x2−6x+y2+10y+34=−(4z−1)2
x^2-6x+9+y^2+10y+25+(4z-1)^2=0x2−6x+9+y2+10y+25+(4z−1)2=0
(x-3)^2+(y+5)^2+(4z-1)^2=0(x−3)2+(y+5)2+(4z−1)2=0
{nghiempt}x-3=0\\y+5=0\\4z-1=0
{nghiempt}x=3\\y=-5\\z={1}{4}

\(x^2-6x+y^2+10y+34=-\left(4z-1\right)^2\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(y^2+10y+34\right)+\left(4z-1\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+5\right)^2+\left(4z-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(y+5\right)^2=0\\\left(4z-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-5\\z=\dfrac{1}{4}\end{matrix}\right.\)

Vậy........

x²-6x+y²+10y+34=-(4z-1)²

=>x²-6x+9+y²+10y+25+(4z-1)²=0=B

=>(x-3)²+(y+5)²+(4z-1)²=0

với mọi x,y,z ta có :

(x-3)²>=0

(y+5)²>=0

(4z-1)²>=0

=>(x-3)²+(y+5)²+(4z-1)²>=0

hay B>=0

dấu bằng xảy ra khi (x-3)²=0 => x-3=0  =>x=3

=>(y+5)²=0 =>y+5=0  =>y=-5

=>(4z-1)²=0 =>4z-1=0  => z=1/4

Vậy y=-5

Giúp mình với nha.

ko pit mà giúp 

Kaitou Kid ơi k mik vs

a: 2x-3y-4z=24

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)

=>x=-6/7; y=-36/7; z=-18/7

b: 6x=10y=15z

=>x/10=y/6=z/4=k

=>x=10k; y=6k; z=4k

x+y-z=90

=>10k+6k-4k=90

=>12k=90

=>k=7,5

=>x=75; y=45; z=30

d: x/4=y/3

=>x/20=y/15

y/5=z/3

=>y/15=z/9

=>x/20=y/15=z/9

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)

=>x=500; y=375; z=225

\(1,\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y+4z}{2-18+12}=\dfrac{24}{-4}=-6\\ \Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-36\\z=-18\end{matrix}\right.\\ 2,\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50-34}{8}=\dfrac{16}{8}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)

\(3,6x=10y=15z\Leftrightarrow\dfrac{6x}{30}=\dfrac{10y}{30}=\dfrac{15z}{30}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{x+y-z}{5+3-2}=\dfrac{90}{6}=15\\ \Leftrightarrow\left\{{}\begin{matrix}x=75\\y=45\\z=30\end{matrix}\right.\)

\(x^2-6x+y^2+10y+34=-\left(4z-1\right)^2\)

\(x^2-6x+9+y^2+10y+25+\left(4z-1\right)^2=0\)

\(\left(x-3\right)^2+\left(y+5\right)^2+\left(4z-1\right)^2=0\)

\(\left[\begin{array}{nghiempt}x-3=0\\y+5=0\\4z-1=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=3\\y=-5\\z=\frac{1}{4}\end{array}\right.\)