a=2.9.8+3.12.10+4.15.12+......+98.297.200 / 2.3.4+3.4.5+4.5.6+.....+98.99.100
cho \(a=\dfrac{2.9.8+3.12.10+4.15.12+...+98.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}\)
tính a2
\(a=\dfrac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297.200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...98\cdot99\cdot100}\\ =\dfrac{2\cdot3\cdot3\cdot4\cdot2+3\cdot3\cdot4\cdot2\cdot5+4\cdot3\cdot5\cdot2\cdot6+...+98\cdot99\cdot3\cdot100\cdot2}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot2\cdot3+...+98\cdot99\cdot100}\\ =\dfrac{3\cdot2+3\cdot2+6+3\cdot2}{0}=\dfrac{24}{0}=0\)
\(a=\dfrac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\\ =\dfrac{\left(2\cdot3\right)\left(2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\right)}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\\ =6\\ a^2=6^2=36\)
tính S= \(\frac{2.9.8+3.12.10+4.15.12+............+98.297.200}{2.3.4+3.4.5+4.5.6+............+98.99.100}\)
GIÚP MÌNH VỚI
\(\frac{2.9.8+3.12.10+4.15.12+...+98.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}=\frac{3.2.\left(2.3.4+3.4.5+4.5.6+...+98.99.100\right)}{2.3.4+3.4.5+4.5.6+...+98.99.100}=6\)
S=3+3/2+3/2²+......+3/2^9
A=2.9.8+3.12.10+4.15.12+....+98.297.200/23.4+3.4.5+4.5.6+…98.99.100 tinh a²
SAI ĐỀ RỒI BẠN. SỬA 23=2.3
\(\frac{2.9.8+3.12.10+4.15.12+...+98.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}\)
\(\frac{1.2.3.\left(2.3.4+3.4.5+4.5.6+...+98.99.100\right)}{\left(2.3.4+3.4.5+4.5.6+...+98.99.10\right)}\)
\(=6\)
VẬYa2=62=36
tính \(a^2\)cho a= \(\frac{2.9.8+3.12.10+4.15.12+...+98.297.200}{23.4+3.4.5+4.5.6+...+98.99.100}\)
Ta có:
\(2.9.8+3.12.10+...+98.297.200\)
\(=2.3.4.3.2+3.4.5.3.2+...+98.99.100.3.2\)
\(=6.\left(2.3.4+3.4.5+...+98.99.100\right)\)
Thế lại bài toán (sửa đề luôn)
\(a=\frac{2.9.8+3.12.10+...+98.297.200}{2.3.4+3.4.5+...+98.99.100}\)
\(=\frac{6.\left(2.3.4+3.4.5+...+98.99.100\right)}{2.3.4+3.4.5+...+98.99.100}=6\)
\(\Rightarrow a^2=6^2=36\)
Mình thấy ý kiến của bạn alibaba nguyễn đúng rồi
cho a=\(\frac{2.9.8+3.12.10+4.15.12+...+98.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}\)
Hỏi a có phải là nghiệm của đa thức P(x)=\(^{x^2}\)-12x+35 không ? Vì sao
- Ai trả lời nhanh mình cho like trước -
Cho \(a=\frac{2.9.8+3.12.10+4.15.12+...+99.297.200}{2.3.4+3.4.5+4.5.6+...+98.99.100}\)
Tính a2
\(A=\frac{2.9.8+12.10.4+4.15.12+.......+98.297.200}{2.3.4+3.4.5+4.5.6+.......+98.99.100}\)
Tính A2
GÍUP MÌNH VỚI NHÉ
in1. So sánh A và B, biết:
A=20112011+2/20112011-1
B=20112011/20112011-3
2.tính
(217+417)(314-312)(24-42)/152+53
3.chứng tỏ rằng:
1/22+1/32+1/42+...+1/20102<1
4. Tính a2, biết:
a= 2.9.8+3.12.10+4.15.12+...+98.297.200/2.3.4+3.4.5+4.5.6+...+98.99.100
Bài 3:
Ta có:
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(...\)+\(\frac{1}{2010^2}\)<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2009.2010}\)
Xét:\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.....+\(\frac{1}{2009+2010}\)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)=\(1-\frac{1}{2010}\)<1
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2010^2}< 1\)
\(\)Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< 1\)
Tính:
1.2.3+2.3.4+3.4.5+4.5.6+...+97.98.99+98.99.100
Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101
4A=98.99.100.101
=>A=98.99.100.101/4
TICK ĐÚNG GIÚP MÌNH Ặ