Mấy chế em xin câu 3 ạ :>>
3. Giải pt :
\(x^2-10x+16=0\)
\(\Leftrightarrow x^2-8x-2x+16=0\)
\(\Leftrightarrow\left(x-8\right)\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy gt của x để bt đạt giá trị bằng 0 là \(x\in\left\{2;8\right\}\)
4. \(2x^2+2xy+y^2+2x+1=0\)
\(\Leftrightarrow y^2+2xy+2x^2+2x+1=0\)
\(\Leftrightarrow y^2+2xy+x^2+x^2+2x+1=0\)
\(\Leftrightarrow\left(y+x\right)^2+\left(x+1\right)^2=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\Rightarrow y+x=0\Leftrightarrow y-1=0\Rightarrow y=1\)
Vậy giá trị của \(x\) là -1. (Nếu kết luận cả y thì giá trị của \(y\) là 1)
4.
\(2x^2+2xy+y^2+2x+1=0\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)=0\)
\(\Rightarrow\left(x+y\right)^2+\left(x+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-1+y=0\\x=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=1\\x=-1\end{matrix}\right.\)
4. \(2x^2+2xy+y^2+2x+1=0\Leftrightarrow x^2+2xy+y^2+x^2+2x+1=0\Leftrightarrow\left(x+y\right)^2+\left(x+1\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-1\end{matrix}\right.\)
Vậy x=-1; y=1.
2. \(x^4-3x^2-2xy+y^2+1=0\Leftrightarrow x^4-2x^2+1-x^2-2xy+y^2=0\Leftrightarrow\left(x^2-1\right)^2+\left(x-y\right)^2-2x^2=0\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-1\right)^2=0\\\left(x-y\right)^2=0\\-2x^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\x=y\\x=0\end{matrix}\right.\)
Đức Minh xem hộ t sai gì vậy?
