Question

Giai cac phuong trinh sau

1) (5x-4).(4x+6) = 0

2) (4x-10).(24+5x) = 0

3) (x-3).(2x+1) = 0

4) (2x+1).(x²+2) = 0

5) (x²+4).(7x-3) = 0

6) (x-5).(3-2x).(3x+4) = 0

7) (x-2).(3x+5) = (2x-4).(x+1)

8) (2x+5).(x-4) = (x-5).(4-x)

9) 9x²-1 = (3x+1).(2x-3)

10) (2x-1)² = 49

11) (5x-3)² - (4x-7)² = 0

12) (2x+7)² = 9.(x+2)²

^{Giúp mình giải gấp .Cám ơn}

Answer 1

1) \(\left(5x-4\right)\left(4x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)

2) \(\left(4x-10\right)\left(24+5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)

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

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

Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)