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a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )
Vậy x = 1
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)
=> x + 100 = 0
=> x = -100
c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)
=> x - 100 = 0
=> x = 100
Chúc bạn học tốt
có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai
Ta có \(1\frac{1}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{6}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{28}{15}x=-\frac{56}{125}\)
<=> \(x=-\frac{2}{15}\)
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
<=> \(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
<=> x + 100 = 0
<=> x = -100
a) Số số hạng: \(\frac{\left(99-1\right)}{1}+1=99\)
Tổng: \(\frac{99+1}{2}\cdot99=4950\)
b) Số số hạng: \(\frac{\left(100-2\right)}{2}+1=50\)
Tổng: \(\frac{100+2}{2}\cdot50=2550\)
c) \(S=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(3\cdot S=1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+3\cdot4\left(5-2\right)+...+99\cdot100\left(101-98\right)\)
\(3\cdot S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+99\cdot100\cdot101-98\cdot99\cdot100\)
\(3\cdot S=99\cdot100\cdot101\)
Vậy, \(S=\frac{1}{3}\cdot99\cdot100\cdot101=333300\)
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
Ta có:
\(A=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(\Leftrightarrow3A=99.100.101\Leftrightarrow A=\frac{99.100.101}{3}=333300\)
\(B=1.2.3+2.3.4+4.5.6+...+98.99.100\)
\(\Rightarrow4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+4.5.6.\left(7-3\right)+...+98.99.100.\left(101-97\right)\)
\(\Rightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100\)
\(\Leftrightarrow4B=98.99.100.101\Leftrightarrow B=\frac{98.99.100.101}{4}=24497550\)
323400 nha !
A = 1.2 + 2.3 + 3.4 + ... + 98.99
3A = 1.2.3 + 2.3.3 + 3.4.3 + ...+ 98.99.3
3A = 1.2.3 + 2.3.(4 -1) + 3.4.(5 - 2) + ...+ 98.99.(100 - 97)
3A = (1.2.3 + 2.3.4 + 3.4.5 + ... + 98.99.100) - (0.1.2 + 1.2.3 + 2.3.4 + ... + 97.98.99)
3A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 98.99.100
3A = 98.99.100
=> A = \(\frac{98.99.100}{3}=323400\)