Bài 3:

Giải:

Ta có: \(\widehat{N_4}=\widehat{N_1}=120^o\) ( đối đỉnh )

Ta thấy \(\widehat{N_1}+\widehat{M_1}=180^o\) và 2 góc này ở vị trí trong cùng phía nên suy ra a // b

Vì a // b nên \(\widehat{M_1}=\widehat{N_3}=60^o\) ( đồng vị )

\(\widehat{N_3}=\widehat{N_2}=60^o\) ( đối đỉnh )

Vậy a // b 

        \(\widehat{N_1}=120^o,\widehat{N_2}=60^o,\widehat{N_3}=60^o\)

Bài 1: 

Ta có: \(\widehat{aOc}=\widehat{bOd}=45^o\) (đối đình)

Lại có: \(\widehat{aOc}+\widehat{aOd}=180^o\) (kề bù)

\(\Rightarrow45^o+\widehat{aOd}=180^o\)

\(\Rightarrow\widehat{aOd}=180^o-45^o=35^o\)

Vì: \(\widehat{aOd}\) và \(\widehat{cOd}\) đối đỉnh

\(\Rightarrow\widehat{aOd}=\widehat{cOd}=135^o\)

I sports day at school. School organizes a lot of games. But I the game more tug. Each class will nominate 20 friends to compete. Hold on to your hands is a large and long rope, in the middle is a red ribbon marks the landmark. There is a white line in the yard to divide the boundaries of the two teams. Once you know your opponent is 3H class, we are happy to play in the field. Everybody expresses great determination. The sound of "start" sounded, both teams put their strength into their hands, their feet clinging to the ground, who leaned back to pull the pull. The strings move to the side of the team, and then to the side of the team. The drumming, cheering cheers of the fans make us even more excited. In the end, after two matches, the victory was for our 3I class. The class hugged each other dancing, excited victory. I to play tug of war because the pull brings us health and solidarity.

Dịch ra hả

1, \(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}\)

 =>   \(\dfrac{a+b}{c}-1=\dfrac{a+c}{b}-1=\dfrac{b+c}{a}-1\)

 =>   \(\dfrac{a+b}{c}=\dfrac{a+c}{b}=\dfrac{b+c}{a}\)

=>    \(\dfrac{a+b}{c}=\dfrac{a+c}{b}=\dfrac{b+c}{a}=\dfrac{a+b+a+c+b+c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

=>  \(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{a+b}{c}\times\dfrac{a+c}{b}\times\dfrac{b+c}{a}=2.2.2=8\)

=>   \(M=8\)

Bài 1:

Nếu $a+b+c=0$ thì đkđb thỏa mãn

$M=\frac{(-c)(-a)(-b)}{abc}=\frac{-(abc)}{abc}=-1$

Nếu $a+b+c\neq 0$. Áp dụng TCDTSBN:

$\frac{a+b-c}{c}=\frac{a+c-b}{b}=\frac{b+c-a}{a}=\frac{a+b-c+a+c-b+b+c-a}{c+b+a}=\frac{a+b+c}{a+b+c}=1$

$\Rightarrow a+b-c=c; a+c-b=b; b+c-a=a$

$\Leftrightarrow a+b=2c; a+c=2b; b+c=2a$

$\Rightarrow a=b=c$

$M=\frac{(a+a)(a+a)(a+a)}{aaa}=\frac{8a^3}{a^3}=8$

Bài 2a

Đặt $2x=3y=4z=t$

$\Rightarrow x=\frac{t}{2}; y=\frac{t}{3}; z=\frac{t}{4}$

Khi đó:

$|x+y+3z|=1$

$\Leftrightarrow |\frac{t}{2}+\frac{t}{3}+\frac{3t}{4}|=1$

$\Leftrightarrow |\frac{19}{12}t|=1$

$\Rightarrow t=\pm \frac{12}{19}$

Nếu $t=\frac{12}{19}$ thì:

$x=\frac{t}{2}=\frac{6}{19}; y=\frac{4}{19}; z=\frac{3}{19}$

Nếu $t=-\frac{12}{19}$ thì:

$x=\frac{t}{2}=\frac{-6}{19}; y=\frac{-4}{19}; z=\frac{-3}{19}$

Bài 2b:

$\frac{x^3}{8}=\frac{y^3}{27}=\frac{z^3}{64}$

$\Leftrightarrow (\frac{x}{2})^3=(\frac{y}{3})^3=(\frac{z}{4})^3$

$\Leftrightarrow \frac{x}{2}=\frac{y}{3}=\frac{z}{4}$

Đặt $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=t$

$\Rightarrow x=2t; y=3t; z=4t$

Khi đó:
$x^2+2y^2-3z^2=-650$

$\Leftrightarrow (2t)^2+2.(3t)^2-3(4t)^2=-650$

$\Leftrightarrow -26t^2=-650$

$\Leftrightarrow t=\pm 5$

Nếu $t=5$ thì:

$x=2t=10; y=3t=15; z=4t=20$

Nếu $t=-5$ thì:

$x=2t=-10; y=3t=-15; z=4t=-20$

Bài 1:

Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\Leftrightarrow M=\dfrac{-abc}{abc}=-1\)

Với \(a+b+c\ne0\Leftrightarrow\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b+c}{a+b+c}=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\Leftrightarrow M=\dfrac{2a\cdot2b\cdot2c}{abc}=8\)

Bài 2:

\(a,TH_1:x+y+3z=1\\ \Leftrightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+3z}{6+4+9}=\dfrac{1}{19}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{6}{19}\\y=\dfrac{4}{19}\\z=\dfrac{3}{19}\end{matrix}\right.\\ TH_2:x+y+3z=-1\\ \Leftrightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+3z}{6+4+9}=\dfrac{-1}{19}\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{6}{19}\\y=-\dfrac{4}{19}\\z=-\dfrac{3}{19}\end{matrix}\right.\)

Bài 2:

\(b,\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+2y^2-3z^2}{4+18-48}=\dfrac{-650}{-26}=25\\ \Leftrightarrow\left\{{}\begin{matrix}x^2=100\\y^2=225\\z^2=400\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=15\\z=20\end{matrix}\right.\)

1. TH1:a+b+c≠0

Áp dụng t/c dtsbn ta có:

\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+a+c-b+b+c-a}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{a+c-b}{b}=1\Rightarrow a+c-b=b\Rightarrow a+c=2b\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\)

\(=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\\ =\dfrac{2c.2a.2b}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)

TH2:a+b+c=0

\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{-c.-a.-b}{abc}=\dfrac{-abc}{abc}=-1\)

n= -1 ; -5 ; 1; 5

\(n+6⋮n+1\)

\(\Rightarrow5⋮n+1\)

\(\Rightarrow\left(n+1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(n\in\left\{-6;-2;0;4\right\}\)

n+6 ⋮ n+1

=> n+1+5 ⋮ n+1 mà n+1 ⋮ n+1=>5 ⋮ n+1

=>n+1∈Ư(5)={±1;±5}

=>n={-6;-2;1;4