\(A=5^4\cdot13^2\cdot17\)

Số ước của A là: \(\left(4+1\right)\left(2+1\right)\left(1+1\right)=5\cdot3\cdot2=25\)

thích

\(\Leftrightarrow x+y-xy=0\\ \Leftrightarrow\left(y-1\right)-x\left(y-1\right)=-1\\ \Leftrightarrow\left(1-x\right)\left(y-1\right)=-1\\ \Leftrightarrow\left(x-1\right)\left(y-1\right)=1=1.1=\left(-1\right)\left(-1\right)\\ TH_1:\left\{{}\begin{matrix}y-1=1\\x-1=1\end{matrix}\right.\Leftrightarrow x=y=2\\ TH_2:\left\{{}\begin{matrix}x-1=-1\\y-1=-1\end{matrix}\right.\Leftrightarrow x=y=0\)

Vậy \(\left(x;y\right)=\left(2;2\right);\left(0;0\right)\)

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

Áp dụng tc dtsbn:

\(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{6}=\dfrac{b-a}{5-3}=\dfrac{18}{2}=9\\ \Rightarrow\left\{{}\begin{matrix}a=27\\b=45\\c=54\end{matrix}\right.\)

Vậy ...

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Câu 4:

\(a,\tan B=\dfrac{AC}{AB}=\dfrac{12}{5}\approx\tan67^0\\ \Rightarrow\widehat{B}\approx67^0\\ b,\text{Áp dụng PTG: }BC=\sqrt{AC^2+AB^2}=13\left(cm\right)\\ \text{Áp dụng HTL: }\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{25}{13}\left(cm\right)\\CH=\dfrac{AC^2}{BC}=\dfrac{144}{13}\left(cm\right)\\AH=\sqrt{BH\cdot CH}=\dfrac{60}{13}\left(cm\right)\end{matrix}\right.\)