ko hiểu chi luôn á trời

1.C

2.B

3.B

4.B

5.B

\(-4x^2-24xy-36y^2\)

\(=-\left(4x^2+24xy+36y^2\right)\)

\(=-\left[\left(2x\right)^2+24xy+\left(6y\right)^2\right]\)

\(=-\left[\left(2x\right)^2+2\cdot2x\cdot6y+\left(6y\right)^2\right]\)

\(=-\left(2x+6y\right)^2\)

\(=-\left[2\left(x+3y\right)\right]^2\)

\(=-4\left(x+3y\right)^2\)

Chịu nhá

\(A=\dfrac{b\left(2a^2+10ab+a+5b\right)}{a-3b}:\dfrac{a^2b+5ab^2}{a^2-3ab}\\ =\dfrac{b\left[2a\left(a+5b\right)+\left(a+5b\right)\right]}{a-3b}.\dfrac{a\left(a-3b\right)}{ab\left(a+5b\right)}\\ =\dfrac{b\left(a+5b\right)\left(2a+1\right)}{a-3b}.\dfrac{a\left(a-3b\right)}{ab\left(a+5b\right)}\\ =2a+1\)

Do : \(a\in Z=>2a+1\)  là số nguyên lẻ (DPCM)

\(\left(a\right):ĐKXĐ:\left\{{}\begin{matrix}x+3\ne0\\3-x\ne0\\x^2-9=\left(x-3\right)\left(x+3\right)\ne0\\x+2\ne0\end{matrix}\right.\\ < =>x\ne\left\{\pm3;-2\right\}\)

\(P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\\ =\left[\dfrac{2x-1}{x+3}+\dfrac{x}{x-3}-\dfrac{3-10x}{\left(x-3\right)\left(x+3\right)}\right].\dfrac{x-3}{x+2}\\ =\dfrac{\left(2x-1\right)\left(x-3\right)+x\left(x+3\right)-\left(3-10x\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+2}\\ =\dfrac{2x^2-x-6x+3+x^2+3x-3+10x}{\left(x+3\right)\left(x+2\right)}\\ =\dfrac{3x^2+6x}{\left(x+3\right)\left(x+2\right)}=\dfrac{3x\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}\\ =\dfrac{3x}{x+3}\)

\(\left(b\right):x^2-4=0< =>x^2=4\\ < =>\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)

Với \(x=2=>P=\dfrac{3.2}{2+3}=\dfrac{6}{5}\)

\(\left(a\right):A=\dfrac{x-5}{x-4}\left(x\ne4\right)\\ x^2-3x=0< =>x\left(x-3\right)=0\\ < =>\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\left(TMDK\right)\)

Với \(x=0=>A=\dfrac{0-5}{0-4}=\dfrac{-5}{-4}=\dfrac{5}{4}\)

Với \(x=3=>A=\dfrac{3-5}{3-4}=\dfrac{-2}{-1}=2\)

\(\left(b\right):\dfrac{x+5}{2x}-\dfrac{x-6}{5-x}-\dfrac{2x^2-2x-50}{2x^2-10x}\left(x\ne\left\{0;5\right\}\right)\)

\(=\dfrac{x+5}{2x}+\dfrac{x-6}{x-5}-\dfrac{2x^2-2x-50}{2x\left(x-5\right)}\\ =\dfrac{\left(x+5\right)\left(x-5\right)+2x\left(x-6\right)-\left(2x^2-2x-50\right)}{2x\left(x-5\right)}\\ =\dfrac{x^2-25+2x^2-12x-2x^2+2x+50}{2x\left(x-5\right)}\\ =\dfrac{x^2-10x+25}{2x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{2x\left(x-5\right)}\\ =\dfrac{x-5}{2x}\)

\(\left(a\right):A=\dfrac{5}{x+3}-\dfrac{2}{3-x}-\dfrac{3x^2-2x-9}{x^2-9}\left(x\ne\left\{\pm3\right\}\right)\\ =\dfrac{5}{x+3}+\dfrac{2}{x-3}-\dfrac{3x^2-2x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{5\left(x-3\right)+2\left(x+3\right)-\left(3x^2-2x-9\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)

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

\(\left(b\right):\left|x-2\right|=1\\ =>\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=3\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\)

Với \(x=1=>A=-\dfrac{3.1}{1+3}=-\dfrac{3}{4}\)

\(\left(b\right):A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right).\dfrac{x+2}{2}\\ =\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right].\dfrac{x+2}{2}\\ =\dfrac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{2}\\ =\dfrac{x-2x-4+x-2}{2\left(x-2\right)}\\ =\dfrac{-6}{2\left(x-2\right)}=\dfrac{3}{2-x}\)

\(\left(a\right):ĐKXĐ:\left\{{}\begin{matrix}x^2-4\ne0\\2-x\ne0\\x+2\ne0\end{matrix}\right.< =>x\ne\pm2\)

\(\left(a\right):\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}=\dfrac{30}{7xy}\)

\(\left(b\right):\dfrac{x^2-36}{2x+10}.\dfrac{3}{6-x}\\ =\dfrac{\left(x-6\right)\left(x+6\right)}{2x+10}.\dfrac{-3}{x-6}\\ =\dfrac{-3\left(x+6\right)}{2x+10}=\dfrac{-3x-18}{2x+10}\)

\(\left(c\right):\dfrac{1-x^2}{x^2-2x}:\dfrac{x+1}{x}\\ =\dfrac{\left(1-x\right)\left(1+x\right)}{x\left(x-2\right)}.\dfrac{x}{1+x}\\ =\dfrac{1-x}{x-2}\)

\(\left(d\right):\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\dfrac{3x}{2\left(1-2x\right)}\\ =\dfrac{3\left(1+2x\right)}{2\left(x+4\right)}=\dfrac{3+6x}{2x+8}\)

Có:

\(a^3+b^3+c^3=3abc\\\Leftrightarrow a^3+b^3+c^3-3abc=0\\\Leftrightarrow (a+b)^3+c^3-3ab(a+b)-3abc=0\\\Leftrightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Leftrightarrow (a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab)=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\\\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0(vì.a+b+c\ne0)\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(a-c\right)^2\ge0\forall a,c\end{matrix}\right.\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)

Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)

Thay \(a=b=c\) vào \(A\), ta được:

\(A=\dfrac{\left(2016+\dfrac{a}{a}\right)+\left(2016+\dfrac{b}{b}\right)+\left(2016+\dfrac{c}{c}\right)}{2017^3}\left(a,b,c\ne0\right)\)

\(=\dfrac{2016+1+2016+1+2016+1}{2017^3}\)

\(=\dfrac{2016\cdot3+1\cdot3}{2017^3}\)

\(=\dfrac{3\cdot\left(2016+1\right)}{2017^3}\)

\(=\dfrac{3}{2017^2}\)

Vậy: ...