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(x+1) + (x+2) +...+(x+99) = 6138
=> 1+2+...+99 + 99x = 6138
=> 100 x 99 : 2 + 99x = 6138
=>4950 + 99x = 6138
=> 99x = 6138 - 4950
=> 99x = 1188
=> x = 1188 :99
=> x = 12
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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1
b;
B=1+ (7-5) + (11-9) + ...+(101-99)
B=1+2+2+..+2
B=1+25.2=51
2.
a.
ĐK : x+2 >=0 => x>=-2
\(\left|x+2\right|-x=2\\ \Rightarrow\left|x+2\right|=2+x\\ \Rightarrow\left[{}\begin{matrix}x+2=x+2\\x+2=-x-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\x=-2\end{matrix}\right.\)
Vậy x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1.2.3+2.3.4+....+99.100.101
4A=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+....+98.99.100.(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-3.4.5.2+....+98.99.100.101-98.99.100.97
4A=98.99.100.101
4A=97990200
A=97990200/4
A=24497550
B=1.2+3.4+5.6+7.8+8.9+...+999.1000
3B=1.2.3+2.3.(4-1)+3.4(5-2)+....+998.999(1001-998)
3B=1.2.3+2.3.4-2.3.1+3.4.5-3.4.2+....+998.999.1001-998.999.998
3B=999.1000.1001
3B=999999000
B=999999000/3
B=333333000
C=1+4+9+16+25+36+.....+10000
C=1^2+2^2+3^2+4^2+5^2+6^2+....+100^2
C=(1^2+3^2+5^2+.....+99^2)+(2^2+4^2+6^2+....+100^2)
C=99.100.101/6 + 100.101.102/6
C=166650 +171700
C=338350
Còn câu d bạn dựa vào câu c là làm được ngay bây h mk mỏi tay rùi ko muốn đánh nữa khi nào rảnh mk gửi công thức cho nha bây h mk bận rùi.
chúc bn học tốt
A=1.2.3+2.3.4+....+99.100.101
4.A=1.2.3.(4-0)+2.3.4.(5-1)+...+99.100.101.(102-98)
4.A=1.2.3.1-0.1.2.3+2.3.4.5-1.2.3.4+....+99.100.101.102-98.99.100.101
4.A=99.100.101.102
A=\(\frac{99.100.101.102}{4}\)
B=1.2+2.3+3.4+...+999.1000
3.B=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+.....+999.1000.(1001-998)
3.B=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+......+999.1000.1001-998.999.1000
3.B=999.1000.1001
=>B=\(\frac{999.1000.1001}{3}\)
C và D dễ lắm bạn tự làm nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)
\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)
\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)
\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)
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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Dấu chấm là nhân
a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)
\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)
CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!
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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, \(S=\frac{1}{3}\cdot99\cdot100\cdot101=333300\)
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