\(PT\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Với \(x=0\Leftrightarrow0+2y=0\Leftrightarrow y=0\)

Với \(y=0\Leftrightarrow x=0\)

Vậy \(\left(x,y\right)=\left(0,0\right)\)

Diện tích phòng là \(\left(60\times60\right)\times75=270000\left(cm^2\right)=27\left(m^2\right)\)

Chiều rộng là \(27:5,4=5\left(m\right)\)

\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{2}=\dfrac{z}{4}\\ \Leftrightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{12}=\dfrac{x+y-z}{4+6-12}=\dfrac{10}{-2}=-5\\ \Leftrightarrow\left\{{}\begin{matrix}x=-20\\y=-30\\z=-60\end{matrix}\right.\)

\(=\dfrac{x^2-6x+9-x^2-9}{x\left(x-3\right)}=\dfrac{-6x}{x\left(x-3\right)}=\dfrac{6}{3-x}\)

\(a,P=x^3-4x-x^3-1+x^2+6x+9=x^2+2x+8\\ b,P=11\Leftrightarrow x^2+2x+8=11\Leftrightarrow x^2+2x-3=0\\ \Leftrightarrow x^2-x+3x-3=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\\ c,P=\left(x^2+2x+1\right)+7=\left(x+1\right)^2+7\ge7\\ P_{min}=7\Leftrightarrow x=-1\)

Bài 1:

Gọi số cây 3 lớp 7A,7B,7C lần lượt là \(a,b,c\in \mathbb{N^*},cây\)

Áp dụng tc dtsbn:

\(\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{5}=\dfrac{a+b+c}{6+4+5}=\dfrac{30}{15}=2\\ \Rightarrow\left\{{}\begin{matrix}a=12\\b=8\\c=10\end{matrix}\right.\)

Vậy ...

Bài 3:

\(a+b\ne-c\Rightarrow a+b+c\ne0\\ \Rightarrow\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b+c}{a+b+c}=1\\ \Rightarrow\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\\ \Rightarrow P=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{2a\cdot2b\cdot2c}{abc}=8\)

\(1,M=\sqrt{3}+4\sqrt{3}-2\sqrt{3}=3\sqrt{3}\\ 2,\\ a,ĐK:x\le3\\ PT\Leftrightarrow3\sqrt{3-x}-\sqrt{3-x}=6\\ \Leftrightarrow\sqrt{3-x}=3\Leftrightarrow3-x=9\Leftrightarrow x=-6\left(tm\right)\\ b,ĐK:x\ge0\\ BPT\Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\Leftrightarrow0\le x< 9\)

\(A=4,32\times1,5+4,32\times4,7+4,32\times2,8+4,32\\ A=4,32\times\left(1,5+4,7+2,8+1\right)=4,32\times10=43,2\)

\(a,\) Áp dụng Pytago, ta có \(EF=\sqrt{DE^2+DF^2}=20\left(cm\right)\)

Vì DN là trung tuyến ứng với cạnh huyền EF nên \(DN=\dfrac{1}{2}EF=10\left(cm\right)\)