Question

[Nguyễn Trâm]

hộ mình bài này đi mai mình nộp rồi ý b,c,d thôi

Answer 1

[Nguyễn Ngọc Lộc]

c, Ta có : \(2x^2+2x+3x+3=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

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

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

Vậy ...

d, Ta có : \(\dfrac{3-2x}{2006}+\dfrac{3-2x}{2007}+\dfrac{3-2x}{2008}=\dfrac{3-2x}{2009}+\dfrac{3-2x}{2010}\)

\(\Leftrightarrow\dfrac{3-2x}{2006}+\dfrac{3-2x}{2007}+\dfrac{3-2x}{2008}-\dfrac{3-2x}{2009}-\dfrac{3-2x}{2010}=0\)

\(\Leftrightarrow\left(3-2x\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{1}{2010}\right)=0\)

\(\Leftrightarrow3-2x=0\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy ...

Answer 2

[Nguyễn Lê Phước Thịnh]

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

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

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

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

\(\Leftrightarrow\left(3x-2\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\5x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{2}{5}\right\}\)

b) Ta có: \(x^2+\left(x+3\right)\left(5x-7\right)=9\)

\(\Leftrightarrow x^2+5x^2-7x+15x-21-9=0\)

\(\Leftrightarrow6x^2+8x-30=0\)

\(\Leftrightarrow6x^2+18x-10x-30=0\)

\(\Leftrightarrow6x\left(x+3\right)-10\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(6x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\6x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{3}\end{matrix}\right.\)

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