Bài 3: Phương trình đưa được về dạng ax + b = 0

\(\text{ 2x² + 6x - 8 }\)

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

Câu 12: A

Câu 13: C

Câu 17: A

Câu 19: A

Câu 16: B

12. A

13. C

16. B

17. A

19. A

\(\dfrac{1}{x-1}+1=\dfrac{x}{x-2}\\ \Rightarrow\dfrac{1+x-1}{x-1}=\dfrac{x}{x-2}\\ \dfrac{x}{x-1}=\dfrac{x}{x-2}\\ \Leftrightarrow x\left(x-2\right)=x\left(x-1\right)\\ x^2-2x=x^2-x\\ x^2-x^2=x-2x\\ -x=0\\ x=0\)

\(\Leftrightarrow x+2+x^2-3x+2=x^2-x\)

=>-2x+4+x=0

=>4-x=0

hay x=4

địt mie bạn

\(\Leftrightarrow x+66=0\)

hay x=-66

\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\\ \Rightarrow\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)=\left(\dfrac{x+5}{61}+1\right)+\left(\dfrac{x+7}{59}+1\right)\\ \Rightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\\ \Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\\ \Rightarrow x+66=0\\ \Rightarrow x=-66\)

\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\\ \dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\\ \left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\\ x+66=0\\ x=-66\)

lỗi

Ảnh đâu pé

check ib mình ạ

=>10y=18-8x

=>y=(9-4x)/5

a: \(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)

\(a,x^2+9x+20=0\\ \Leftrightarrow\left(x^2+4x\right)+\left(5x+20\right)=0\\ \Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\\ \Leftrightarrow\left(x+4\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)

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

\(b,\Rightarrow\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)+\left(\dfrac{x+9}{2012}+1\right)=0\\ \Rightarrow\dfrac{x+2021}{2020}+\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}+\dfrac{x+2021}{2012}=0\\ \Rightarrow\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2018}+\dfrac{1}{2017}+\dfrac{1}{2012}\right)=0\\ \Rightarrow x+2021=0\\ \Rightarrow x=-2021\)

\(c,\Rightarrow\dfrac{x+2}{4}-\left(\dfrac{x-1}{3}+1\right)=0\\ \Rightarrow\dfrac{x+2}{4}-\dfrac{x+2}{3}=0\\ \Rightarrow\left(x+2\right)\left(\dfrac{1}{4}-\dfrac{1}{3}\right)=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ d,\Rightarrow\dfrac{5x+9}{6}=\dfrac{5x+1}{3}\\ \Rightarrow3\left(5x+9\right)=6\left(5x+1\right)\\ \Rightarrow15x+27=30x+6\\ \Rightarrow15x=21\\ \Rightarrow x=\dfrac{7}{5}\)

B

B

Chọn B