Question

 

        Tìm X

  2X(3X+5)-6X-10=0

Mn làm giúp mk vs ạ❤️

 

 

 

Answer 1

\(\Leftrightarrow6x^2+10x-6x-10=0\)

\(\Leftrightarrow6x^2+4x-10=0\)

Ta có \(\Delta=4^2+4.6.10=256,\sqrt{\Delta}=16\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-4+16}{12}=1\\x=\frac{-4-16}{12}=\frac{-5}{3}\end{cases}}\)

Answer 2

e mới lớp 5 nên chưa chắc ạ >:

\(2x\left(3x+5\right)-6x-10=0\)

\(=>6x^2+10x-6x-10=0\)

\(=>6x.\left(x-1\right)+10.\left(x-1\right)=0\)

\(=>\left(6x+10\right)\left(x-1\right)=0\)

\(=>\orbr{\begin{cases}6x+10=0\\x-1=0\end{cases}=>\orbr{\begin{cases}x=\frac{-10}{6}\\x=1\end{cases}}}\)

Answer 3

Bài làm:

2x( 3x+ 5 ) -6x -10 =0

=> 6x^2 + 10x - 6x -10 =0

=> 6x^2 +4x -10 =0

=> 6x^2 - 6x + 10x -10 =0

=> 6x(x -1) + 10( x-1) =0

=> (6x +10)( x -1) =0

ĐẾN ĐÂY BẠN TỰ GIẢI NHÉ

# hỌC TỐT

Answer 4

\(2x\left(3x+5\right)-6x-10=0\)

\(6x^2+10x-6x=-10\)

\(6x^2+4x+10=0\)

\(6x\left(x-1\right)+10\left(x-1\right)=0\)

\(\left(x+1\right)\left(6x+10\right)=0\)

\(Th1:x+1=0\Leftrightarrow x=-1\)

\(Th2:6x+10=0\Leftrightarrow6x=-10\Leftrightarrow x=-\frac{5}{3}\)