tại sao cậu toàn cho bài khó thế tớ chịu

Kẻ đường cao AH của tam giác ABC, ta có:

\(MI^2+MJ^2+MK^2=MI^2+MA^2=\left(MI+MA\right)^2-2MI.MA\ge\frac{\left(MI+MA\right)^2}{2}\)

Lại có: \(MI+MA\ge AI\ge AH\), cho nên: \(MI^2+MJ^2+MK^2\ge\frac{AH^2}{2}\)(không đổi)

Dấu "=" xảy ra <=> M là trung điểm AH.

dùng hằng đẳng thức nhé bạn

\(N=2x^4+4x^2y^2+2y^4-y^4-x^2y^2+y^2\)

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

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

mà ta có: \(x^2+y^2=1\)

\(\Rightarrow N=2-y^2+y^2=2\)

chúc bạn học tốt

Vì \(\left(x+2y-3\right)^{2016}\ge0;\left|2x+3y-5\right|\ge0\forall x;y\)

\(\Rightarrow\left(x+2y-3\right)^{2016}+\left|2x+3y-5\right|\ge0\forall x;y\)

Mà \(\left(x+2y-3\right)^{2016}+\left|2x+3y-5\right|=0\) \(\Leftrightarrow\left(x+2y-3\right)^{2016}=0\) ; \(\left|2x+3y-5\right|=0\)

\(\Rightarrow x+2y-3=0;2x+3y-5=0\)

\(\Leftrightarrow x+2y=3;2x+3y=5\)

\(\Rightarrow x=3-2y\)

\(\Rightarrow2\left(3-2y\right)+3y=5\Leftrightarrow6-4y+3y=5\Leftrightarrow6-y=5\Rightarrow y=1\)

\(\Rightarrow x=3-2.1=1\)

Vậy \(x=1;y=1\)

Ta có:\(3^{4n+2}+2.4^{3n+1}\)

\(=3^{4n}.3^2+2.4^{3n}.4\)

\(=64^n.9+64^n.8\)

\(=64^n.\left(9+8\right)\)

\(=64^n.17\)

\(vì\) \(17⋮17\)nên  \(64^n.17⋮17\)

Vậy \(3^{4n+2}\)\(+2.4^{3n+1}⋮17\)

=\(3^{4n}.3^2+2.4^{3n}.4\)

\(=81^n\cdot9+64^n\cdot8\)

\(=\left(64+17\right)^n.3^2+64^n\cdot8\)

\(=64^n.17^n.9+64^n\cdot8\)

\(64^n\left(17^n+8+9\right)⋮17\)

\(m-1⋮2m-1\)

\(\Leftrightarrow2\left(m-1\right)⋮2m-1\)

\(\Leftrightarrow2m-2⋮2m-1\)

\(\Leftrightarrow\left(2m-1\right)-1⋮2m-1\)

\(\Rightarrow1⋮2m-1\) \(\Rightarrow2m-1\inƯ\)(1) = {\(-1;1\)}

Với \(2m-1=-1\Rightarrow2m=0\Rightarrow n=0\) (TM)

Với \(2m-1=1\Leftrightarrow2m=2\Rightarrow m=1\)(TM)

Vậy \(m=\left[0;1\right]\) thì \(m-1⋮2m-1\)

\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{9}\right)\left(1+\frac{1}{10}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{10}{9}\cdot\frac{11}{10}\)

\(=\frac{3.4.5.....10.11}{2.3.4....10}=\frac{11}{2}\)

cảm ơn anh