解方程组x+2y+3z=根号(x+2y+3z+2)+4

zzzz2022-10-04 11:39:542条回答

解方程组x+2y+3z=根号(x+2y+3z+2)+4
x/2=y/3=z/4

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pererperer 共回答了15个问题 | 采纳率93.3%
令k= x/2=y/3=z/4
x=2k,y=3k,z=4k
令a=√(x+2y+3z+2)
则a²=x+2y+3z+2
x+2y+3z=a²-2
所以a²-2=a+4
a²-a-6=0
(a-3)(a+2)=0
a=√(x+2y+3z+2)>=0
所以a=3
x+2y+3z=a²-2=7
把x=2k,y=3k,z=4k代入
2k+6k+12k=7
20k=7
k=7/20
x=2k=7/10
y=3k=21/20
z=4k=7/5
1年前
一路小泡 共回答了89个问题 | 采纳率
1/sin10-√3/cos10
=(cos10-√3sin10)/sin10cos10
=2(1/2*cos10-√3/2*sin10)/sin10cos10
=2(sin30cos10-cos30sin10)/sin10cos10
=4sin(30-10)/2sin10cos10
=4sin20/sin20
=4 1/sin10-√3/...
1年前

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