\(A=a\left(b+c\right)-b\left(a-c\right)=ab+ac-ab+bc=ac+bc\\ B=\left(a+b\right)c=ac+bc\\ =>A=B\)

\(9876543.9876545=\left(9876544-1\right).\left(9876544+1\right)=9876544^2-1^2< 9876544^2\)

\(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\\ < =>\left(6x-2\right)^2-2.\left(6x-2\right).\left(5x-2\right)+\left(5x-2\right)^2=0\\ < =>\left[\left(6x-2\right)-\left(5x-2\right)\right]^2=0\\ < =>x^2=0\\ < =>x=0=>S=\left\{0\right\}\)

\(Bổ\ sung\ đề\ bài : Tìm\ n\in Z\)

\(ĐK:n\ne-3\)

\(n^2+6n-3=\left(n^2+3n\right)+\left(3n+9\right)-12\\ =n\left(n+3\right)+3\left(n+3\right)-12\\ =\left(n+3\right)^2-12\)

\(Theo\ đề\ bài :\) \(n^2+6n-3⋮\left(n+3\right)\)

\(=>\left(n+3\right)^2-12⋮\left(n+3\right)\\ =>12⋮\left(n+3\right)\)

\(=>n+3\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

\(=>n\in\left\{-4;-5;-6;-7;-9;-15;-2;-1;0;1;3;9\right\}\left(TMDK\right)\)

\(Bổ\ sung\ đề \ bài :a,b,c>0\)

\(Nhân\ cả\ 2\ vế\ cho\ a+b+c>0,\ ta\ được :\)

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\\< =>1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\ge9\)

\(< =>\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge6\) \((*)\)

\(Áp\ dụng\ BĐT\ Cô\ Si\ cho\ các\ số\ dương, ta\ được :\)

\(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}+2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\sqrt{1}+2\sqrt{1}+2\sqrt{1}=6\)

\(=>(*)\ luôn\ đúng\)

\(=>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) \((Dấu''=''\ xảy\ ra \ khi :a=b=c>0)\)

\(\sqrt{10-\sqrt{84}}-\sqrt{34+2\sqrt{189}}\\ =\sqrt{7-2\sqrt{21}+3}-\sqrt{27+2\sqrt{189}+7}\\ =\sqrt{\sqrt{7}^2-2.\sqrt{7}.\sqrt{3}+\sqrt{3}^2}-\sqrt{\sqrt{27}^2+2.\sqrt{27}.\sqrt{7}+\sqrt{7}^2}\\ =\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{27}+\sqrt{7}\right)^2}\\ =\left|\sqrt{7}-\sqrt{3}\right|-\left|\sqrt{27}+\sqrt{7}\right|\\ =\sqrt{7}-\sqrt{3}-\left(3\sqrt{3}+\sqrt{7}\right)=-4\sqrt{3}\)

\(a,15^2-10.15+5^2=15^2-2.15.5+5^2\\ =\left(15-5\right)^2=10^2=100\)

\(b,99^2-1=99^2-1^2\\ =\left(99-1\right)\left(99+1\right)\\ =98.100=9800\)

\(a^2+2ab+b^2+5=\left(a+b\right)^2+5\ge5>0\forall a,b\left(DPCM\right)\)

\(x^2+y^2+4x+4=\left(x^2+4x+4\right)+y^2\\ =\left(x+2\right)^2+y^2\ge0\forall x,y\left(DPCM\right)\)

\(\left(x-3\right)\left(x-5\right)+3=x^2-8x+15+3\\ =\left(x^2-8x+16\right)+2\\ =\left(x-4\right)^2+2\ge2>0\forall x\left(DPCM\right)\)

\(a+b\ge2\sqrt{ab}\) \((BĐT\ Cô\ Si)\)

\(< =>1\ge2\sqrt{ab}\\ < =>\sqrt{ab}\le\dfrac{1}{2}\\ < =>0< ab\le\dfrac{1}{4}\)

\(A=\dfrac{a}{b}+\dfrac{b}{a}+ab\)

\(=\left(\dfrac{a}{b}+4ab\right)+\left(\dfrac{b}{a}+4ab\right)-7ab\ge2\sqrt{\dfrac{a}{b}.4ab}+2\sqrt{\dfrac{b}{a}.4ab}-7ab\) \((BĐT\ Cô\ Si)\)

\(=>A\ge2\sqrt{4a^2}+2\sqrt{4b^2}-7ab\ge2.2a+2.2b-7.\dfrac{1}{4}=4\left(a+b\right)-\dfrac{7}{4}=4.1-\dfrac{7}{4}=\dfrac{9}{4}\)

\(Dấu''=''\ xảy\ ra\ khi:\) \(\left\{{}\begin{matrix}a=b>0\\\dfrac{a}{b}=4ab\\\dfrac{b}{a}=4ab\\a+b=1\end{matrix}\right.< =>a=b=\dfrac{1}{2}\)

\(Vậy\ GTNN\ của\ A\ là :\) \(\dfrac{9}{4}< =>a=b=\dfrac{1}{2}\)

\(Áp\ dụng\ BĐT\ Cô\ Si\ cho\ 2\ số\ thực\ dương,\ ta\ được :\)

\(a^2+b^2\ge2\sqrt{a^2.b^2}=2ab\\ < =>4\ge2ab\\ < =>ab\le2\)

\(Ta\ có :\)

\(a^2+b^2=4\\ < =>\left(a+b\right)^2=4+2ab\le4+2.2=8\\ < =>0< a+b\le2\sqrt{2}\)

\(Lại\ có:\)

 \(a^2+b^2=4\\ < =>\left(a+b\right)^2=4+2ab\\ < =>2ab=\left(a+b\right)^2-4\\ < =>ab=\dfrac{\left(a+b-2\right)\left(a+b+2\right)}{2}\)

\(=>T=\dfrac{\dfrac{\left(a+b-2\right)\left(a+b+2\right)}{2}}{a+b+2}=\dfrac{a+b-2}{2}\le\dfrac{2\sqrt{2}-2}{2}=\sqrt{2}-1\)

\(Dấu ''=''\ xảy\ ra\ khi :\) \(\left\{{}\begin{matrix}a^2=b^2\\a^2+b^2=4\\a,b>0\end{matrix}\right.< =>a=b=\sqrt{2}\)

\(Vậy\ GTLN\ của\ T\ là: \) \(\sqrt{2}-1< =>a=b=\sqrt{2}\)

\(A=5+5^3+5^5+5^7+...+5^{49}+5^{51}\\ =>5^2.A=5^3+5^5+5^7+5^9+...+5^{51}+5^{53}\\ =>25A-A=\left(5^3+5^5+5^7+5^9+...+5^{51}+5^{53}\right)-\left(5+5^3+5^5+5^7+...+5^{49}+5^{51}\right)\\ =>24A=5^{53}-5\\ =>A=\dfrac{5^{53}-5}{24}\)