5050,đúng thì tick cho mik

\(•B=4x-9x^2=-\left(9x^2-4x\right)\\ =-\left(9x^2-3x.2.\dfrac{2}{3}+\dfrac{4}{9}\right)+\dfrac{4}{9}\\ =-\left(3x-\dfrac{2}{3}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\\dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{2}{9}\\ vậy\: MAX_B=\dfrac{4}{9}\: tại\: x=\dfrac{2}{9}\\ •C=5-2x-4x^2=-\left(4x^2+2x-5\right)\\ =-\left(4x^2+2.2x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{21}{4}\\ =-\left(2x+\dfrac{1}{2}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\\ dấu\: "="\: xảy\: ra\: khi\: x=-\dfrac{1}{4}\\ vậy\: MAX_C=\dfrac{21}{4}\: tại\: x=\dfrac{-1}{4}\\ •D=7+3x-x^2=-\left(x^2-3x-7\right)\\ =-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)+\dfrac{37}{4}\\ =-\left(x-\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\\ dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{3}{2}\\ vậy\: MAX_D=\dfrac{37}{4}\: tại\: x=\dfrac{3}{2}\)\(•E=1+x-x^2=-\left(x^2-x-1\right)\\ =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\\ dấu\:"="\:xảy\:ra\:khi\:x=\dfrac{1}{2}\\ vậy\:MAX_E=\dfrac{5}{4}\:tại\:x=\dfrac{1}{2}\\ •F=-5x-6x^2\\ -\dfrac{F}{6}=x^2+\dfrac{5}{6}x=x^2+2.x.\dfrac{5}{12}+\dfrac{25}{144}-\dfrac{25}{144}\\ -\dfrac{F}{6}=\left(x+\dfrac{5}{12}\right)^2-\dfrac{25}{144}\\ F=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{25}{24}\le\dfrac{25}{24}\\ dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{-5}{12}\\ vậy\: MAX_F=\dfrac{25}{24}\: tại\: x=\dfrac{-5}{12}\)

\(a^2+b^2+c^2=ab+ac+bc\\ a^2+b^2+c^2-ab-ac-bc=0\\ 2a^2+2b^2+2c^2-2ab-2ac-2bc=0\\ a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\\ \left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\\ \Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\left(đpcm\right)\)

1

\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\\ 10x-4x^2-14x+35=4x^2-25\\ 8x^2+4x-60=0\\ x^2+\dfrac{1}{2}x-\dfrac{60}{8}=0\\ x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}=\dfrac{1}{16}+\dfrac{60}{8}\\ \left(x+\dfrac{1}{4}\right)^2=\dfrac{121}{16}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{11}{4}\\x+\dfrac{1}{4}=-\dfrac{11}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{4}\\x=-\dfrac{12}{4}=-3\end{matrix}\right.\)

2

\(x^2-4x+5=x^2-4x+4+1\\ =\left(x-2\right)^2+1>0\)

\(2x^3+3x^2+2x+3=0\\ 2x\left(x^2+1\right)+3\left(x^2+1\right)=0\\ \left(x^2+1\right)\left(2x+3\right)=0\\ 2x+3=0\\ 2x=-3\\ x=-\dfrac{3}{2}\)

\(\left[{}\begin{matrix}25^{10}=5^{20}\\125^7=5^{21}\end{matrix}\right.\)

vì 20<21 nên\(25^{10}< 125^7\)

chọn A

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\\ 4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ 2x+255=0\\ 2x=-255\\ x=-\dfrac{255}{2}\)

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

đặt: \(\left[{}\begin{matrix}n=5k+1\\m=5k+2\end{matrix}\right.\)

khi đó:

\(n^2+m^2=\left(5k+1\right)^2+\left(5k+2\right)^2\\ =25k^2+10k+1+25k^2+20k+4\\ =50k^2+30k+5=5\left(10k^2+6k+1\right)⋮5\)

vậy \(n^2+m^2⋮5\)