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dài lắm
\(\left(x+y\right):\left(8-z\right):\left(y+z\right):\left(10+z\right)=2:5:3:4\\ < =>\dfrac{x+y}{2}=\dfrac{8-z}{5}=\dfrac{y+z}{3}=\dfrac{10-z}{4}\left(1\right)\)
\(\left(1\right)=>\dfrac{8-z}{5}=\dfrac{10+z}{4}\\ < =>4\left(8-z\right)=5\left(10+z\right)\\ < =>32-4z=50+5z\\ < =>-9z=18\\ < =>z=-2\left(2\right)\)
\(\left(1\right)=>\dfrac{y+z}{3}=\dfrac{8-z}{5}\left(3\right)\)
thay (2) vào (3)
\(=>\dfrac{y-2}{3}=\dfrac{8+2}{5}\\ < =>\dfrac{y-2}{3}=2\\ < =>y=8\left(4\right)\)
\(\left(1\right)=>\dfrac{x+y}{2}=\dfrac{8-z}{5}\left(5\right)\)
thay 4 và 2 vào 5
\(=>\dfrac{x+8}{2}=\dfrac{8+2}{5}\\ < =>\dfrac{x+8}{2}=2\\ < =>x=-4\left(6\right)\)
\(=>\dfrac{xyz}{x+y+z}\\ =\dfrac{\left(-2\right).8.\left(-4\right)}{\left(-4\right)+8+\left(-2\right)}\\ =\dfrac{64}{2}\\ =32\)
vậy ...
bài dễ nhưng dài quá @@
chúc may mắn
Đáp án:
P=±36P=±36
Giải thích các bước giải:
Ta có:
x2+y2+z2=16xy−yz+zx=−10⇒(x2+y2+z2)−2.(xy−yz+zx)=16−2.(−10)⇔x2+y2+z2−2xy+2yz−2zx=36⇔(x2−2xy+y2)+z2+2yz−2zx=36⇔(x−y)2+2z(y−x)+z2=36⇔(x−y)2−2.(x−y).z+z2=36⇔(x−y−z)2=36⇔x−y−z=±6P=x3−y3−z3−3xyz=(x3−3x2y+3xy2−y3)−z3+3x2y−3xy2−3xyz=(x−y)3−z3+3x2y−3xy2−3xyz=[(x−y)−z].[(x−y)2+(x−y).z+z2]+3xy(x−y−z)=(x−y−z).(x2−2xy+y2+xz−yz+z2+3xy)=(x−y−z).(x2+y2+z2+xy−yz+zx)Trường hợp 1: x−y−z=6⇒P=6.(16+(−10))=36Trường hợp 2: x−y−z=−6⇒P=(−6).(16+(−10))=−36x2+y2+z2=16xy−yz+zx=−10⇒(x2+y2+z2)−2.(xy−yz+zx)=16−2.(−10)⇔x2+y2+z2−2xy+2yz−2zx=36⇔(x2−2xy+y2)+z2+2yz−2zx=36⇔(x−y)2+2z(y−x)+z2=36⇔(x−y)2−2.(x−y).z+z2=36⇔(x−y−z)2=36⇔x−y−z=±6P=x3−y3−z3−3xyz=(x3−3x2y+3xy2−y3)−z3+3x2y−3xy2−3xyz=(x−y)3−z3+3x2y−3xy2−3xyz=[(x−y)−z].[(x−y)2+(x−y).z+z2]+3xy(x−y−z)=(x−y−z).(x2−2xy+y2+xz−yz+z2+3xy)=(x−y−z).(x2+y2+z2+xy−yz+zx)Trường hợp 1: x−y−z=6⇒P=6.(16+(−10))=36Trường hợp 2: x−y−z=−6⇒P=(−6).(16+(−10))=−36
Vậy P=±36P=±36.
(x+y):(8-z):(y+z):(10+z)=2:5:3:4
\(\Rightarrow\)\(\dfrac{x+y}{2}=\dfrac{8-z}{5}=\dfrac{y+z}{3}=\dfrac{10+z}{4}\)
giải ra x=-4;y=8;z=-2