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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)\(219-7\left(x+1\right)=100\)
⇔\(7\left(x+1\right)=119\)
⇔\(x+1=17\)
⇔\(x=16\)
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a) ( 2x + 1 ) . 2907 = 8 721
2x + 1 = 3
2x = 2
x = 1
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x + 1234 - 532 = 2907
x + 1234 = 2907 + 532
x + 1234 = 3439
x = 3439 - 1234
x = 2205
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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
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S= 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
S x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
S x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
S x 3 = 99x100x101 A = 99x100x101 : 3 A = 333300
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a: =>3x+27-2x+8=-13*100=-1300
=>x+35=-1300
=>x=-1335
b: =>100x-5050=2750
=>100x=2750+5050=7800
=>x=78
c: =>-2x-8-64:16=198
=>-2x=210
=>x=-105
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\(c,\)\(\left(x-1\right)+\left(x-2\right)+....+\left(x-100\right)=50\)
\(\left(x+x+...+x\right)-\left(1+2+...+100\right)=50\)
\(100x-5050=50\)
\(100x=50+5050\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
\(a,\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=7\)
\(b,x+\left(1+2+3+...+50\right)=2000\)
\(x+\frac{\left[1+50\right]\cdot\left[\left(50-1\right)\div1+1\right]}{2}=2000\)
\(x+1275=2000\)
\(\Rightarrow x=2000-1275=725\)
x=357