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Bài 3:
a) Đặt f(x)=0
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
b) Đặt f(x)=0
\(\Leftrightarrow x^2-7x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Bài 3:
c) Đặt f(x)=0
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
d) Đặt f(x)=0
\(\Leftrightarrow x^4+2=0\)
\(\Leftrightarrow x^4=-2\)(Vô lý)
Bài 1:
Để E nguyên thì \(x+5⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{3;1;9;-5\right\}\)
Ta có: \(1-\frac{x-\frac{1+3x}{5}}{3}=\frac{x}{2}-\frac{2x-\frac{10-6x}{7}}{2}\)
\(\Leftrightarrow1-\frac{5x-1-3x}{15}=\frac{x}{2}-\frac{14x-10+6x}{14}\)
\(\Leftrightarrow\frac{15-2x+1}{15}=\frac{7x-20x+10}{14}\)
\(\Leftrightarrow\frac{16-2x}{15}=\frac{10-13x}{14}\)
\(\Leftrightarrow224-28x=150-195x\)
\(\Leftrightarrow167x=-74\Rightarrow x=-\frac{74}{167}\)
\(1-\frac{x-\frac{1+3x}{5}}{3}=\frac{x}{2}-\frac{2x-\frac{10-6x}{7}}{2}\)
\(\frac{3-\left(\frac{5x-\left(1+3x\right)}{5}\right)}{3}=\frac{x-\left(\frac{14x-\left(10-6x\right)}{7}\right)}{2}\)
\(\frac{3-\frac{2x-1}{5}}{3}=\frac{x-\frac{20x-10}{7}}{2}\)
\(\frac{\frac{15-\left(2x-1\right)}{5}}{3}=\frac{\frac{7x-\left(20x-10\right)}{7}}{2}\)
\(\frac{16-2x}{15}=\frac{10-13x}{14}\)
\(\left(16-2x\right)\cdot14=\left(10-13x\right)\cdot15\)
\(224-28x=150-195x\)
\(195x-28x=150-224\)
\(167x=-74\)
\(x=-74:167\)
\(x=\frac{-74}{167}\)