x+(x+2)+(x+4)+........+(x+100)=2907
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g)\(2907\left(2x+1\right)=8721\)
⇔\(2x+1=3\)
⇔\(2x=2\)
⇔\(x=1\)
h)\(\left(4x-16\right):1905=60\)
⇔\(4x-16=114300\)
⇔\(4x=114316\)
⇔\(x=28579\)
i)\(23+3x=5^6:5^3\)
⇔\(23+3x=5^3\)
⇔\(23+3x=125\)
⇔\(3x=102\)
⇔\(x=34\)
k)\(219-7\left(x+1\right)=100\)
⇔\(7\left(x+1\right)=119\)
⇔\(x+1=17\)
⇔\(x=16\)
a) ( 2x + 1 ) . 2907 = 8 721
2x + 1 = 3
2x = 2
x = 1
x + 1234 - 532 = 2907
x + 1234 = 2907 + 532
x + 1234 = 3439
x = 3439 - 1234
x = 2205
756 = 700 + 50 + 6 hoặc 7 x 100 + 5 x 10 + 6 x 1
862 = 8 x 100 + 6 x 10 + 2
HT
b) 200 x 4 = .800..... 300 x 2 = ..600....
200 x 2 = .400..... 300 x 3 = ...900...
400 x 2 = ...800... 500 x 1 = ..500....
100 x 4 = ..400.... 100 x 3 = .300.....
a: S=1(1+1)+2(1+2)+...+100(1+100)
=1+2+...+100+1^2+2^2+...+100^2
\(=\dfrac{100\cdot102}{2}+\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)
\(=100\cdot51+\dfrac{100\cdot101\cdot201}{6}\)
=343450
b: \(A=1\cdot2\cdot3+2\cdot3\cdot4+...+100\cdot101\cdot102\)
=>\(4\cdot A=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+...+100\cdot101\cdot102\left(103-99\right)\)
=>4*A=100*101*102*103
=>A=25*101*102*103
A=100/1 x 2 + 100/2 x 3 + 100/3 x 4 +...+100/99 x 100
A/100=1/1 x 2 + 1/2 x 3 + 1/3 x 4 +...+1/99 x 100
A/100=2-1/1x2 + 3-2/2x3 + ... + 100-99/99x100
A/100=1-1/2 + 1/2-1/3+...+1/99-1/100
A/100=1-1/100
A/100=99/100
A=99/100x100=99
Vậy A=99.
Ta có:
\(\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+...+\frac{100}{99.100}\)
\(\Rightarrow100.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{100}\right)\Leftrightarrow100.\frac{99}{100}=99\)
x=357