Đề 1

Câu 1:

\(\text{a) }8x-4y=4\left(2x-y\right)\)

\(\text{b) }5y\left(x-1\right)-3x\left(x-1\right)=\left(5y-3x\right)\left(x-1\right)\)

\(\text{c) }-6x-6y+x\left(x+y\right)\\ =-\left(6x+6y\right)-x\left(x+y\right)\\ =-6\left(x+y\right)+x\left(x+y\right)\\ =\left(x-6\right)\left(x+y\right)\)

\(\text{d) }x^2-49=\left(x-7\right)\left(x+7\right)\)

Câu 2:

\(\text{a) }5x^2-x=0\\ \Leftrightarrow5x\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ \text{Vậy }x=0\text{ hoặc }x=1\)

\(\text{b) }x^2-25=0\\ \Leftrightarrow x^2=25\\ \Leftrightarrow x=\pm5\\ \text{Vậy }x=\pm5\)

\(\text{c) }x^4+2x^3+6x-9=0\\ \Leftrightarrow x^4+3x^3-x^3+9x-3x-9+3x^2-3x^2=0\\ \Leftrightarrow\left(x^4+3x^3+3x^2+9x\right)-\left(x^3+3x^2+3x+9\right)=0\\ \Leftrightarrow x\left(x^3+3x^2+3x+9\right)-\left(x^3+3x^2+3x+9\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+3x^2+3x+9\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(x^3+3x^2\right)+\left(3x+9\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x^2=-3\left(Vô\text{ }lí\right)\\x=-3\left(TM\right)\end{matrix}\right.\\ \text{Vậy }x=1\text{ }\text{hoặc }x=-3\)

Câu 2:

\(a^2-2a+b^2+4b+4c^2-4c+6=0\\ \Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\\ \Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\\ Do\text{ }\left(a-1\right)^2\ge0\forall x\\ \left(b+2\right)^2\ge0\forall x\\ \left(2c-1\right)^2\ge0\forall x\\ \Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\forall x\\ \text{Dấu }"="\text{ xảy ra khi }:\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-1=0\\b+2=0\\2c-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-2\\c=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(a=1;b=-2;c=\dfrac{1}{2}\)

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

Câu 1:

\(a^2+b^2+c^2+3=2\left(a+b+c\right)\\ \Leftrightarrow a^2+b^2+c^2+3-2\left(a+b+c\right)=0\\ \Leftrightarrow a^2+b^2+c^2-2a-2b-2c+3=0\\ \Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\\ \Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\\ Do\text{ }\left(a-1\right)^2\ge0\forall x\\\left(b-1\right)^2\ge0\forall x\\ \left(c-1\right)^2\ge0\forall x\\ \Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\forall x\)

\(\text{Dấu }"="\text{ xảy ra khi : }\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\Leftrightarrow a=b=c=1\)

Vậy \(a=b=c=1\text{ }khi\text{ }\left(a+b+c\right)^2=2\left(a+b+c\right)\)