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Ta có A= (1.2).(2.2) + (2.2).(2.3)+(2.3).(2.4)+...+(2.49).(2.50)+(2.50).(2.51)
A=4. (1.2) + 4.(2.3) + 4. (3.4) + ... + 4.(49.50) + 4. (50.51)
A=4. (1.2+2.3+3.4+4.5+ ...+ 49.50+50.51)
Tính B=1.2+2.3+3.4+4.5+ ...+ 49.50+50.51
3.B= 3. (1.2+2.3+3.4+4.5+ ...+ 49.50+50.51)
3B= 1.2.3+2.3.3+3.4.3+4.5.3+ ...+ 49.50.3+50.51.3
3B= 1.2(3-0)+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+ ...+ 49.50.(51-48)+50.51.(52-49)
3B=1.2.3-1.2.0+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+50.51.52-49.50.51
3B=50.51.52
B= 44200
A=4.44200=176800
\(\frac{x+110}{100}+\frac{x+8}{102}+\frac{x+6}{104}=\frac{x+4}{106}+\frac{x+2}{108}\)
\(\Leftrightarrow\frac{x+110}{100}+\left(\frac{x+8}{102}+1\right)+\left(\frac{x+6}{104}+1\right)=\left(\frac{x+4}{106}+1\right)+\left(\frac{x+2}{108}+1\right)\)
\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}=\frac{x+110}{106}+\frac{x+110}{108}\)
\(\Leftrightarrow\frac{x+110}{100}+\frac{x+110}{102}+\frac{x+110}{104}-\frac{x+110}{106}-\frac{x+110}{108}=0\)
\(\Leftrightarrow\left(x+110\right)\left(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}+\frac{1}{106}+\frac{1}{108}\right)=0\)
\(\Leftrightarrow x+110=0\) (vì \(\frac{1}{100}+\frac{1}{102}+\frac{1}{104}-\frac{1}{106}-\frac{1}{108}>0\))
\(\Leftrightarrow x=-110\)
Vậy \(x=-110\)
Ta có : A = 1.2 + 2.3 + 3.4 + ...... + 100.101
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 100.101.102
=> 3A = 100.101.102
=> A = 100.101.102/3
=> A = 343400
S= (1+2-3-4)-(5+6-7-8)-...-(97+98-99-100)+101+102 S= (-4 -4 -... -4) +101+102 S=(-4).25+101+102 S=-100+101+102 S=103
\(\text{B = 2.4 + 4.6 + 6.8 + .... +100.102}\)
\(\text{B = 2( 1.2 + 2.3 + 3.4 + ... + 50.51 )}\)
Đặt \(\text{A = 1.2 + 2.3 + 3.4 + ... + 50.51}\)
\(3A=1.2\left(3-0\right)+2.3\left(4-1\right)+...+50.51\left(52-49\right)\)
\(3A=1.2.3-0.1.2+2.3.4-1.2.3+...+50.51.52-49.50.51\)
\(3A=0.1.2+50.51.52\)
\(3A=50.51.52\)
\(3A=132600\)
\(A=\frac{132600}{3}=44200\)
\(\Rightarrow B=2.44200\)
\(\Rightarrow B=88400\)