Question

a) (4x-10)(24+3x)=0
b)7x-21+x(x-3)=0
c)x^2-1=2x(x+1)

Answer 1

a, \(\left(4x-10\right)\left(24+3x\right)=0\)

⇔\(\left[{}\begin{matrix}4x-10=0\\24+3x=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}4x=10\\3x=-24\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=\frac{5}{2}\\x=-8\end{matrix}\right.\)

Vậy...

b,\(7x-21+x\left(x-3\right)=0\)

⇔\(7\left(x-3\right)+x\left(x-3\right)=0\)

⇔\(\left(7+x\right)\left(x-3\right)=0\)

⇔\(\left[{}\begin{matrix}7+x=0\\x-3=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=-7\\x=3\end{matrix}\right.\)

Vậy...

c,Mình bận quá.Xin lỗi mình xin không làm!

Gợi ý:

Phân tích vế trái sang hằng đẳng thức số 3 rồi tính nhé!

Answer 2

a) Ta có: \(\left(4x-10\right)\left(24+3x\right)=0\)

\(\Leftrightarrow6\left(2x-5\right)\left(8+x\right)=0\)

mà 6≠0

nên \(\left[{}\begin{matrix}2x-5=0\\8+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-8\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{5}{2};-8\right\}\)

b) Ta có: \(7x-21+x\left(x-3\right)=0\)

\(\Leftrightarrow7\left(x-3\right)+x\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(7+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\7+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)

Vậy: S={3;-7}

c) Ta có: \(x^2-1=2x\left(x+1\right)\)

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

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

\(\Leftrightarrow\left(x+1\right)\left(-x-1\right)=0\)

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

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: S={-1}

Answer 3

c,\(x^2-1=2x\left(x+1\right)\)

⇔\(\left(x+1\right)\left(x-1\right)=2x\left(x+1\right)\)

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

⇔\(\left(x+1\right)\left(-x-1\right)=0\)

⇔\(\left[{}\begin{matrix}x+1=0\\-x-1=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\)

Vậy...

Answer 4

a) (4x-10)(24+3x)=0

=> \(\begin{matrix}4x-10=0\\24+3x=0\end{matrix}\)

=>[\(\begin{matrix}4x=10\\3x=-24\end{matrix}\)

=> [\(\begin{matrix}x=\frac{5}{2}\left(n\right)\\x=-8\left(n\right)\end{matrix}\)

b) 7x-21+x(x-3)=0

=> \(7x-21+x^2-3x=0\)

=> \(x^2-4x-21=0\)

=>[\(\begin{matrix}x-7=0\\x+3=0\end{matrix}\)

=>[\(\begin{matrix}x=7\left(n\right)\\x=-3\left(n\right)\end{matrix}\)