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\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)
\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)
\(P=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(P=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
\(P\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(x^2+y^2+z^2\right)}\)
\(P\ge\dfrac{x^2+y^2+z^2}{10}\ge\dfrac{1}{30}\)
\(P_{min}=\dfrac{1}{30}\) khi \(x=y=z=\dfrac{1}{3}\)
đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
BBDT AM-GM
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)
vì \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)
\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)
dấu"=" xảy ra<=>x=y=z=1/3
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
a. 2x\(^2\)-8=0
2x\(^2\)=8
x\(^2\)=4
x=2
b.3x\(^3\)-5x=0
x(3x\(^2\)-5)=0
\(\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=0\\x^2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=^+_-\sqrt{5}\end{matrix}\right.\)
c.x\(^4\)+3x\(^2\)-4=0\(^{\left(\cdot\right)}\)
đặt t=x\(^2\) (t>0)
ta có pt: t\(^2\)+3t-4=0 \(^{\left(1\right)}\)
thấy có a+b+c=1+3+(-4)=0 nên pt\(^{\left(1\right)}\) có 2 nghiệm
t\(_1\)=1; t\(_2\)=\(\dfrac{c}{a}\)=-4
khi t\(_1\)=1 thì x\(^2\)=1 ⇒x=\(^+_-\)1
khi t\(_2\)=-4 thì x\(^2\)=-4 ⇒ x=\(^+_-\)2
vậy pt đã cho có 4 nghiệm x=\(^+_-\)1; x=\(^+_-\)2
d)3x\(^2\)+6x-9=0
thấy có a+b+c= 3+6+(-9)=0 nên pt có 2 nghiệm
x\(_1\)=1; x\(_2\)=\(\dfrac{c}{a}=\dfrac{-9}{3}=-3\)
e. \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\) (ĐK: x#5; x#2 )
⇔\(\dfrac{\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}\)=\(\dfrac{6\left(x-5\right)}{\left(x-5\right)\left(2-x\right)}\)
⇒2x - x\(^2\) + 4 - 2x + 6x - 6x\(^2\) + 12 - 6x - 6x +30 = 0
⇔-7x\(^2\) - 6x + 46=0
Δ'=b'\(^2\)-ac = (-3)\(^2\) - (-7)\(\times\)46= 9+53 = 62>0
\(\sqrt{\Delta'}=\sqrt{62}\)
vậy pt có 2 nghiệm phân biệt
x\(_1\)=\(\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{3+\sqrt{62}}{-7}\)
x\(_2\)=\(\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{3-\sqrt{62}}{-7}\)
vậy pt đã cho có 2 nghiệm x\(_1\)=.....;x\(_2\)=......
câu g làm tương tự câu c
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)
Yêu cầu là gì v c.
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