Diện tích dùng để trồng cây ăn quả bằng số phần diện tích của vườn là:

\(1-\dfrac{1}{3}-\dfrac{5}{12}=\dfrac{12}{12}-\dfrac{4}{12}-\dfrac{5}{12}=\dfrac{3}{12}=\dfrac{1}{4}\) (diện tích)

Đáp số: \(\dfrac{1}{4}\) diện tích của vườn

\(5^5-5^4+5^3\\ =5^3\cdot5^2-5^3\cdot5+5^3\cdot1\\ =5^3\cdot\left(5^2-5+1\right)\\ =5^3\cdot21\\ =5^3\cdot7\cdot3⋮7\left(đpcm\right)\)

\(5^7\cdot5=5^7\cdot5^1=5^{7+1}=5^8\)

\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-5\right)\left(x-2\right)\)

\(f,x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\\ g,x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\\ h,x^2-7x+12=x^2-3x-4x+12=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)

\(a,x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-3\right)\left(x-2\right)\\ b,3x^2+9x-30=3x^2+15x-6x-30=3x\left(x+5\right)-6\left(x+5\right)=\left(3x-6\right)\left(x+5\right)=3\left(x-2\right)\left(x+5\right)\\ c,x^2-3x+2=x^2-x-2x+2=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\\ d,x^2-9x+18=x^2-3x-6x+18=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\\ e,x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\)

\(\dfrac{x-4}{10}+\dfrac{x-3}{11}=\dfrac{x-2}{12}+\dfrac{x-1}{13}\)

\(\Rightarrow\dfrac{x-4}{10}+\dfrac{x-3}{11}-\dfrac{x-2}{12}-\dfrac{x-1}{13}=0\)

\(\Rightarrow\left(\dfrac{x-4}{10}-1\right)+\left(\dfrac{x-3}{11}-1\right)-\left(\dfrac{x-2}{12}-1\right)-\left(\dfrac{x-1}{13}-1\right)=0\)

\(\Rightarrow\dfrac{x-14}{10}+\dfrac{x-14}{11}-\dfrac{x-14}{12}-\dfrac{x-14}{13}=0\)

\(\Rightarrow\left(x-14\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)=0\)

\(\Rightarrow x-14=0\) (Mà \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\ne0)\)

\(\Rightarrow x=0+14\\ \Rightarrow x=14\)

Vậy: \(x=14\)

\(A=10\cdot11+11\cdot12+12\cdot13+...+499\cdot500\\ 3A=10\cdot11\cdot\left(12-9\right)+11\cdot12\cdot\left(13-10\right)+12\cdot13\cdot\left(14-11\right)+...+499\cdot500\cdot\left(501-498\right)\\ 3A=10\cdot11\cdot12-9\cdot10\cdot11+11\cdot12\cdot13-10\cdot11\cdot12+12\cdot13\cdot14-11\cdot12\cdot13+...+499\cdot500\cdot501-498\cdot499\cdot500\\ 3A=\left(10\cdot11\cdot12-10\cdot11\cdot12\right)+\left(11\cdot12\cdot13-11\cdot12\cdot13\right)+...+\left(498\cdot499\cdot500-498\cdot499\cdot500\right)+499\cdot500\cdot501-9\cdot10\cdot11\\ 3A=499\cdot500\cdot501-9\cdot10\cdot11\\ 3A=124999500-990\\ 3A=124998510\\ A=41666170\)

\(\left(x+3y\right)^2-\left(x-y\right)^2\\ =\left[\left(x+3y\right)-\left(x-y\right)\right]\left[\left(x+3y\right)+\left(x-y\right)\right]\\ =\left[x+3y-x+y\right]\left[x+3y+x-y\right]\\ =4y\cdot\left(2x+2y\right)\\ =8xy+8y^2\)

\(\left(x+3y\right)^2-\left(x-y\right)^2\\ =\left(x^2+6xy+9y^2\right)-\left(x^2-2xy+y^2\right)\\ =x^2+6xy+9y^2-x^2+2xy-y^2\\ =\left(x^2-x^2\right)+\left(6xy+2xy\right)+\left(9y^2-y^2\right)\\ =8xy+8y^2\)