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Bài 1 câu g bạn kia làm sai mình sửa lại nhá
\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2\right)-12c^2\)
\(=3\left(a-b\right)^2-12c^2\)
\(=3\left[\left(a-b\right)^2-4c^2\right]\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
Để mình làm tiếp cho :))
Bài 2 :
Câu a : \(37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5\)
\(=\left(37,5.8,5+1,5.37,5\right)-\left(7,5.3,4+6,6.7,5\right)\)
\(=37,5\left(8,5+1,5\right)-7,5\left(3,4+6,6\right)\)
\(=37,5.10-7,5.10\)
\(=10.30=300\)
Câu b : \(35^2+40^2-25^2+80.35\)
\(=\left(35^2+80.35+40^2\right)-25^2\)
\(=\left(30+45\right)^2-25^2\)
\(=75^2-25^2\)
\(=\left(75+25\right)\left(75-25\right)\)
\(=100.50=5000\)
Bài 3 :
Câu a : \(x^3-\dfrac{1}{9}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{9}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{1}{3}\end{matrix}\right.\)
Câu b : \(2x-2y-x^2+2xy-y^2=0\)
\(\Leftrightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\2-x+y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=2\Rightarrow x=2-y\end{matrix}\right.\)
Câu c :
\(x\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(x^2\left(x-3\right)+27-9x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-9\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\pm3\end{matrix}\right.\)
Bài 4 :
Câu a :
\(x^2-4x+3\)
\(=x^2-x-3x+3\)
\(=\left(x^2-x\right)-\left(3x-3\right)\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
Câu b :
\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+3\right)\)
Câu c :
\(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
Câu d :
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)