已知x-y=2,y-z=2,x+z=14,求x²-z²的值
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已知x-y=2,y-z=2,x+z=14,求x²-z²的值
已知x-y=2,y-z=2,x+z=14,求x²-z²的值
已知x-y=2,y-z=2,x+z=14,求x²-z²的值
x-y-(y-z)=4所以x-z=4所以x=8,z=4,x²-z²=48
原式=(x+z)×(x-z)
把x-y=2与y-z=2相加得:x-z=4
∴x²-z²=(x+z)×(x-z)=4×14=56
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