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\(\frac{a+3-3}{5-3}=\frac{b-2+2}{3+2}=\frac{c-1+1}{7+1}=\frac{a}{2}=\frac{b}{5}=\frac{c}{8}=\frac{3a}{6}=\frac{5b}{25}=\frac{7c}{56}=\frac{86}{37}\)
\(\Rightarrow\frac{a}{2}=\frac{86}{37}\Rightarrow a=\frac{172}{37};\frac{b}{5}=\frac{86}{37}\Rightarrow\frac{430}{37};\frac{c}{8}=\frac{86}{37}\Rightarrow\frac{688}{37}\)
\(\dfrac{a+3}{5}=\dfrac{b-2}{3}=\dfrac{c-1}{7}=\dfrac{3a+9}{15}=\dfrac{5b-10}{15}=\dfrac{7c-7}{49}\)
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:
\(\dfrac{3a+9}{15}=\dfrac{5b-10}{15}=\dfrac{7c-7}{49}=\dfrac{3a+9-\left(5b-10\right)+\left(7c-7\right)}{15-15+49}=\dfrac{\left(3a-5b+7c\right)+9+10-7}{49}=\dfrac{86+12}{49}=\dfrac{98}{49}=2\)
\(\Rightarrow\left\{{}\begin{matrix}3a+9=30\\5b-10=30\\7c-7=98\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a=21\\5b=40\\7c=105\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=7\\b=8\\c=15\end{matrix}\right.\)
\(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\\ 5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\\ \Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}=\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a-7b+5c}{63-98+50}=\dfrac{-30}{15}=-2\\ \Rightarrow\left\{{}\begin{matrix}a=-42\\b=-28\\c=-20\end{matrix}\right.\)
\(x:y:z=3:4:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow x=3k;y=4k;z=5k\)
\(2x^2+2y^2-3z^2=-100\\ \Rightarrow18k^2+32k^2-75k^2=-100\\ \Rightarrow-25k^2=-100\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=6;y=8;z=10\\x=-6;y=-8;z=-10\end{matrix}\right.\)
\(3\left(a+3\right)=5\left(b-2\right)
\)
\(3a+9=5b-10\)
\(3a-5b=-9-10=-19\)
\(3a-5b=-19\)
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
Còn một cách khác mà làm biếng tí
\(M=\left|x-2021\right|+\left|x-2020\right|=\left|2021-x\right|+\left|x-2020\right|\)
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: \(\hept{\begin{cases}2021-x\ge0\\x-2020\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\le2021\\x\ge2020\end{cases}}\Rightarrow2020\le x\le2021\)
Ta có \(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}\)\(=\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}=\frac{3a+9-5b+10+7c-7}{15-15+49}\)\(=\frac{\left(3a-5b+7c\right)+\left(9+10-7\right)}{49}=\frac{86+12}{49}=\frac{98}{49}=2\)
\(=>\hept{\begin{cases}a+3=10\\b-2=6\\c-1=14\end{cases}\left(=\right)\hept{\begin{cases}a=7\\b=8\\c=15\end{cases}}}\)
học tốt
\(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}\) => 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