ko thấy hết bài đâu

toichua hoc

Sửa đề : Giải phương trình

\(3x^2+10x+2=10\)

\(\Leftrightarrow3x^2+10x-8=0\)

\(\Leftrightarrow3x^2+12x-2x-8=0\)

\(\Leftrightarrow3x\left(x+4\right)-2\left(x+4\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-4\end{cases}}\)

Vậy \(S=\left\{-4;\frac{2}{3}\right\}\)

Yêu cầu đề sai nhé.

\(3x^2+10x+2=10\)

\(\Rightarrow3x^2+10x-8=0\)

\(\Rightarrow3\left(x^2+\frac{10x}{3}-\frac{8}{3}\right)=0\)

\(\Rightarrow3\left(x+4\right)\left(x-\frac{2}{3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-4\\x=\frac{2}{3}\end{cases}}\)

ủa ủa 

sao lên chửi nhau vậy bạn 

???

Thi sao lien quan den may a

b) \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Rightarrow5x-2x^2+2x^2-2x=13\)

\(\Rightarrow3x=13\)

\(\Rightarrow x=13:3=\frac{13}{3}\)

c) \(\left(x-2\right)\left(x+5\right)-\left(x-3\right)\left(x+3\right)=5\)

\(\Rightarrow x^2+5x-2x-10-\left(x^2+3x-3x-9\right)=5\)

\(\Rightarrow x^2+5x-2x-10-x^2-3x+3x+9=5\)

\(\Rightarrow3x-1=5\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=6:3=2\)

b) \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow13\div3=\frac{1}{3}\)