có 6 cách

chắc là 6 cách

\(Q=x^2(x+1)-3xy(x-y+1)-y^2(y-1)+xy\\=x^3+x^2+3xy(y-x)-3xy-y^3+y^2+xy\\=-(y^3-x^3)+3xy(y-x)+x^2-2xy+y^2\\=-(y-x)^3-3xy(y-x)+3xy(y-x)+(y-x)^2\\=-11^3+11^2=-1210\)

thiếu dữ kiện rồi. 2 số chẵn hay 2 số lẻ?

Hiệu 2 số đó là :

\(50x2+1=101\)

Số bé là :

\(\left(1221-101\right):2=560\)

Số lớn là :

\(1221-560=761\)

x=23

\(x-12=11\)

\(x=11+12\)

\(x=23\)

sai đề

x - 12 + 13 = ??? đề thiếu dữ kiện cậu ơi!

\(Z=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{49\times51}\)

\(=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{49\times51}\right)\)

\(=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}\times\dfrac{16}{51}=\dfrac{8}{17}\)

\(Z=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+...+\dfrac{3}{49x51}\\ =\dfrac{3}{2}x\left(\dfrac{2}{3x5}+\dfrac{2}{5x7}+\dfrac{2}{7x9}+...+\dfrac{2}{49x51}\right)\\ =\dfrac{3}{2}x\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\\ =\dfrac{3}{2}x\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\\ =\dfrac{3}{2}x\dfrac{16}{51}=\dfrac{8}{17}\)

$(17+7).\{460-[10.(64-4^3):2]\}$

$=24.\{460-[10.(64-64):2]\}$

$=24.(460-0)$

$=24.460=11040$

11040

\(S=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+...+\dfrac{2}{99\times100}\)

\(=2\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\right)\)

\(=2\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\times\left(1-\dfrac{1}{100}\right)\)

\(=2\times\dfrac{99}{100}=\dfrac{99}{50}\)

CT: \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a};\left(n\ne0;n\ne-a\right)\)

\(S=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+...+\dfrac{2}{99\times100}\\ \dfrac{S}{2}=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\\ \dfrac{S}{2}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \dfrac{S}{2}=1-\dfrac{1}{100}=\dfrac{99}{100}\\ S=\dfrac{99}{100}\times2=\dfrac{99}{50}\)

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

CT: \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\) (\(n\ne0;n\ne-a\))

\(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{9x10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{9}{10}\)