a) Tứ giác BHCK có:

MC=MB(gt)

MH=MK(H và K đối xứng nhau qua tâm M)

Nên tứ giác BHCK là hình bình hành

b) suy ra BK//HC

mà HC vg với AB

nên BK vg AB

c) tg HIK có D là trung điểm HI và M là trung điểm của HK

nên DM là đường trung bình của tg HIK

suy ra DM//IK hay BC//IK suy ra BCKI là hình thang
mặt khác BC là đường trung trực của HI

nên CH=CI

mà CH=BK(tứ giác BKCH là hình bình hành)

do đó CI=BK
Hình thang BCKI có hai đường chéo CI=BK nên là hình thang cân

Ghi lại đề ^_^

\(A=\left(2x-y+1\right)^2+\left(x-3\right)^2-4y+2019=\frac{1}{5}\left[\left(5x-2y-1\right)^2+\left(y-17\right)^2\right]+1971\ge1971\)

Dấu "=" xảy ra khi x=7 , y=17

( 7x - 5 )² - ( 3x + 25 )² = 0

⇔ [ ( 7x - 5 ) - ( 3x + 25 ) ][ ( 7x - 5 ) + ( 3x + 25 ) ] = 0

⇔ ( 7x - 5 - 3x - 25 )( 7x - 5 + 3x + 25 ) = 0

⇔ ( 4x - 30 )( 10x + 20 ) = 0

⇔ 2( 2x - 15 ).10( x + 2 ) = 0

⇔ 20( 2x - 15 )( x + 2 ) = 0

⇔ 2x - 15 = 0 hoặc x + 2 = 0

⇔ x = 15/2 hoặc x = -2

\(\left(7x-5\right)^2-\left(3x+25\right)^2=0\)

\(\left(7x-5-3x-25\right)\left(7x-5+3x+25\right)=0\)

\(\left(4x-30\right)\left(10x+20\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-30=0\\10x+20=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{15}{2}\\x=-2\end{cases}}\)

Vì \(ab+bc+ca=2020\)

\(\Rightarrow a^2+2020=a^2+ab+bc+ca\)

\(=\left(a^2+ab\right)+\left(bc+ca\right)=a\left(a+b\right)+c\left(a+b\right)\)

\(=\left(a+b\right)\left(a+c\right)\)

Tương tự ta có: \(b^2+2020=\left(b+a\right)\left(b+c\right)\)

                          \(c^2+2020=\left(c+b\right)\left(c+a\right)\)

\(\Rightarrow\frac{a^2-bc}{a^2+2020}+\frac{b^2-ca}{b^2+2020}+\frac{c^2-ab}{c^2+2020}\)

\(=\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ca}{\left(b+a\right)\left(b+c\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)

\(=\frac{\left(a^2-bc\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}+\frac{\left(b^2-ca\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}+\frac{\left(c^2-ab\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(=\frac{\left(a^2-bc\right)\left(b+c\right)+\left(b^2-ca\right)\left(c+a\right)+\left(c^2-ab\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(=\frac{\left(a^2b+a^2c-b^2c-bc^2\right)+\left(b^2c+b^2a-c^2a-ca^2\right)+\left(c^2a+c^2b-a^2b-ab^2\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(=\frac{a^2b+a^2c-b^2c-bc^2+b^2c+b^2a-c^2a-ca^2+c^2a+c^2b-a^2b-ab^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(=\frac{0}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)( đpcm )

Ta có 

  \(a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\)

\(b^2+ab+bc+ac=\left(a+b\right)\left(b+c\right)\)

\(c^2+ab+bc+ac=\left(a+c\right)\left(b+c\right)\)

   Thay ab + bc + ac = 2020 vào biểu thức \(\frac{a^2-bc}{a^2+2020}+\frac{b^2-ca}{b^2+2020}+\frac{c^2-ab}{c^2+2020}\)có

       \(\frac{a^2-bc}{a^2+2020}+\frac{b^2-ca}{b^2+2020}+\frac{c^2-ab}{c^2+2020}\)

\(=\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ca}{\left(a+b\right)\left(b+c\right)}+\frac{c^2-ab}{\left(b+c\right)\left(a+c\right)}\)

\(=\frac{\left(a^2-bc\right)\left(b+c\right)+\left(b^2-ca\right)\left(a+c\right)+\left(c^2-ab\right)\left(b+a\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)

\(=\frac{a^2b+a^2c-b^2c-bc^2+ab^2+b^2c-a^2c-ac^2+ac^2-a^2b+bc^2-ab^2}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(=\frac{0}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=0\)

Xét tứ giác MNPQ có :

\(\widehat{M}+\widehat{N}+\widehat{P}+\widehat{Q}=360^o\)

\(35^o+67^o+127^o+\widehat{Q}=360^o\)

\(229^o+\widehat{Q}=360^o\)

                 \(\widehat{Q}=360^o-229^o\)

                 \(\widehat{Q}=131^o\)

Vậy \(\widehat{Q}=131^o\)

Sửa đề : a² ( b - c ) + b² ( c - a ) + c² ( a - b ) = 0

Ta có

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

\(\Leftrightarrow a^2b-a^2c+b^2c-b^2a-c^2b+c^2a=0\)

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

\(\Leftrightarrow ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(ab-ac-bc+c^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(a-c\right)=0\)

\(\Leftrightarrow\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}\Leftrightarrow\begin{cases}a=b\\b=c\\a=c\end{cases}\)

  Vậy a = b hoặc b = c hoặc a=c

\(3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)\)

\(=\left(3.5\right)^4-\left[\left(15^2\right)^2-1\right]\)

\(=15^4-\left(15^4-1\right)=15^4-15^4+1=1\)