Theo đề bài : A^=B^+10*

B^=C^+10*

C^=D^+10*

<=> A^- 10* =C^ +10*

B^- 10* = D^ + 10*

Mà A^+B^+C^+D^=360*

=> A^ - 10*+C^ +10* <=> B^ (+10*-10*)+D^+10*+10* 

=>2(A^-10*+C^+10*)=360*

=>A^-10*+C^+10*=180*

Ta có : A^-10*=C^+10* =>A^=C^+10*+10*

=> A^=(180*+10*+10*):2=100*

=>C^=180*-100*=80*

=>B^=80*+10*=90*

=>D^=360*-100*-90*-80*=90*

Vậy ....

a) Theo bài ra, ta có:

\(\widehat{A}\):\(\widehat{B}\): \(\widehat{C}\) : \(\widehat{D}\) = 1 : 2 : 3 : 4 => \(\frac{\widehat{A}}{1}=\frac{\widehat{B}}{2}=\frac{\widehat{C}}{3}=\widehat{\frac{D}{4}}\) và \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=180^0\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

\(\frac{\widehat{A}}{1}=\widehat{\frac{B}{2}}=\frac{\widehat{C}}{3}=\frac{\widehat{D}}{4}=\frac{\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}}{1+2+3+4}=\frac{360^0}{10}=36^0\)

=> \(\frac{\widehat{A}}{1}=36^0\) => \(\widehat{A}=36^0\)

    \(\widehat{\frac{B}{2}}=36^0\)=> \(\widehat{B}=72^0\)

   \(\widehat{\frac{C}{3}}=36^0\) => \(\widehat{C}=108^0\)

\(\widehat{\frac{D}{4}}=36^0\) => \(\widehat{D}=144^0\)

Vậy ...

b) Xét tứ giác ABCD có góc A + góc B + góc C + góc D = 360⁰

hay góc  A + (góc A + 10⁰) + góc C + (góc C + 10⁰) = 360⁰

=> 2.(góc A + góc C) = 340⁰

=> góc A + góc C = 170⁰ => góc B + góc D = 360⁰ - 170⁰ = 190⁰

Ta có: góc B = góc A + 10⁰ (1)

     góc C = góc B + 10⁰ (2)

   góc D = góc C + 10⁰ (3)

Từ (1) và (2) cộng vế cho vế :

góc B + góc C = góc A + 10⁰ + góc B + 10⁰

=> góc C = góc A + 20⁰ => góc C - A = 20⁰

Mà góc  A + góc C = 170⁰

=> 2. góc C = 190⁰ => góc C = 95⁰

                           => góc A = 95⁰ - 20⁰ = 75⁰

Từ (2) và (3) cộng vế cho vế :

góc C + góc D = góc B + 10⁰ + góc C + 10⁰ 

=> góc D = góc B + 20⁰ => góc D - góc B = 20⁰

Mà góc D + góc B = 190⁰

=> 2. góc D  = 210⁰ => góc D = 105⁰

                     => góc B = 105⁰ - 20⁰ = 85⁰

c) Xét tứ giác ABCD góc A + góc B + góc C + góc D = 360⁰

=> góc A + góc B = 360⁰ - góc C - góc D = 360⁰ - 60⁰ - 80⁰ = 220⁰

Mà góc A - góc B = 10⁰

=> 2. góc A = 230⁰ => góc A = 115⁰

            => góc B = 115 - 10⁰ = 105⁰

Vậy ... 

ứ giác ABCD có :

ˆB=ˆA+10B^=A^+10(1)(1)

ˆC=ˆB+10C^=B^+10

Thay (1) vào ( 2) ⇒ˆC−10=ˆA+10⇒ˆC=200+ˆA⇒C^−10=A^+10⇒C^=200+A^(2)

ˆD=ˆC+10=200+A+10=300+AD^=C^+10=200+A+10=300+A(3)

(1),(2),(3) =>A+B+C+D=360=>ˆA+10+ˆA+20+ˆA+30+ˆA=360=>4ˆA+60=360=>ˆA=750A+B+C+D=360=>A^+10+A^+20+A^+30+A^=360=>4A^+60=360=>A^=750

=>ˆB=85.;ˆC=950;ˆD=1050=>B^=85.;C^=950;D^=1050.

Ta có :

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10}\)

\(\Rightarrow2017\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017.\frac{1}{10}\)

\(\Rightarrow\frac{2017}{a+b}+\frac{2017}{b+c}+\frac{2017}{c+a}=201,7\)

Mà \(2017=a+b+c\)nên :

\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=201,7\)

\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+b}+\frac{b}{a+c}\right)=201,7\)

\(3+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=201,7\)

\(\Leftrightarrow M=\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=201,7-3\)

\(\Leftrightarrow M=198,7\)

Vậy ...

\(a^2-b^2-c^2-2bc-2ac-2ab\)

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

\(=\left(a-b-c\right)^2=10^2=100\)

Sửa đề tí nha

a-b-c = 10 ms đúng chớ