2x = 1+3+5+.....+99
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b, \(\left(2x-3\right)\left(x+1-x-5\right)=0\Leftrightarrow x=\dfrac{3}{2}\)
c, \(x^2-4x+1=2x-22\Leftrightarrow x^2-6x+23=0\Leftrightarrow\left(x-3\right)^2+14=0\left(voli\right)\)
pt vô nghiệm
d, \(\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1=\dfrac{205-x}{95}+1\)
\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}=\dfrac{300-x}{95}\)
\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}-\dfrac{1}{95}\ne0\right)=0\Leftrightarrow x=300\)
a) \(3\left(2x-5\right)+125=134\)
\(\Leftrightarrow3\left(2x-5\right)=9\)
\(\Leftrightarrow2x-5=3\)
\(\Leftrightarrow2x=8\Leftrightarrow x=4\)
b) \(\left(2x+5\right)+\left(2x+3\right)+\left(2x+1\right)=27\)
\(\Leftrightarrow6x+9=27\)
\(\Leftrightarrow6x=18\Leftrightarrow x=3\)
d) \(27\left(x-27\right)-27=0\)
\(\Leftrightarrow27\left(x-27\right)=27\)
\(\Leftrightarrow x-27=1\Leftrightarrow x=28\)
(x+1)+(2x+3)+(3x+5)+...+(50x+99)=3775
=> (x + 2x + 3x + ... + 50x) + (1 + 3 + 5 + ... + 99) = 3775
=> x . (1 + 2 +3 + ...+ 50) + (1 + 3 + 5 + ... + 99) = 3775
Áp dụng công thức tính dãy số ta có :
\(1+2+3+...+50=\frac{\left[\left(50-1\right):1+1\right].\left(50+1\right)}{2}=\frac{50.51}{2}=25.51=1275\)
\(1+3+5+...+99=\frac{\left[\left(99-1\right):2+1\right].\left(99+1\right)}{2}=\frac{50.100}{2}=50.50=2500\)
=> x . 1275 + 2500 = 3775
=> 1275x = 1275
=> x = 1
2x=1+3+5+...+99
2x = (1+99) x 25 (Có 25 cặp)
2x = 100 x 25
2x = 2500
x=2500:2
x=1250
2x=1+3+5+...+99=\(\dfrac{\left(99-1\right):2+1\cdot\left(1+99\right)}{2}\)=2500
=>x=2500:2=1250.