Question

bài 1

a (4x-1) (x-3)-(x-3) (5x+2)=0

b (x+3)(x-5)+(x+3)(3x-4)=0

c(x+6) (3x -1) + \(x^2-36=0\)

Answer 1

Bài 1 :

a, \(\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)

=> \(\left(x-3\right)\left(4x-1-5x-2\right)=0\)

=> \(\left(x-3\right)\left(-x-3\right)=0\)

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

Vậy phương trình có nghiệm là \(x=\pm3\) .

b, \(\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\)

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

=> \(\left(x+3\right)\left(4x-9\right)=0\)

=> \(\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)

Vậy phương trình có nghiệm là \(x=-3,x=\frac{9}{4}\) .

c, \(\left(x+6\right)\left(3x-1\right)+x^2-36=0\)

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

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

=> \(\left(x+6\right)\left(4x-7\right)=0\)

=> \(\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)

Vậy phương trình có nghiệm là \(x=-6,x=\frac{7}{4}\) .

Answer 2

a) ( 4x - 1 ) ( x - 3 ) - ( x - 3 ) ( 5x + 2 ) = 0

⇔ ( x - 3 ) ( 4x - 1 - 5x - 2 ) = 0

⇔ ( x - 3 ) ( -x - 3 ) = 0

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

Ý b) tương tự ý a) thôi.

c) ( x + 6 ) ( 3x - 1 ) + x² - 36 = 0

⇔ ( x + 6 ) ( 3x - 1 ) + ( x + 6 ) ( x - 6 ) = 0

⇔ (x+6)(3x-1+x-6)=0

⇔ (x+6)(4x-7)=0

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)

Answer 3

a, (4x-1)(x-3)-(x-3)(5x+2)=0

<=> (x-3)(4x-1-5x-2)=0

<=> (x-3)(-x-3)=0

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

b, (x+3)(x+5)+(x+3)(3x-4)=0

<=> (x+3)(x+5+3x-4)=0

<=> (x+3)(4x+1)=0

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

c, (x+6)(3x-1)+\(x^2\) -36=0

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

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

<=> (x+6)(4x-7)=0

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)