x,y,z biết xy/4 =yz/6 =zx/10 và xy+ yz +zx=60
Những câu hỏi liên quan
tìm x,y,z biết \(\frac{xy}{4}=\frac{yz}{6}=\frac{zx}{10}\)
và\(xy+yz+zx=60\)
theo tính chất dãy tỉ số = ta có ;
xy\4+yz/6+zx/10=xy+yz+zx/4+6+10=60/16=3,75
do đó: xy/4=3,75 suy ra xy=3,75.4=15
yz/6=3,75 suy ra yz=3,75.6=22,5
zx/10=3,75 suy ra zx=3,75.10=37,5
Đúng 0
Bình luận (0)
tìm x, y, z chứ ko phải xy, yz, zx đâu bn ơi với lại 4+6+10=20 mà bn, sao 4+6+10=16 đc
Xem thêm câu trả lời
11. xyz - xy - yz - zx + x + y + z - 1
12. xy(x + y) + yz(y + z) + zx(z + x) + 2xyz
13. xy(x + y) + yz(y + z) + zx(z + x) + 3xyz
giúp mik vs mik đang cần gấp =(((
13:
xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z²)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)
Đúng 0
Bình luận (0)
Cho các số dương x;y;z. CMR:
\(\dfrac{xy}{x^2+yz+zx}+\dfrac{yz}{y^2+zx+xy}+\dfrac{zx}{z^2+xy+yz}\le\dfrac{x^2+y^2+z^2}{xy+yz+zx}\)
cho x,y,z>0 và x+y+z=1 chứng minh\(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge1+\sqrt{xy}\sqrt{yz}+\sqrt{zx}\)
chứng minh A=(xy+zx+1)/(xy+x+y+1)+(yz+zy+1)/(yz+y+z+1)+(zx+zx+1)/(zx+x+z+1) không thuộc x, y, z
làm nhanh giùm mình nha ! đang cần gấp <:)
cho x y z > 0 và xyz=1. tìm gtln của \(P=\frac{xy}{x^4+y^4+xy}+\frac{yz}{y^4+z^4+yz}+\frac{zx}{z^4+x^4+zx}\)
TÌM X,Y,Z BIẾT XY+1/9=YZ+2/15=ZX+3/27 BVAF XY+YZ+ZX=11
\(\dfrac{xyz-xy-yz-zx+x+y+z-1}{xyz+xy+yz-zx-x+y-z-1}\) với x = 5001;y=5002;z=5003
\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)
Đúng 2
Bình luận (0)
chứng minh rằng: (x-y)/(1+xy) + (y-z)/(1+yz) +(z-x)/(1+zx) = (x-y)(y-z)(z-x)/(1+xy)(1+yz)(1+zx)
Ta có:
\(\dfrac{x-y}{1+xy}\)+\(\dfrac{y-z}{1+yz}\)+\(\dfrac{z-x}{1+xz}\) = \(\dfrac{x-y}{1+xy}\)+\(\dfrac{-\left(x-y\right)-\left(z-x\right)}{1+yz}\)+\(\dfrac{z-x}{1+xz}\)
=\(\dfrac{x-y}{1+xy}\)\(-\dfrac{x-y}{1+yz}\) \(-\dfrac{z-x}{1+yz}\)+\(\dfrac{z-x}{1+xz}\)
= \(\left(x-y\right)\)\(\left(\dfrac{\left(1+yz\right)-\left(1+xy\right)}{\left(1+yz\right)\left(1+xy\right)}\right)\)+(\(z-x\))\(\left(\dfrac{\left(1+yz\right)-\left(1+zx\right)}{\left(1+yz\right)\left(1+zx\right)}\right)\)
=\(\left(x-y\right)\)\(\dfrac{y\left(z-x\right)}{\left(1+yz\right)\left(1+xy\right)}\)+(\(z-x\))\(\dfrac{-z\left(x-y\right)}{\left(1+yz\right)\left(1+zx\right)}\)
=\(\left(\dfrac{\left(x-y\right)\left(z-x\right)}{1+yz}\right)\)\(\left(\dfrac{y\left(1+xz\right)-z\left(1+xy\right)}{\left(1+xz\right)\left(1+xy\right)}\right)\)
=đpcm
Đúng 0
Bình luận (0)