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Sửa đề : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+....+\frac{\left(x+1\right)-x}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{1}-\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{50}\)
\(\Leftrightarrow x=50-1=49\)
Sửa đề: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
(=) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
(=)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{98}{100}\)
(=)\(1-\frac{1}{x+1}=\frac{98}{100}\)
(=)\(\frac{1}{x+1}=1-\frac{98}{100}\)
(=)\(\frac{1}{x+1}=\frac{1}{50}\)=> \(x+1=50\)
\(x=50-1\)
\(x=49\)
T_i_c_k cho mình nha,thanks you so much!

20 - 2.(x+4) = 4
2.(x+4) = 20 - 4
2.(x+4) = 16
x+4 = 16 : 2
x+4 = 8
x = 8 - 4
x = 4
2(x-1):3=43 + 23
2(x-1):3= 64 + 8
2(x-1):3= 72
2(x-1) = 72 x 3
2(x-1) = 216
x - 1 = 216 : 2
x - 1 = 108
x = 108 + 1
x = 109
Hok tốt !
20-2(x+4)=4
2(x+4)=20-4
2(x+4)=16
x+4=16:2
x+4=8
x=8:4
x=2
Vậy x=2
12(x-1):3=43+23
12(x+1):3=64+8
12(x+1):3=72
12(x+1)=72.3
12(x+1)=216
x+1=216:12
x+1=18
x=18-1
x=17
Vậy x=17

1.
2|x-6|+7x-2=|x-6|+7x
2|x-6| - |x-6|=7x-(7x-2)
|x-6| = 2
=>x-6 = +2
*x-6=2 *x-6 = -2
x =2+6 x = (-2)+6
x =8 x = 4
2.
|x-5|-7(x+4)=5-7x
|x-5|-7x-28 =5-7x
|x-5|-28 =5-7x+7x
|x-5|-28 = 5
|x-5| = 5+28
|x-5| = 33
=>x-5 = +33
*x-5=33 *x-5=-33
x =38 x = -28
3.
3|x+4|-2(x-1)=7-2x
3|x+4|-2x+2 =7-2x
3|x+4|-2 =7-2x+2x
3|x+4|-2 =7
3|x+4| =7+2
3|x+4| = 9
|x+4| =9:3
|x+4| = 3
=>x+4 = +3
*x+4=3 *x+4=-3
x =-1 x = -7

S=30+32+34+36+...+3200
6S=32+34+36+...+3202
6S-S=(32+34+36+...+3202)-(1+32+34+...+3200)
5S=1+(32-32)+(34-34)+...+(3200-3200)+3202
S=(3200+1):5\(\frac{ }{ }\)

\(\left(158-x\right):7=20\)
\(158-x=20\cdot7\)
\(158-x=140\)
\(x=158-140\)
\(x=18\)
\(231-\left(x-6\right)=\frac{1339}{13}\)
\(231-\left(x-6\right)=103\)
\(231-103=x-6\)
\(x-6=128\)
\(x=134\)
\(\frac{81+2x}{5}=25\)
\(\left(81+2x\right):5=25\)
\(81+2x=25\cdot5\)
\(81+2x=125\)
\(2x=125-81\)
\(2x=44\)
\(x=22\)
1.(158-x):7=20
158-x =20.7
158-x =140
x =158-140
x =18
2.231-(x-6)=1339/13
231-(x-6)=103
x-6=231-103
x-6=128
x =128+6
x =134
3.(81+2.x)/.5=25
81+2.x =25.5
81+2.x =125
2.x =125-81
2.x =44
x =44:2
x =22
k mik nha

bạn giải thích đầy đủ hộ mình r mình tk bn nha. Nếu ko thì chịu nha

\(A=7+9+11+...+99\)(có (99-7):2+1=47 số hạng)
\(A=\frac{47\left(99+7\right)}{2}=\frac{47\cdot106}{2}=47\cdot53=2491\)
\(B=10+15+20+...+200\)(có (200-10):5+1=39 số hạng)
\(B=\frac{39\left(200+10\right)}{2}=\frac{39\cdot210}{2}=39\cdot105=18585\)
Bài giải
\(x\left(x+1\right)=2+4+6+...+20=\frac{\left[\left(20-2\right)\text{ : }2+1\right]\left(20+2\right)}{2}=10\cdot11=\left(-11\right)\cdot\left(-10\right)\)
\(\Rightarrow\text{ }x\in\left\{-11\text{ ; }10\right\}\)
Ta có : x.(x+1)=2+4+6+...+20
<=>x.(x+1)=(20+2).[(20-2):2+1]:2
x.(x+1)=22.[18:2+1]:2
x.(x+1)=22.[9+1]:2
x.(x+1)=22.10:2
x.(x+1)=110
x.(x+1)=2.5.11
x.(x+1)=10.11
=> x=10
Vậy x cần tìm là 10