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\(\cdot\cdot\cdot+36x^2+54x+\cdot\cdot\cdot=\left(\cdot\cdot\cdot+\cdot\cdot\cdot\right)^3\)
Khôi phục các đa thức sau:
\(1,\left(2x-\cdot\cdot\cdot\right)^2=\cdot\cdot\cdot-6xy+\cdot\cdot\cdot\)
\(2,\left(\cdot\cdot\cdot+2y\right)^3=\cdot\cdot\cdot+6x^2y+12xy^2+\cdot\cdot\cdot\)
\(3,\left(\cdot\cdot\cdot+5y\right)^2=9x^2+\cdot\cdot\cdot+\cdot\cdot\cdot\)
\(4,\left(x-2y\right)\left(x^2+\cdot\cdot\cdot+4y^2\right)=x^3-8y^3\)
Khôi phục các đa thức sau:
1,\(\left(2x-\dfrac{3}{2}y\right)^2=4x^2-6xy+\dfrac{9}{4}y^2\)
2,\(\left(x+2y\right)^3=x^3+6x^2y+12xy^2+8y^3\)
3,\(\left(3x+5y\right)^2=9x^2+30xy+25y^2\)
4,\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=x^3-8y^3\)
cho f(n)=(n2 + n +1 )2 +1 với n thuộc N* . Đặt \(p_n=\frac{f_{\left(1\right)}\cdot f_{\left(3\right)}\cdot f_{\left(5\right)}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot f_{\left(2n-1\right)}}{f_{\left(2\right)}\cdot f_{\left(4\right)}\cdot f_{\left(6\right)}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot f_{\left(2n\right)}}\)
chứng minh rằng : P1 + P2 +P3 +................+ Pn <1/2
\(B=\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{1+2+3+....+100}\right)\)
Thu gọn các đơn thức sau, xác định hệ số, phần biến và bậc của đơn thức
A=\(\left(\dfrac{-3}{7}\cdot x^3\cdot y^2\right)\cdot\left(\dfrac{-7}{9}\cdot y\cdot z^2\right)\cdot\left(6\cdot x\cdot y\right)\)
B= \(-4\cdot x\cdot y^3\cdot\left(-x^2\cdot y\right)^3\cdot\left(-2\cdot x\cdot y\cdot z^3\right)^2\)
HELP ME
\(A=\left(\dfrac{-3}{7}.x^3.y^2\right).\left(\dfrac{-7}{9}.y.z^2\right).\left(6.x.y\right)\)
\(A=\left(\dfrac{-3}{7}x^3y^2\right).\left(\dfrac{-7}{9}yz^2\right).6xy\)
\(A=\left(\dfrac{-3}{7}.\dfrac{-7}{9}.6\right).\left(x^3.x\right)\left(y^2.y.y\right).z^2\)
\(A=2x^4y^4z^2\)
\(B=-4.x.y^3\left(-x^2.y\right)^3.\left(-2.x.y.z^3\right)^2\)
\(B=\left[\left(-4\right).\left(-2\right)\right].\left(x.x^6.x^2\right)\left(y^3.y^3.y^2\right)\left(z^6\right)\)
\(B=8x^7y^{y^8}z^6\)
tính \(a=1+\frac{1}{2}\cdot\left(1+2\right)+\frac{1}{3}\cdot\left(1+2+3\right)+\cdot\cdot\cdot+\frac{1}{32}\cdot\left(1+2+3+\cdot\cdot\cdot+32\right)\)
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{32}\left(1+2+3+...+32\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{32}.\frac{32.\left(32+1\right)}{2}\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+....+\frac{32+1}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{33}{2}\)
\(\frac{2+3+4+....+33}{2}\)
\(=\frac{\frac{33\left(33+1\right)}{2}-1}{2}=280\)
B=\(\left(1-\dfrac{1}{1+2}\right)\). \(\left(1-\dfrac{1}{1+2+3}\right)\).....\(\left(1-\dfrac{1}{1+2+...+100}\right)\)
B=\(\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot...\cdot\left(1-\dfrac{1}{\left(1+100\right)\cdot100:2}\right)\)
B=\(\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{101\cdot100:2-1}{101\cdot100:2}\)
B=\(\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot...\cdot\dfrac{\left(101.100:2-1\right).2}{101.100}\)
B=\(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}\cdot...\cdot\dfrac{99.102}{100.101}\)
B=\(\dfrac{1.2.3.4.....99}{3.4.5....100}.\dfrac{4.5.6.....102}{3.4.5.....101}\)
B=\(\dfrac{2}{100}\).\(\dfrac{102}{3}\)
B=\(\dfrac{17}{25}\)
G=\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\cdot\frac{50^2}{49.51}\)
H=\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{10}{7}\right)\)
Giúp mình vs
G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)
=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)
=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)
=> G = \(\frac{2.50}{1.51}\)
=> G = \(\frac{100}{51}\)
\(G=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}\)
\(=\frac{\left(2.3.4.....50\right).\left(2.3.4.....50\right)}{\left(1.2.3.....49\right).\left(3.4.5.....51\right)}\)
\(=\frac{50.2}{51}=\frac{100}{51}\)
\(H=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{7}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....0.....\left(1-\frac{10}{7}\right)\)
\(=0\)
TÍNH
\(C=\left(1+\frac{2}{3}\right)\cdot\left(1+\frac{2}{5}\right)\cdot\left(1+\frac{2}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1+\frac{2}{2015}\right)\cdot\left(1+\frac{2}{2017}\right)\)
\(D=\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{6}\right)\cdot\left(1-\frac{1}{10}\right)\cdot\left(1-\frac{1}{15}\right)\cdot\cdot\cdot\cdot\left(1-\frac{1}{780}\right)\)
\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)
\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)
Nhận xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)nên:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)
Tính nhanh :\(\left(3-\frac{1}{18}\cdot\right)\cdot\left(3-\frac{2}{18}\right)\cdot\left(3-\frac{3}{18}\right)\cdot\cdot\cdot\cdot\left(3-\frac{2019}{18}\right)\)