Tìm x,y biet\(\frac{x}{y}=\frac{2}{3}\)và \(3x^2-y^2=75\)
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x/2=y/3=z/4=(x-y+z)/(2-3+4)=10/3
x/2=10/3 => x=10/3.2=20/3
y/3=10/3=> y=10/3.3=10
z/4=10/3=> z=10/3.4=40/3
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Áp dụng bất đẳng thức svác sơ ta có
\(A\ge\frac{\left(x+y+z\right)^2}{y+3z+z+3x+x+3y}=\frac{\left(x+y+z\right)^2}{4\left(x+y+z\right)}=\frac{x+y+x}{4}=\frac{3}{4}\)
Đặt \(P=\frac{x^2}{y+3z}+\frac{y^2}{z+3x}+\frac{z^2}{x+3y}\)
Áp dụng bất đẳng thức Canchy Schwarz dạng Engel :
\(P=\frac{x^2}{y+3z}+\frac{y^2}{z+3x}+\frac{z^2}{x+3y}>\frac{\left(x+y+z\right)^2}{y+3y+z+3z+x+3x}=\frac{\left(x+y+z\right)^2}{4x+4y+4z}=\frac{\left(x+y+z\right)^2}{4.\left(x+y+z\right)}=\frac{3^2}{4}=\frac{3}{4}\)
Dấu " = " xảy ra khi x=y=z=1.
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a)
Ta có
\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{3x}{6}=\frac{y}{5}\)
Áp dụng tc của dãy tỉ só bằng nhau
\(\Rightarrow\frac{3x}{6}=\frac{y}{5}=\frac{3x-y}{6-5}=\frac{10}{1}=10\)
=> x=2.10=20
y=5.10=50
Ta có
\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{25}=\frac{xy}{10}=\frac{30}{10}=3\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{12}\\x=-\sqrt{12}\end{array}\right.\)
\(\left[\begin{array}{nghiempt}y=\sqrt{75}\\y=-\sqrt{75}\end{array}\right.\)
Mà 2;5 cùng dấu
=> x; y cùng dấu
Vậy \(\left(x;y\right)=\left(\sqrt{12};\sqrt{75}\right);\left(-\sqrt{12};-\sqrt{75}\right)\)
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a, x/3=y/4 b, 2x=5y
=> 2x/6=y/4=2x-4/6-4=2/2=1 => x/5=y/2 => 3x/15=y/2=3x-y/15-2=22/13
+, x/3=1 => x=3 +,2x=22/13 => x=11/13
+, y/4=1 => y=4 +,5y=22/13 => y=22/65
Vậy .... Vậy ...........
c, x/y=3/5 d, x/2=y/5
=> x/3=y/5 => 2x/4=y/5
=>3x/9=2y/10 => 2x+y/4+5=18/9=2
=> 3x+2y/9+10=38/19=2 +,x/2=2 => x=4
+,x/3=2 => x=6 +,y/5=2 => y=10
Vậy ........... Vậy ............
+,y/5=2 => y=10
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a) x = 6 ; y = 15.
x = -6 ; y = -15.
b) x = 2 ; y = 2.
x = -2 ; y = -2.
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B)ĐỀ BÀI \(\Leftrightarrow\left(\frac{X}{2}\right)^3=\frac{X}{2}.\frac{Y}{3}.\frac{Z}{5}=\frac{810}{30}=27\\ \)
\(\Leftrightarrow\frac{X}{2}=3\Rightarrow X=6\)
TỪ ĐÓ SUY RA Y=9;Z=15
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1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)
\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)
=>\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)
=>\(x=3\cdot20=60\)
\(y=3\cdot24=72\)
\(z=3\cdot21=63\)
3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)
=> \(x=1\cdot15=15\)
\(y=1\cdot7=7\)
\(z=1\cdot3=3\)
\(t=1\cdot1=1\)
\(\frac{x}{y}=\frac{2}{3}\Rightarrow\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}\Rightarrow\frac{3x^2}{12}=\frac{y^2}{9}=\frac{3x^2-y^2}{12-9}=\frac{75}{3}=25\)
=> \(\frac{x^2}{4}=25\Rightarrow x^2=100\Rightarrow x=\pm10\)
\(\frac{y^2}{9}=25\Rightarrow y^2=225\Rightarrow y=\pm15\)
Vậy...
\(\frac{x}{y}=\frac{2}{3}\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{3x^2-y^2}{12-9}=\frac{75}{3}=25\)
\(\Rightarrow\)\(x=25.2=50\)
\(x=25.3=75\)
Vậy x = 50 và y = 75