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1) Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{2010}=\dfrac{2010}{a}=\dfrac{a+b+c+2010}{b+c+2010+a}=1\)
\(\dfrac{2010}{a}=1\Rightarrow a=2010\);
\(\dfrac{c}{2010}=1\Rightarrow c=2010\);
\(\dfrac{b}{c}=1\Rightarrow\dfrac{b}{2010}=1\Rightarrow b=2010\).
Vậy (a, b, c) = (2010; 2010; 2010)
3)
a) \(A=\sqrt{x+24}+\dfrac{4}{7}\)
Có: \(\sqrt{x+24}\ge0\forall x\in R\)
\(\Rightarrow\sqrt{x+24}+\dfrac{4}{7}\ge\dfrac{4}{7}\forall x\in R\)
\(\Rightarrow A\ge\dfrac{4}{7}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x+24}=0\Rightarrow x+24=0\Rightarrow x=-24\)
Vậy GTNN của \(A=\dfrac{4}{7}\Leftrightarrow x=-24\)
b) \(B=\sqrt{2x+\dfrac{4}{13}}-\dfrac{13}{191}\)
Có: \(\sqrt{2x+\dfrac{4}{13}}\ge0\forall x\in R\)
\(\Rightarrow\sqrt{2x+\dfrac{4}{13}}-\dfrac{13}{191}\ge-\dfrac{13}{191}\forall x\in R\)
\(\Rightarrow B\ge-\dfrac{13}{191}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{2x+\dfrac{4}{13}}=0\)
\(\Rightarrow2x+\dfrac{4}{13}=0\)
\(\Rightarrow2x=-\dfrac{4}{13}\)
\(\Rightarrow x=-\dfrac{2}{13}\)
Vậy GTNN của \(B=-\dfrac{13}{191}\Leftrightarrow x=-\dfrac{2}{13}\)
4)
a) \(A=-\sqrt{x+\dfrac{5}{41}}+\dfrac{7}{12}\)
Có: \(\sqrt{x+\dfrac{5}{41}}\ge0\forall x\in R\)
\(\Rightarrow-\sqrt{x+\dfrac{5}{41}}\le0\forall x\in R\)
\(\Rightarrow-\sqrt{x+\dfrac{5}{41}}+\dfrac{7}{12}\le\dfrac{7}{12}\forall x\in R\)
\(\Rightarrow A\le\dfrac{7}{12}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x+\dfrac{5}{41}}=0\)
\(\Rightarrow x+\dfrac{5}{41}=0\)
\(\Rightarrow x=-\dfrac{5}{41}\)
Vậy GTLN của \(A=\dfrac{7}{12}\Leftrightarrow x=-\dfrac{5}{41}\)
b) \(B=\dfrac{-5}{13}-\sqrt{x-\dfrac{2}{3}}\)
Có: \(\sqrt{x-\dfrac{2}{3}}\ge0\forall x\in R\)
\(\Rightarrow-\sqrt{x-\dfrac{2}{3}}\le0\forall x\in R\)
\(\Rightarrow\dfrac{-5}{13}-\sqrt{x-\dfrac{2}{3}}\le\dfrac{-5}{13}\forall x\in R\)
\(\Rightarrow B\le\dfrac{-5}{13}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x-\dfrac{2}{3}}=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\)
\(\Rightarrow x=\dfrac{2}{3}\)
Vậy GTLN của \(B=\dfrac{-5}{13}\Leftrightarrow x=\dfrac{2}{3}\)
a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)
\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)
\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)
c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)
\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)
Đặt \(S=\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{c}{c+a}\)
\(\Rightarrow S=\dfrac{2010-\left(a+b\right)}{a+b}+\dfrac{2010-\left(b+c\right)}{b+c}+\dfrac{2010-\left(c+a\right)}{c+a}\)\(\Rightarrow S=\dfrac{2010}{a+b}+\dfrac{2010}{b+c}+\dfrac{2010}{c+a}-3\)
\(\Rightarrow S=2010\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3\)
\(\Rightarrow S=2010.\dfrac{1}{3}-3\)
\(\Rightarrow S=670-3\)
\(\Rightarrow S=667\)
Ta có: \(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=\left(\dfrac{c}{a+b}+1\right)+\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)-3\)
\(=\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3\)
\(=2010.\dfrac{1}{3}-3\)
\(=670-3\)
\(=667\)
Vậy \(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=667\)
4.a
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Câu 2 :
\(x-y=7\)
\(\Rightarrow x=7+y\)
*)
\(B=\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)
\(=\dfrac{21+3y-7}{14+3y}-\dfrac{3y+7}{3y+7}\)
\(=\dfrac{14y+3y}{14y+3y}-1\)
\(=1-1\)
\(=0\)
Vậy B = 0
2/ Ta có :
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)
\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)
\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)
\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)
\(=1-1=0\)
a,
\(a+b=-9\\ b+c=2\\ c+a=-3\\ \Rightarrow a+b+b+c+c+a=\left(-9\right)+2+\left(-3\right)\\ 2a+2b+2c=-10\\ 2\left(a+b+c\right)=-10\\ a+b+c=-5\\ a+b=-9\\ \Rightarrow a+b+c=-5\Leftrightarrow\left(-9\right)+c=-5\Rightarrow c=4\\ b+c=2\\ \Rightarrow a+b+c=-5\Leftrightarrow a+2=-5\Rightarrow a=-7\\ c+a=-3\\ \Rightarrow a+b+c=-5\Leftrightarrow\left(-3\right)+b=-5\Rightarrow b=-2\)
Vậy \(a=-7;b=-2;c=5\)
b,
\(a+b=\dfrac{1}{2}\\ b+c=\dfrac{3}{4}\\ c+a=\dfrac{-5}{6}\\ \Rightarrow a+b+b+c+c+a=\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{-5}{6}\\ 2a+2b+2c=\dfrac{6}{12}+\dfrac{9}{12}+\dfrac{-10}{12}\\ 2\left(a+b+c\right)=\dfrac{5}{12}\\ a+b+c=\dfrac{5}{24}\\ a+b=\dfrac{1}{2}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow\dfrac{1}{2}+c=\dfrac{5}{24}\Rightarrow c=\dfrac{-7}{24}\\ b+c=\dfrac{3}{4}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow a+\dfrac{3}{4}=\dfrac{5}{24}\Rightarrow a=\dfrac{-13}{24}\\ a+c=\dfrac{-5}{6}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow b+\dfrac{-5}{6}=\dfrac{5}{24}\Rightarrow b=\dfrac{25}{24}\)
Vậy \(a=\dfrac{-13}{24};b=\dfrac{25}{24};c=\dfrac{-7}{24}\)
c,
\(a+b=2\\ b+c=6\\ c+a=3\\ \Rightarrow a+b+b+c+c+a=2+6+3\\ 2a+2b+2c=11\\ 2\left(a+b+c\right)=11\\ a+b+c=5,5\\ a+b=2\\ \Rightarrow a+b+c=5,5\Leftrightarrow2+c=5,5\Rightarrow c=3,5\\ b+c=6\\ \Rightarrow a+b+c=5,5\Leftrightarrow a+6=5,5\Rightarrow a=-0,5\\ c+a=3\\ \Rightarrow a+b+c=5,5\Leftrightarrow b+3=5,5\Rightarrow b=2,5\)
Vậy \(a=-0,5;b=2,5;c=3,5\)
d,
\(a+b=\dfrac{5}{6}\\ b+c=\dfrac{3}{4}\\ c+a=\dfrac{5}{3}\\ \Rightarrow a+b+b+c+c+a=\dfrac{5}{6}+\dfrac{3}{4}+\dfrac{5}{3}\\ 2a+2b+2c=\dfrac{10}{12}+\dfrac{9}{12}+\dfrac{20}{12}\\ 2\left(a+b+c\right)=\dfrac{13}{4}\\ a+b+c=\dfrac{13}{8}\\ a+b=\dfrac{5}{6}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow\dfrac{5}{6}+c=\dfrac{13}{8}\Rightarrow c=\dfrac{19}{24}\\ b+c=\dfrac{3}{4}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow a+\dfrac{3}{4}=\dfrac{13}{8}\Rightarrow a=\dfrac{7}{8}\\ c+a=\dfrac{5}{3}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow b+\dfrac{5}{3}=\dfrac{13}{8}\Rightarrow b=\dfrac{-1}{24}\)
Vậy \(a=\dfrac{7}{8};b=\dfrac{-1}{24};c=\dfrac{19}{24}\)
\(\left\{{}\begin{matrix}a+b=-9\\b+c=2\\c+a=-3\end{matrix}\right.\)
\(\Rightarrow a+b+b+c+c+a=\left(-9\right)+2+\left(-3\right)\)
\(\Rightarrow2a+2b+2c=-10\)
\(\Rightarrow2\left(a+b+c\right)=-10\)
\(\Rightarrow a+b+c=-5\)
\(\Rightarrow\left\{{}\begin{matrix}c=-5-9=-14\\a=-5-2=-7\\b=-5-\left(-3\right)=-2\end{matrix}\right.\)
\(M=\dfrac{3^{14}\cdot5^4-3^{12}\cdot5^4}{3^{12}\cdot5^6+7\cdot3^{12}\cdot5^6}=\dfrac{3^{12}\cdot5^4\left(3^2-1\right)}{3^{12}\cdot5^6\left(1+7\right)}=\dfrac{1}{25}\)