Question

giải phương trình:

a)(x+6)(3x-1)=(x-6)(x+6)

b)(x+1)^2=(2x+3)^2

 

Answer 1

`a)(x+6)(3x-1)=(x-6)(x+6)`

`<=>(x+6)(3x-1+6-x)=0`

`<=>(x+6)(2x+5)=0`

`<=>[(x=-6),(x=-5/2):}`

`b)(x+1)^2=(2x+3)^2`

`<=>(x+1)^2-(2x+3)^2=0`

`<=>(x+1-2x-3)(x+1+2x+3)=0`

`<=>(-x-2)(3x+4)=0`

`<=>[(x=-2),(x=-4/3):}`

Answer 2

a)

`(x+6)(3x-1)=(x-6)(x+6)`

`<=> (x+6)(3x-1)-(x-6)(x+6)=0`

`<=> (x+6)(3x-1-x+6)=0`

`<=> (x+6)(2x+5)=0`

\(< =>\left[{}\begin{matrix}x+6=0\\2x+5=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-6\\x=-\dfrac{5}{2}\end{matrix}\right.\)

b)

`(x+1)^2 =(2x+3)^2`

`<=> (x+1)^2 -(2x+3)^2 =0`

`<=> (x+1-2x-3)(x+1+2x+3)=0`

`<=> (-x-2)(3x+4)=0`

\(< =>\left[{}\begin{matrix}-x-2=0\\3x+4=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Answer 3

`a)(x+6)(3x-1)=(x-6)(x+6)`

`<=> (x+6)(3x-1-x +6) =0`

`<=> (x+6)(2x+5)=0`

`<=> [(x+6=0),(2x=-5):}`

`<=>[(x=-6),(x=-5/2):}`

`b)(x+1)^2=(2x+3)^2`

`<=> (x+1)^2 - (2x+3)^2 =0`

`<=> (x+1 -2x -3)(x+1+2x +3) =0`

`<=> (-x - 2)(3x +4) =0`

`<=> (x+2)(3x +4) =0`

`<=> [(x =-2),(x =-4/3):}`

Answer 4

\(a,\left(x+6\right)\left(3x-1\right)=\left(x-6\right)\left(x+6\right)\)

\(\Leftrightarrow\left(x+6\right)\left(3x-1\right)-\left(x-6\right)\left(x+6\right)\)

\(\Leftrightarrow\left(x+6\right)\left(3x-1-x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\2x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-\dfrac{5}{2}\end{matrix}\right.\)

\(b,\left(x+1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow x^2+2x+1=4x^2+12x+9\)

\(\Leftrightarrow-3x^2-10x-8=0\)

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

Answer 5

\(a,\left(x+6\right)\left(3x-1\right)=\left(x-6\right)\left(x+6\right)\)

\(\Leftrightarrow\left(x+6\right)\left(3x-1\right)-\left(x-6\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(3x-1-x-6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(2x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{7}{2}\end{matrix}\right.\)

\(b,\left(x+1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow x^2+2x+1=4x^2+12x+9\)

\(\Leftrightarrow x^2+2x+1-4x^2-12x-9=0\)

\(\Leftrightarrow-3x^2-10x-8=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-4}{3}\end{matrix}\right.\)