\(a.\left(x-2\right)^2-x\left(2-x\right)=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(b.x\left(x-1\right)+1-x=0\)

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

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

\(x\sqrt{x}-\sqrt{x}+x-1=\sqrt{x}\left(x-1\right)+\left(x-1\right)=\left(\sqrt{x}+1\right)\left(x-1\right)\)

Để đường thẳng \(y=\left(m^2+1\right)x+m\) song song với đường thẳng \(y=5x+2\) ta cần:

\(\left\{{}\begin{matrix}m^2+1=5\\m\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=2\left(loại\right)\\m=-2\left(tm\right)\end{matrix}\right.\\m\ne2\end{matrix}\right.\)

Vậy m = -2 thì ...

Thay x = 3 vào biểu thức, ta được:

\(\left(2m-3\right).3-2=\left(m+1\right)-3\)

\(\Leftrightarrow6m-9-2=m+1-3\)

\(\Leftrightarrow5m=9\)

\(\Leftrightarrow m=\dfrac{9}{5}\)

Vậy để phương trình có nghiệm x = 3 thì m phải có giá trị 9/5

x - 1 cũng là HĐT thường gặp cx phân tích đc nhé

Đề là gì e ?

Anh ko bt a nói đúng hay sai nhưng mà \(\left(\pm1\right)^2=1\) đúng ko nhỉ ? Nếu v viết 1 luôn thì đúng hơn

a) \(\left\{{}\begin{matrix}3x+y=3\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=10\\3x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}2x+5y=8\\2x-3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8y=8\\2x-3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=\dfrac{3}{2}\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}4x+3y=6\\2x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\4x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\2x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)

\(A=\left(\dfrac{2x}{x^2-x-6}+\dfrac{x}{x-3}\right):\dfrac{x}{x-3}\left(x\ne3;x\ne-2\right)\)

\(=\left[\dfrac{2x}{\left(x+2\right)\left(x-3\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}\right].\dfrac{x-3}{x}\)

\(=\dfrac{x^2+4x}{\left(x+2\right)\left(x-3\right)}.\dfrac{x-3}{x}\)

\(=\dfrac{x+4}{x+2}\)