Với \(a+b+c+d=0\)

\(\Rightarrow a+b=-\left(c+d\right);b+c=-\left(d+a\right);c+d=-\left(a+b\right);d+a=-\left(b+c\right)\)

Khi đó \(M=-1-1-1-1=-4\)

Với \(a+b+c+d\ne0\)

Áp dụng dãy tỉ số bằng nhau

\(\frac{2019a+b+c+d}{a}=\frac{a+2019b+c+d}{b}=\frac{a+b+2019c+d}{c}=\frac{a+b+c+2019d}{d}\)

\(=\frac{2022\left(a+b+c+d\right)}{a+b+c+d}=2022\)

\(\Rightarrow a=b=c=d\)

\(\Rightarrow M=4\)

Làm giúp mik với,ngày mai mik phải nộp bài cho cô rồi:(

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{a+b+c+d}=1\\ \Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Rightarrow a=b=c=d\\ \Rightarrow VT=\left(\dfrac{2019a+2020a-2021a}{2019a+2020a-2021a}\right)^3=1^3=1=\dfrac{a^2}{a\cdot a}=VP\)

2.

\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2a+2b+2c+2d}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)

\(\Rightarrow a=\frac{2b}{2}=b;b=\frac{2c}{2}=c;c=\frac{2d}{2}=d;d=\frac{2a}{2}=a\)

\(\Rightarrow a=b=c=d\)

Ta có : \(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}\)

\(=\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}\)

\(=\frac{4a}{2a}=2\)

3.

\(\left(x-1\right)\left(x-3\right)< 0\)

\(\Rightarrow\hept{\begin{cases}x-1< 0\\x-3>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-1>0\\x-3< 0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x< 1\\x>3\end{cases}}\)( loại ) hoặc \(\hept{\begin{cases}x>1\\x< 3\end{cases}}\)

Vậy \(1< x< 3\)

Đặt \(A=\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{44\times49}\)

Ta có : \(5\times A=\frac{5}{4\times9}+\frac{5}{9\times14}+\frac{5}{14\times19}+...+\frac{5}{44\times49}=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}=\frac{1}{4}-\frac{1}{49}\)

\(=\frac{49}{196}-\frac{4}{196}=\frac{45}{196}\)

\(\Rightarrow A=\frac{9}{196}\)

Đặt \(B=1-3-5-7-...-49=1-\left(3+5+...+49\right)\)

Đặt \(C=3+5+...+49\) ( khoảng cách là 2 )

Số số hạng là : \(\left(49-3\right):2+1=24\)

Tổng C là : \(\left(49+3\right)\times24:2=624\)

\(\Rightarrow B=1-264=-623\)

Vậy \(A=\frac{9}{196}\times\frac{-623}{89}=\frac{-9}{28}\)

Dòng cuối cùng mình không chắc là đúng nhé !

\(\left(x-1\right)\left(x-3\right)< 0\)

=> x-1 và x-3 trái dấu

mà x-1>x-3 nên ta có:

\(\hept{\begin{cases}x-1>0\\x-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< 3\end{cases}\Rightarrow}-1< x< 3}\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

vậy x \(\in\left\{-2;-1;0;1;2\right\}\)

Theo tính chất tỉ dãy số bằng nhau thì:

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

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

\(\Rightarrow M\Leftrightarrow1+1+1+1=4\)

Ps: Cách mình nhanh hơn nè!

bạn trừ đi một rồi áp dụng tính chất dãy tỉ số bằng nhau nhé

Ta có:

\(\frac{a+b+c-d}{d}=\frac{b+c+d-a}{a}=\frac{c+d+a-b}{b}=\frac{d+a+b-c}{c}\)

\(=\frac{2\left(b+c\right)}{d+a}=\frac{2\left(a+b\right)}{c+d}=\frac{2\left(c+d\right)}{a+b}=\frac{2\left(d+a\right)}{b+c}=\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)

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

\(\Rightarrow M=\frac{b+c}{d+a}+\frac{a+b}{c+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)

\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\)

\(=\frac{a+b+2c+d+a+b+c+2d}{c+d}=\frac{2\left(a+b\right)}{c+d}+3=\)

Tương tự

\(=\frac{2\left(b+c\right)}{d+a}+3=\)

\(=\frac{2\left(c+d\right)}{a+b}+3=\)

\(=\frac{2\left(d+a\right)}{b+c}+3\)

\(\Rightarrow\frac{2\left(a+b\right)}{c+d}+3=\frac{2\left(b+c\right)}{d+a}+3=\frac{2\left(c+d\right)}{a+b}+3=\frac{2\left(d+a\right)}{b+c}+3\)

\(\Rightarrow\frac{2\left(a+b\right)}{c+d}=\frac{2\left(b+c\right)}{d+a}=\frac{2\left(c+d\right)}{a+b}=\frac{2\left(d+a\right)}{b+c}=\)

\(=\frac{2\left(a+b\right)+2\left(b+c\right)+2\left(c+d\right)+2\left(d+a\right)}{c+d+d+a+a+b+b+c}=\frac{4\left(a+b+c+d\right)}{2\left(a+b+c+d\right)}=2\)

\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)