\(y=f\left(x\right)=3x-2\)

a)

\(y=f\left(1\right)=3.1-2=3-2=1\)

\(y=f\left(x\right)=3.\dfrac{1}{3}-2=1-2=-1\)

b) \(y=f\left(x\right)=3x-2=4\)

\(\Rightarrow3x=4+2\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=6:3\)

\(\Rightarrow x=2\)

Vậy \(x=2\) thì \(y=4\)

Gọi ba cạnh của tam giác ABC lần lượt là a, b, c. \(\left(a< b< c\right)\)

Theo đề ta có:

\(\dfrac{a}{5}=\dfrac{b}{6}=\dfrac{c}{7}\) và \(a+c-b=12\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{a}{5}=\dfrac{b}{6}=\dfrac{c}{7}=\dfrac{a+c-b}{5+7-6}=\dfrac{12}{6}=2\)

\(\dfrac{a}{5}=2\Rightarrow a=5.2=10\)

\(\dfrac{b}{6}=2\Rightarrow b=6.2=12\)

\(\dfrac{c}{7}=2\Rightarrow c=7.2=14\)

Vậy ba cạnh của tam giác ABC lần lượt là \(10;12;14\)

1) \(\left(x^2-16\right):\left(x+4\right)\)

\(=\left(x^2-4^2\right):\left(x+4\right)\)

\(=\left(x-4\right)\left(x+4\right):\left(x+4\right)\)

\(=x-4\)

2) \(8x^3-1\)

\(=\left(2x\right)^3-1^3\)

\(=\left(2x-1\right)\left[\left(2x\right)^2+2x.1+1^2\right]\)

\(=\left(2x-1\right)\left(4x^2+2x+1\right)\)

\(\left(2x+1\right)\left(2x-1\right)\)

\(=\left(2x\right)^2-1^2\)

\(=4x^2-1\)

\(5x^2+6x-4xy-2y+2+y^2=0\)

\(\Leftrightarrow4x^2+x^2+2x+4x-4xy-2y+1+1+y^2=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(4x-2y\right)+\left(x^2+2x+1\right)+1=0\)

\(\Leftrightarrow\left(2x-y\right)^2+2\left(2x-y\right)+1+\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(2x-y+1\right)^2+\left(x+1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y+1\right)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\)

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

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

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

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

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

Vậy \(x=-1\) và \(y=-1\)

\(B=3^5+3^6+3^7+3^8+....+3^{100}\)

\(\Rightarrow3B=3^6+3^7+3^8+3^9+....+3^{101}\)

\(\Rightarrow3B-B=\left(3^6+3^7+3^8+3^9+....+3^{101}\right)-\left(3^5+3^6+3^7+3^8+....+3^{100}\right)\)

\(\Rightarrow2B=3^{101}-3^5\)

\(\Rightarrow B=\dfrac{3^{101}-3^5}{2}\)

\(\Rightarrow B=\dfrac{3^{101}-3^5}{2}\)

Ta có:

\(4n^2+n+2⋮n+1\)

\(\Rightarrow\left(4n^2+4n\right)-3n+2⋮n+1\)

\(\Rightarrow4n\left(n+1\right)-3n+2⋮n+1\)

\(\Rightarrow-3n+2⋮n+1\)

\(\Rightarrow-\left(3n+3\right)+5⋮n+1\)

\(\Rightarrow-3\left(n+1\right)+5⋮n+1\)

\(\Rightarrow5⋮n+1\)

\(\Rightarrow n+1\in\left\{1;5\right\}\) ( Vì \(n\in N\) )

\(\Rightarrow\left\{{}\begin{matrix}n+1=1\Rightarrow n=0\\n+1=5\Rightarrow n=4\end{matrix}\right.\)

Vậy \(n\in\left\{0;4\right\}\)

Ta có:

\(x+3⋮x-1\)

\(\Rightarrow\left(x-1\right)+4⋮x-1\)

\(\Rightarrow4⋮x-1\)

\(\Rightarrow x-1\in U\left(4\right)=\left\{-1;1;-2;2;-4;4\right\}\)

\(\Rightarrow x\in\left\{0;2;-1;3;-3;5\right\}\)

Vậy \(x\in\left\{0;2;-1;3;-3;5\right\}\)

\(2012^0.\left(1,25-\dfrac{1}{4}\right)+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-\left|-5^2\right|\)

\(=1,25-\dfrac{1}{4}+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-\left|-25\right|\)

\(=1,25-\dfrac{1}{4}+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-25\)

\(=1,25-0,25+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-25\)

\(=1+\left(\dfrac{3}{5}\right)^8.\left(0,6\right)^7-25\)

\(=1+\left(\dfrac{3}{5}\right)^8.\left(\dfrac{3}{5}\right)^7-25\)

\(=1+\left(\dfrac{3}{5}\right)^{8+7}-25\)

\(=1+\left(\dfrac{3}{5}\right)^{15}-25\)

\(=1+\dfrac{3^{15}}{5^{15}}-25\)

\(=1,000470185-25\)

\(=-23,99952982\)