tìm x biết:1+3+5+.....+[2x-1]=1069
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56.x=1069-1013+1-1
56.x=56+1-1
56.x=55+1
56.x=56
x=56:56
x=1
chuc bn hoc gioi1
\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)
a) \(\Leftrightarrow x^3-8-x^3+9x=1\)
\(\Leftrightarrow9x=9\Leftrightarrow x=1\)
b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)
\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)
a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)
b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))
\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)
\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)
d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)
\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)
\(\Leftrightarrow2x^2+2x=2x^2+1\)
\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).
Ta có: \(1+3+5+...+\left(2x-1\right)=1069\)
\(\Leftrightarrow\frac{\left(2x-1+1\right)\cdot\left[\left(2x-1-1\right)\div2+1\right]}{2}=1069\)
\(\Leftrightarrow\frac{2x\cdot\left(x-1+1\right)}{2}=1069\)
\(\Leftrightarrow x^2=1069\)
\(\Rightarrow x=\sqrt{1069}\)
Số số hạng
\(\left(2x-1-1\right):2+1=x\)
Tổng
\(\left(2x-1+1\right)\cdot x:2=x^2\)
\(x^2=1069\left(x\ge0\right)\)
\(x=\sqrt{1069}\)
\(x=33\)