\(\left\{{}\begin{matrix}x^2-xy-2y^2=0\\3x+y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x\left(1-3x\right)-2\left(1-3x\right)^2=0\\y=1-3x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-14x^2+11x-2=0\\y=1-3x\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{7}\end{matrix}\right.\\y=1-3x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{2}{7}\\y=\dfrac{1}{7}\end{matrix}\right.\end{matrix}\right.\)

Vậy...

(4x - 3)² - (2x + 1)² = 0

\(\Leftrightarrow\) (4x - 3 - 2x - 1)(4x - 3 + 2x + 1) = 0

\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy ...

3x - 12 - 5x(x - 4) = 0

\(\Leftrightarrow\) 3x - 12 - 5x² + 20x = 0

\(\Leftrightarrow\) -5x² + 23x - 12 = 0

\(\Leftrightarrow\) 5x² - 23x + 12 = 0

\(\Leftrightarrow\) 5x² - 20x - 3x + 12 = 0

\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0

\(\Leftrightarrow\) (x - 4)(5x - 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy ...

(8x + 2)(x² + 5)(x² - 4) = 0

\(\Leftrightarrow\) (8x + 2)(x² + 5)(x - 2)(x + 2) = 0

Vì x² \(\ge\) 0 \(\forall\) x nên x² + 5 > 0 \(\forall\) x

\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=2\\x=-2\end{matrix}\right.\)

Vậy ...

Chúc bn học tốt!

a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3-2x-1\right)\left(4x-3+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)

b) Ta có: \(3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)

c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)

mà \(2>0\)

và \(x^2+5>0\forall x\)

nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-1\\x=2\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=2\\x=-2\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)

Đáp án là C

\(=2x^3y^2+y-3x\)

a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)

b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)

c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)

\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)

d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)

hay \(N=y^2-x^2\)