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243 + (345 - x) = 500
345 - x = 500 - 243
345 - x = 257
x = 345 - 257
x = 88
a) (2x-6)3 = (2x-6)2018
=> (2x-6)3 - (2x-6)2018 = 0
(2x-6)3.[1-(2x-6)2015 ] = 0
=> (2x-6)3 = 0 =>...
1 - (2x-6)2015 = 0 => (2x-6)2015 = 1 => ...
b) (2x-1)3 = 27 = 33
=> 2x - 1 = 3
=> ...
c) (x + 1) + (x+2) + (x+3) + ...+ (x+100) = 5750
x.100 + (1+2+3+...+100) = 5750
x.100 + [(1+100).100:2] = 5750
x.100 + 5050 = 5750
x.100 = 700
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\(\frac{ }{ }\)
\(=100x+\left(1+2+3+4..+100\right)=5750\)
\(=100x+5050=5750\)
\(=100x=5750-5050=700\)
\(=x=700:100=7\)
\(x=7\)
(x + 1) + (x + 2) + (x + 3) + .... + (x + 100) = 5750
=> 100x + (1 + 2 + 3 + ... + 100) = 5750
=> 100x + 5050 = 5750
=> 100x = 5750 - 5050
=> 100x = 700
=> x = 700 : 100
=> x = 7
a, 52x - 3 - 2.52 = 52.3
=> 52x - 3 - 50 = 75
=> 52x - 3 = 125
=> 52x - 3 = 53
=> 2x - 3 = 3
=> 2x = 0
=> x = 0
vậy_
b, (x + 1) + (x + 2) + ....+ (x + 100) = 5750
=> x + 1 + x + 2 + ... + x + 100 = 5750
=> (x + x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 5750
=> 100x + 5050 = 5750
=> 100x = 700
=> x = 7
vậy_
Bài giải
a, \(1075\cdot\left(x-3\right)\cdot\left(x-1\right)=0\)
\(\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{3\text{ ; }1\right\}\)
b, \(2\cdot\left(x-7\right)+3\cdot\left(x+1\right)\)
\(=2x-14+3x+3\)
\(=5x-11\)
c, \(x+1+x+2+...+x+100=5750\)
\(\left(x+x+...+x\right)+\left(1+2+...+100\right)=5750\)
\(100x+\left(100-1+1\right)\cdot\left(100+1\right)\text{ : }2=5750\)
\(100x+100\cdot101\text{ : }5=5750\)
\(100x+50\cdot101=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(x=700\text{ : }100\)
\(x=7\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
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!
\(a,345-x=257\)
\(x=88\)
\(b,7.x=115\)
\(x=\frac{115}{7}\)
\(c,\left(x+x+...+x\right)+\left(1+2+...+100\right)=5750\)
\(x\times100+5050=5750\)
\(x\times100=700\)
\(x=7\)