b) (AMS 2007) Tìm x biết: (𝑥 + 4) + (𝑥 + 6) + (𝑥 + 8) + ⋯ + (𝑥 + 26) = 210
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\(\dfrac{25}{3}.x=\dfrac{5}{6}+\dfrac{4}{3}\)
\(\dfrac{25}{3}.x=\dfrac{13}{6}\)
\(x=\dfrac{13}{3}:\dfrac{25}{3}\)
\(x=\dfrac{39}{75}\)
a) \(\Rightarrow x^3-3x^2+3x-1+3x^2-12x+1=0\)
\(\Rightarrow x^3-9x=0\)
\(\Rightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x^3-1=x^3-9x^2+2x^2+6\)
\(\Rightarrow7x^2=7\)
\(\Rightarrow x^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Leftrightarrow\left(x^2+6x+9\right)-\left(x^2-4x+8x-32\right)=1\\ \Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\\ \Leftrightarrow2x+40=0\\ \Leftrightarrow2x=-40\\ \Leftrightarrow x=-20\)
\(\Leftrightarrow x^2+6x+9-x^2+4x-32=1\)
=>10x=22
hay x=11/5
a: \(x\in\left\{25;30;35\right\}\)
b: \(x\in\left\{24;32;40;48;56;64\right\}\)
c: \(x\in\left\{3;4;6\right\}\)
\(b,x=ƯCLN\left(45,30\right)=15\\ c,x=BCNN\left(6,8\right)=24\\ d,x\in\left\{10;25;50\right\}\)
\(\left(x+4\right)+\left(x+6\right)+\left(x+8\right)+...+\left(x+26\right)=210\)
Số các chữ số x: \(\left(26-4\right):2+1=12\left(số\right)\)
\(\Rightarrow12x+\left(4+6+8+...+26\right)=210\)
\(\Rightarrow12x+\dfrac{\left(26+4\right)\left(\dfrac{26-4}{2}+1\right)}{2}=210\)
\(\Leftrightarrow12x+180=210\Leftrightarrow12x=30\Leftrightarrow x=\dfrac{5}{2}\)
(x + 4) + (x + 6) + (x + 8) + ... + (x + 26) = 210
<=> x[(26 - 4) : 2 + 1] + (26 + 4)[(26 - 4) : 2 + 1] : 2 = 210
<=> 12x + 180 = 210
<=> 12x = 210 - 180
<=> 12x = 30
<=> x = 2,5