\(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}\)  => 3(a+3)=5(b-2) <=> 3a+9=5b-10 => 3a=5b-19 (1)

và 7(b-2)=3(c-1) <=> 7b-14=3c-3 => \(c=\frac{7b-11}{3}\) (2)

Mà: 3a-5b-7c=86 => Thay (1) và (2) vào ta được:

\(5b-19-5b-7.\frac{7b-11}{3}=86\)=> 7b-11=-(86+19).3/7=-45 => 7b=-34 => b=-34/7

\(c=\frac{7\left(-\frac{34}{7}\right)-11}{3}=-\frac{45}{3}=-15\)

\(a=\frac{5\left(-\frac{34}{7}\right)-19}{3}=-\frac{303}{3}=-101\)

Đáp số: a=-101; b=-34/7; c=-15

\(\frac{a+3-3}{5-3}=\frac{b-2+2}{3+2}=\frac{c-1+1}{7+1}=\frac{a}{2}=\frac{b}{5}=\frac{c}{8}=\frac{3a}{6}=\frac{5b}{25}=\frac{7c}{56}=\frac{86}{37}\)

\(\Rightarrow\frac{a}{2}=\frac{86}{37}\Rightarrow a=\frac{172}{37};\frac{b}{5}=\frac{86}{37}\Rightarrow\frac{430}{37};\frac{c}{8}=\frac{86}{37}\Rightarrow\frac{688}{37}\)

\(\dfrac{a+3}{5}=\dfrac{b-2}{3}=\dfrac{c-1}{7}=\dfrac{3a+9}{15}=\dfrac{5b-10}{15}=\dfrac{7c-7}{49}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\dfrac{a+3}{5}=\dfrac{b-2}{3}=\dfrac{c-1}{7}=\dfrac{3a+9}{15}=\dfrac{5b-10}{15}=\dfrac{7c-7}{49}\\ =\dfrac{3a+9-5b+10+7c-7}{15+15+49}=\dfrac{\left(3a-5b+7c\right)+\left(9+10-7\right)}{79}\\ =\dfrac{86+12}{79}=\dfrac{98}{79}\\ a+3=5\cdot\dfrac{98}{79};b-2=3\cdot\dfrac{98}{79};c-1=7\cdot\dfrac{98}{79}\\ a+3=\dfrac{490}{79};b-2=\dfrac{294}{79};c-1=\dfrac{686}{79}\\ a=\dfrac{253}{79};b=\dfrac{452}{79};c=\dfrac{765}{79}\)

Đặt a+35=b−23=c−17=k⇒⎧⎩⎨a=5k−3b=3k+2c=7k+1a+35=b−23=c−17=k⇒{a=5k−3b=3k+2c=7k+1

Vì 3a - 5b + 7c = 86 => 5k - 3 - 3k - 2 + 7k + 1 = 86

=>9k + -4 = 86 => 9k = 90 => k = 10

=> ⎧⎩⎨a=47b=32c=71

\(\dfrac{a+3}{5}=\dfrac{b-2}{3}=\dfrac{c-1}{7}\)

\(\Leftrightarrow\dfrac{3a+9}{15}=\dfrac{5b-10}{15}=\dfrac{7c-7}{49}\)

Theo tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{3a+9}{15}=\dfrac{5b-10}{15}=\dfrac{7c-7}{49}=\dfrac{3a+9-\left(5b-10\right)+\left(7c-7\right)}{15-15+49}=\dfrac{\left(3a-5b+7c\right)+9+10-7}{49}=\dfrac{86+12}{49}=\dfrac{98}{49}=2\)

\(\Rightarrow\left\{{}\begin{matrix}3a+9=30\\5b-10=30\\7c-7=98\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a=21\\5b=40\\7c=105\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=7\\b=8\\c=15\end{matrix}\right.\)

\(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}=\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}=\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}\)

\(=\frac{3a+9-5b+10+7c-7}{15-15+49}=\frac{86+12}{49}=2\)

=>a=2.5-3=7;b=2.3+2=8;c=2.7+1=15

\(3\left(a+3\right)=5\left(b-2\right) \)
\(3a+9=5b-10\)

\(3a-5b=-9-10=-19\)
\(3a-5b=-19\)

thay -19 vào 3a -5b ta có: -19+7c =86  suy ra c= (86+19)/7=15
thay 15 vào c ta có \(\frac{14}{7}=\frac{a+3}{5}=\frac{b-2}{3}\)hay \(2=\frac{a+3}{5}=\frac{b-2}{3}\)vậy a= 7, b=8 và c=15
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Bạn kia làm đúng rồi

\(M=\left|x-2021\right|+\left|x-2020\right|=\left|2021-x\right|+\left|x-2020\right|\)

Ta có: \(\hept{\begin{cases}\left|2021-x\right|\ge2021-x\\\left|x-2020\right|\ge x-2020\end{cases}}\Rightarrow M\ge2021-x+x-2020=1\)

Dấu '' = '' xảy ra khi: \(\hept{\begin{cases}2021-x\ge0\\x-2020\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\le2021\\x\ge2020\end{cases}}\Rightarrow2020\le x\le2021\)