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Áp dụng BĐT Cauchy-Schwarz và BĐT AM-GM ta có:
\(S=\dfrac{1}{xy}+\dfrac{1}{xz}\ge\dfrac{4}{xy+xz}=\dfrac{4}{x\left(y+z\right)}\)
\(\ge\dfrac{4}{\dfrac{\left(x+y+z\right)^2}{4}}=\dfrac{16}{\left(x+y+z\right)^2}=1\)
Xảy ra khi \(x=2;y=z=1\)
\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
Ta có :\(\dfrac{x}{y+z}=\dfrac{123-\left(y+z\right)}{y+z}\)
\(\dfrac{y}{x+z}=\dfrac{123-\left(x+z\right)}{x+z}\)
\(\dfrac{z}{y+x}=\dfrac{123-\left(y+x\right)}{y+x}\)
\(\Rightarrow P=\dfrac{123-\left(y+z\right)}{y+z}+\dfrac{123-\left(z+x\right)}{z+x}+\dfrac{123-\left(y+x\right)}{y+x}\)\(\Rightarrow P=123\left(\dfrac{1}{y+z}+\dfrac{1}{x+y}+\dfrac{1}{z+x}\right)-3\)
\(\Rightarrow P=123.\dfrac{1}{45}-3\)
\(\Rightarrow P=-\dfrac{4}{15}\)
a. Có \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}\) => \(\dfrac{x}{4}=\dfrac{3x}{9}=\dfrac{4z}{36}\) và x-3y+4z=62
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{4}=\dfrac{3y}{9}=\dfrac{4z}{36}\)= \(\dfrac{x-3y+4z}{4-9+36}=\dfrac{62}{31}=2\)
=> x=8
3y=18=>y=6
4z=72=>z=18
Vậy x=8 ; y=6 ; z=18
b, Ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\\ =\dfrac{2x+3y-5z}{4+9-20}=\dfrac{-21}{-7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=3\cdot2=6\\y=3\cdot3=9\\z=3\cdot4=12\end{matrix}\right.\\ vậy...\)
Câu c bạn làm tương tự nhé!
d, Ta có : \(\left|x+y-z\right|=95\Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\)
\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\\ =\dfrac{x+y-z}{15+10-6}=\dfrac{x+y-z}{19}\\ \Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=15\cdot5=75\\y=10\cdot5=50\\z=6\cdot5=30\end{matrix}\right.\\\left\{{}\begin{matrix}x=-5\cdot15=-75\\y=-5\cdot10=-50\\z=-5\cdot6=-30\end{matrix}\right.\end{matrix}\right.\)
Vậy...
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{y}{2y+x+z}\le\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right);\dfrac{z}{2z+y+x}\le\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)+\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{x+z}{x+z}\right)=\dfrac{1}{4}\left(1+1+1\right)=\dfrac{3}{4}\)
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
a, \(\dfrac{x}{y}=\dfrac{4}{9}\Rightarrow\dfrac{x}{4}=\dfrac{y}{9}\)
Theo tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{4}=\dfrac{y}{9}=\dfrac{x+y}{4+9}=\dfrac{-30}{13}\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.\left(-\dfrac{30}{13}\right)=\dfrac{-120}{13}\\y=9.\left(-\dfrac{30}{13}\right)=\dfrac{-270}{13}\end{matrix}\right.\)
Vậy....
b, \(\dfrac{4}{x}=\dfrac{7}{y}\Rightarrow\dfrac{x}{4}=\dfrac{y}{7}\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{4}=\dfrac{y}{7}=\dfrac{2x-y}{2.4-7}=\dfrac{10}{1}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.10=40\\y=7.10=70\end{matrix}\right.\)
Vậy......
c, Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{9}=\dfrac{x-zy+z}{4-9.6+9}=\dfrac{-30}{-41}=\dfrac{30}{41}\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.\dfrac{30}{41}=\dfrac{120}{41}\\y=6.\dfrac{30}{41}=\dfrac{180}{41}\\z=9.\dfrac{30}{41}=\dfrac{270}{41}\end{matrix}\right.\)
Vậy....
Câu 1: Mình chỉnh sửa lại đầu bài của bạn nha. Không biết có đúng không. Nếu để đầu bài như bạn thì mình không làm ra được. Mog góp ý !!!!
Áp dụng t/c DTSBN ta có:
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\)
\(=\dfrac{x+y+x}{y+z+1+x+z+1+x+y-2}=\dfrac{x+y+x}{2x+2y+2z}=\dfrac{1}{2}\)
=>\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\left(1\right)\)
=>\(\dfrac{y}{x+z+1}=\dfrac{1}{2}\left(2\right)\)
=>\(\dfrac{z}{x+y-2}=\dfrac{1}{2}\left(3\right)\)
=> x+y+z = 1/2 (4)
Ta có : Từ (1) => 2x = y+z+1 kết hợp (4)
=> 2x = 1/2-x+1
=> 3x = 3/2 => x=1/2
Ta có: Từ (2) => 2y = x+z+1
=> 2y + y = x+y+z+1
=> 3y = 1/2+1 (theo 4) => 3y=3/2
=> y=1/2
Ta có : Từ (4) => x+y+z=1/2
=>1/2 + 1/2 +z = 1/2
=> z=-1/2
Vậy ( x;y;z)=(1/2;1/2;-1/2)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(P=\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}=\dfrac{1^2}{x}+\dfrac{2^2}{y}+\dfrac{3^2}{z}\)
\(\ge\dfrac{\left(1+2+3\right)^2}{x+y+z}=\dfrac{36}{4}=9\)
Xảy ra khi \(x=\dfrac{2}{3};y=\dfrac{4}{3};z=2\)
xét dấu = thì thế nào bn Ace Legona