Tìm xyz biết
3x=2y; 5y=2x và x.y=24
Giúp mk nhé
tìm đa thức m biết
3x^2y^3 - x^2y - M=x^2y^3 + x^2y
\(3x^2y^3-x^2y-M=x^2y^3+x^2y\\ \Rightarrow M=3x^2y^3-x^2y-x^2y^3-x^2y\\ \Rightarrow M=2x^2y^3-2x^2y\)
\(\Leftrightarrow M=3x^2y^3-x^2y-x^2y^3-x^2y=2x^2y^3-2x^2y\)
Tìm x,biết
3x-4-63=18
Tìm x biết
3x .32 .3 = 243.3
\(3^x.3^2.3=243.3\\ \Rightarrow3^x.3^2=243\\ \Rightarrow3^x.3^2=3^5\\ \Rightarrow3^x=3^5:3^2\\ \Rightarrow3^x=3^3\\ \Rightarrow x=3\)
=>3^x*3^3=3^5*3
=>x+3=6
=>x=3
Tìm x,y biết
3x=4y và x+y=58
tìm x,biết
3x(x+4)-3x^2-4=0
Among us:)
\(3x\left(x+4\right)-3x^2-4=0\\ \Rightarrow3x^2+12x-3x^2-4=0\\ \Rightarrow12x-4=0\\ \Rightarrow12x=4\\ \Rightarrow x=\dfrac{1}{3}\)
Tìm x,y
x+2y-3z=5^xyz và (x-2y)(y+7)-x=19^2 (xyz>0)
tìm x,y biết
3x+1.5y-2=152x-y
e đang cần gấp, cảm ơn mn
Cho xy+yz+xz=2xyz (x,y,z>0). Tìm Max P= \(\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2z^2x^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)
Cho xy+yz+zx=2xyz ; x,y,z>0 Tìm max \(A=\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2x^2z^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)
\(A=\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2x^2z^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)
\(A=\sqrt{\frac{x^2}{2xyz.yz+xz.xy}}+\sqrt{\frac{y^2}{2xyz.xz+xy.yz}}+\sqrt{\frac{z^2}{2xyz.xy+xz.yz}}\)
\(A=\sqrt{\frac{x^2}{yz\left(xy+yz+xz\right)+xz.xy}}+\sqrt{\frac{y^2}{xz\left(xy+yz+xz\right)+xy.yz}}+\sqrt{\frac{z^2}{xy\left(xy+yz+xz\right)+xz.yz}}\)
\(A=\sqrt{\frac{x^2}{\left(yz+xy\right)\left(yz+xz\right)}}+\sqrt{\frac{y^2}{\left(xz+xy\right)\left(xz+yz\right)}}+\sqrt{\frac{z^2}{\left(xy+yz\right)\left(xy+xz\right)}}\)
Áp dụng bđt \(\sqrt{ab}\le\frac{a+b}{2}\) ta có:
\(2A\le\frac{x}{yz+xy}+\frac{x}{yz+xz}+\frac{y}{xz+xy}+\frac{y}{xz+yz}+\frac{z}{xy+yz}+\frac{z}{xy+xz}\)
\(=\frac{x+z}{yz+xy}+\frac{x+y}{yz+xz}+\frac{y+z}{xz+xy}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Mà: \(xy+yz+xz=2xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Rightarrow2A\le2\Rightarrow A\le1."="\Leftrightarrow a=b=c=\frac{3}{2}\)
Tìm x, y,z: x+ 2y- 3z= 5xyz; (x- 2y).(y+7)- x= 192. (xyz>0)