24(15-4)+4(24-15)

=24*15-24*4+4*24-4*15

=24*15-4*15=15(24-4)=15*20=300

199(15-17)-199(-17+5)

=199(15-17+17-5)

=199*10=1990

a) Đặt A=\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{98\cdot99\cdot100}\)

\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+....+\frac{2}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+.....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)

2A=\(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}=\frac{4949}{9900}\) =>A=\(\frac{4949}{9900}\div2=\frac{4949}{19800}\)

Đặt B=\(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29\cdot30}\)

=>3B=\(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+....+\frac{3}{27\cdot28\cdot29\cdot30}\)

3B=\(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\)

3B=\(\frac{1}{1\cdot2\cdot3}-\frac{1}{28\cdot29\cdot30}=\frac{1353}{8120}\)

=>B=\(\frac{1353}{8120}\div3=\frac{451}{8120}\)

Ta có : A-3x=B=>3x=A-B=\(\frac{4949}{19800}\)-\(\frac{451}{8120}\)\(\approx\frac{1}{5}\)=>x=\(\frac{1}{5}\div3\)=\(\frac{1}{15}\)

Ta có : \(1^3+2^3+3^3+....+n^3\)

=\(\left(1+2+3+4+...+n\right)^2\)

=\(\left(\frac{n\left(n+1\right)}{2}\right)^2\) (đpcm)

Do x>y>0 nên x+y\(\ne0\)

Ta có \(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(x+y\right)}=\frac{x^2-y^2}{x^2+2xy+y^2}\) (1)

Mặt khác ,do x,y>0 nên \(x^2+2xy+y^2>x^2+y^2\)

Vậy: \(\frac{x^2-y^2}{x^2+2xy+y^2}< \frac{x^2-y^2}{x^2+y^2}\) (2)

Từ (1),(2) ta suy ra : \(\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)

1. Ta có \(x^3+3x^2+x+3=0\)

\(\Leftrightarrow\left(x^3+3x^2\right)+\left(x+3\right)=0\)

\(\Leftrightarrow x^2\left(x+3\right)+\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)

Nếu x+3=0 =>x=-3

Nếu \(x^2+1=0\) =>x\(=\varnothing\) (vì \(x^2+1>0\))

Vậy x=-3

Ta có \(x+\frac{1}{x}=5\)

\(\Rightarrow\left(x+\frac{1}{x}\right)^{^{ }2}=5^2=25\)

\(\Leftrightarrow x^{^{ }2}+2\cdot x\cdot\frac{1}{x}+\frac{1}{x^2}=25\)

\(\Leftrightarrow x^2+\frac{1}{x^2}=25-2\cdot x\cdot\frac{1}{x}=25-2=23\)

Vậy \(x^2+\frac{1}{x^2}=23\)

Ta có h(h+1)(h+2)(h+3)=h(h+3)(h+2)(h+1)=(\(h^2+3h\))(\(h^2+3h+2\))

Đặt \(h^2+3h=x\) ta được:

A=x(x+2)=\(x^2+2x=x^{^{ }2}+2x+1-1=\left(x+1\right)^2-1\ge-1\)

Vậy GTNN của A là -1

Bác hồ sinh năm 1890. Bác hồ sinh vào thế kỉ 19.Bác ra đi tìm đường cứu nước 1911 năm đó thuộc thế kỉ 20

b) Đặt B=25-(-37)*(-29)*(-154)*21

=25+37*(-29)*(-154)*21

=25+37*29*154*21 (tích 2 dấu - bằng dấu trừ

Vậy B>0