12\times\left\lbrack\left(1\times1+2\times2+\cdots+2014\times2014\right)+\left(1+2+\cdots+2014\right)\right\rbrack\)

\(=\frac12\times\left\lbrack\frac{2014\times\left(2014+1\right)\times\left(2\times2014+1\right)}{6}+\frac{2014\times2015}{2}\right\rbrack\)

\(=\frac12\times\left\lbrack1007\times2015\times1343+1007\times2015\right\rbrack=\frac12\times1007\times2015\times\left(1343+1\right)\)

=1007x2015x672

=4x168x2015x1007

Do đó: M=N

\(P=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\)

\(P=\frac{\left(1+2\right).2:2-1}{\left(1+2\right).2:2}.\frac{\left(1+3\right).3:2-1}{\left(1+3\right).3:2}.\frac{\left(1+4\right).4:2-1}{\left(1+4\right).4:2}...\frac{\left(1+2014\right).2014:2-1}{\left(1+2014\right).2014:2}\)

\(P=\frac{2}{2.3:2}.\frac{5}{3.4:2}.\frac{9}{4.5:2}...\frac{2029104}{2014.2015:2}\)

\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2013.2016}{2014.2015}\)

\(P=\frac{1.2.3...2013}{2.3.4...2014}.\frac{4.5.6...2016}{3.4.5...2015}\)

\(P=\frac{1}{2014}.\frac{2016}{3}=\frac{1}{2014}.672=\frac{336}{1007}\)

biết vậy mà vẫn đòi lấy ảnh! ok!

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