vì: | x - 1 | \(\ge\)0

Mà | x - 1 | . ( y + 2 ) = -6 

=> | x - 1 | \(\in\left\{1;2;3;6\right\}\)

\(\Rightarrow x+1\in\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Mà | x - 1 | . ( y + 2 ) = -6 

\(\Rightarrow y-2\in\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Rồi bạn kẻ bảng là được

 Vi /x-1/.(y+20)=-6 nen x-1 va y+2 thuoc U(-6)={1;-1;2-2;3-3;6-6}

Ta co 2 truong hop: x-1<0 ;y+2 >0

                                    x-1>0;y+2 <0

ta xet TH1: x-1 <0 nen x-1=(-x-1) ;y+2>0 nen y+2= y+2

VAY THOI CU TU DO MA PHAT TRIEN LEN

Bài 5: 

a: BC=10cm

b: HA=4,8cm

HB=3,6(cm)

HC=6,4(cm)

Bài 6:

\(x^3=6+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow x^3=6+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=6\\ y^3=34+3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3y\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=34\\ \Leftrightarrow P=x^3-3x+y^3-3y+1980=6+34+1980=2020\)

gfrưerrrrrrrrrrr

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc hiểu đề và hỗ trợ tốt hơn nhé.

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:

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