ĐÂY KHÔNG PHẢI TOÁN LỚP 9 MÀ LÀ LỚP 5 VÒNG 7 VIOLYMPIC.VN

Câu 3:

\(x+3y=xy+3\\ \Leftrightarrow x+3y-\left(xy+3\right)=0\\\Leftrightarrow x+3y-xy-3=0\\ \Leftrightarrow\left(x-xy\right)+\left(3y-3\right)=0\\ \Leftrightarrow x\left(1-y\right)-\left(3-3y\right)=0\\ \Leftrightarrow x\left(1-y\right)-3\left(1-y\right)=0\\ \Leftrightarrow\left(x-3\right)\left(1-y\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-3=0\\1-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Vậy \(x=3;y=1\)

Câu 2 : \(\dfrac{\text{Giải.}}{ }\)

\(\text{Ta có : }n^3-n\\ =n\left(n^2-1\right)\\ =n\left(n-1\right)\left(n+1\right)\)

Vì \(n\left(n-1\right)\left(n+1\right)\) là 3 số tự nhiên liên tiếp.

nên \(\Rightarrow n\left(n-1\right)\left(n+1\right)⋮2;3\)

Mà \(2\) và \(3\) là 2 số nguyên tố cùng nhau.

\(\Rightarrow n\left(n-1\right)\left(n+1\right)⋮2\cdot3\)

\(\Rightarrow n\left(n-1\right)\left(n+1\right)⋮6\left(đpcm\right)\)

Vậy \(n^3-n⋮6\)

Câu 1:

\(\text{ a) }\left(x+2\right)^2-2\left(x+2\right)\left(x-8\right)+\left(x-8\right)^2\\ =\left[\left(x+2\right)-\left(x-8\right)\right]^2\\ =\left(x+2-x+8\right)^2\\ =10^2=100\\ \text{Vậy giá trị của biểu thức không phụ thuộc vào biến.}\\ \)

\(\text{b) }\left(x+y-z-t\right)^2-\left(z+t-x-y\right)^2\\ =\left[\left(x+y-z-t\right)+\left(z+t-x-y\right)\right]\left[\left(x+y-z-t\right)-\left(z+t-x-y\right)\right]\\ =\left(x+y-z-t+z+t-x-y\right)\left(x+y-z-t-z-t+x+y\right)\\ =\left[\left(x-x\right)+\left(y-y\right)-\left(z-z\right)-\left(t-t\right)\right]\left(x+y-z-t-z-t+x+y\right)\\ =0\cdot\left(x+y-z-t-z-t+x+y\right)\\ =0\\ \text{Vậy giá trị của biểu thức không phụ thuộc vào biến.}\)

\(\text{a) }\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(6x+1\right)\left(6x-1\right)\\ =\left[\left(6x+1\right)-\left(6x-1\right)\right]^2\\ =\left[6x+1-6x+1\right]^2\\ =2^2\\ =4\)

\(\text{b) }3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^{16}-1\right)\left(2^{16}+1\right)\\ =2^{32}-1\)

\(\text{a) }x^3+x^2+4\\ =x^3+2x^2-x^2+2^2\\ =\left(x^3+2x^2\right)-\left(x^2-2^2\right)\\ =x^2\left(x+2\right)-\left(x+2\right)\left(x-2\right)\\ =\left(x^2-x+2\right)\left(x+2\right)\)

\(\text{b) }4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)\\ =4\left(x^2-6x+9\right)-\left(4x^2-1\right)\\ =4x^2-24x+36-4x^2+1\\ =\left(4x^2-4x^2\right)-24x+\left(36+1\right)\\ =-24x+37\)

\(\text{c) }\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\\ \Leftrightarrow\left(x^2-8x+16\right)-\left(x^2-4\right)=6\\\Leftrightarrow x^2-8x+16-x^2+4=6\\ \Leftrightarrow\left(x^2-x^2\right)-8x+\left(16+4\right)=6\\ \Leftrightarrow-8x+20=6\\ \Leftrightarrow-8x=-14\\ \Leftrightarrow x=\dfrac{7}{4} \)

Vậy \(x=\dfrac{7}{4}\)