(2x+1)(x-1)=0
b , (x+2/3) ( x-1/2)=0
c, (3x-1)(2x-3)(x+5)=0
d,3x-15=2x(x-5)
e x^2-x=0
f.x^2-2x=0
g,x^2-3x=0
h, (x+1)(x+2)=(2-x)(x+2)
Mn giúp em với ạ !
Ngoài ra em cũng nói lun : Happy birthday to Ho Chi Minh♥
(2x+1)(x-1)=0 b , (x+2/3) ( x-1/2)=0 c, (3x-1)(2x-3)(x+5)=0 d,3x-15=2x(x-5) e x^2-x=0 f.x^2-2x=0 g,x^2-3x=0 h, (x+1)(x+2)=(2-x)(x+2)
By Margaret
Đáp án:
`a)(2x+1)(x-1)=0`
`<=>`\(\left[ \begin{array}{l}2x-1=0\\x-1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=1\end{array} \right.\)
`b)` `(x+2/3)(x-1/2)=0`
`<=>`\(\left[ \begin{array}{l}x+\dfrac{2}{3}=0\\x-\dfrac{1}{2}=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{array} \right.\)
`c)(3x-1)(2x-3)(x+5)=0`
`<=>`\(\left[ \begin{array}{l}3x-1=0\\2x-3=0\\x+5=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=\dfrac{3}{2}\\x=-5\end{array} \right.\)
`e)x^2-x=0`
`<=>x(x-1)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
`f)x^2-2x=0`
`<=>x(x-2)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
`g)x^2-3x=0`
`<=>x(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
`h)(x+1)(x+2)=(2-x)(x+2)`
`<=>x^2+2x+x+2=2^2-x^2`
`<=>(x^2+x^2)+3x+2-4=0`
`<=>2x^2+3x-2=0`
Đáp án:+Giải thích các bước giải:
`a)(2x_1)(x-1)=0`
`<=>`\(\left[ \begin{array}{l}2x-1=0\\x-1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=1\end{array} \right.\)
`b)` `(x+2/3)(x-1/2)=0`
`<=>`\(\left[ \begin{array}{l}x+\dfrac{2}{3}=0\\x-\dfrac{1}{2}=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{array} \right.\)
`c)(3x-1)(2x-3)(x+5)=0`
`<=>`\(\left[ \begin{array}{l}3x-1=0\\2x-3=0\\x+5=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=\dfrac{3}{2}\\x=-5\end{array} \right.\)
`d)3x-15=2x(x-5)`
`<=>3x-15=2x^2-10x`
`<=>-2x^2+3x+10x-15=0`
`<=>-2x^2+13x-15=0`
`e)x^2-x=0`
`<=>x(x-1)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
`f)x^2-2x=0`
`<=>x(x-2)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
`g)x^2-3x=0`
`<=>x(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
`h)(x+1)(x+2)=(2-x)(x+2)`
`<=>x^2+2x+x+2=2^2-x^2`
`<=>(x^2+x^2)+3x+2-4=0`
`<=>2x^2+3x-2=0`