代数式的恒等变形(1).已知x²+y²+z²-2x+4y-6z+14=0,求(x-y-z)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/14 15:16:06
代数式的恒等变形
(1).已知x²+y²+z²-2x+4y-6z+14=0,求(x-y-z)^2010
(2)设m、n满足m²n²+m²+n²+10mn+16=0,求m+n的值
(3)已知14(a²+b²+c²)=(a+2b+3c)² 求证a:b:c=1:2:3
(4)已知实数x、y、z满足x+y=5 z²=xy+y-9,求x+2y+3z
(1).已知x²+y²+z²-2x+4y-6z+14=0,求(x-y-z)^2010
(2)设m、n满足m²n²+m²+n²+10mn+16=0,求m+n的值
(3)已知14(a²+b²+c²)=(a+2b+3c)² 求证a:b:c=1:2:3
(4)已知实数x、y、z满足x+y=5 z²=xy+y-9,求x+2y+3z
1) 原式左边 = x² - 2x +1 +y²+4y +4 + z²-6z +9 = (x-1)² + (y+2)² + (z-3)² = 0
于是 x=1 y=-2 z=3,x-y-z = 0
(x-y-z)^2010 = 0
2) 原式左边 = m²n²+m²+n²+2mn +8mn+16 = m²n²+8mn+16 + m²+n²+2mn
= (mn+4)² + (m+n)² = 0
于是 mn=-4,m+n=0
3) 14(a²+b²+c²)=(a+2b+3c)²
14a² + 14b² + 14c² = a² +4b² +9c² +4ab + 6ac + 12bc
右边移到左边得到
13a² + 10b² + 5c² -4ab -6ac -12bc = 0
配方:
(4a² -4ab +b²) + (9a²-6ac+c²) + (9b² -12bc+4c²) = 0
(2a-b)² + (3a-c)² + (3b-2c)²= 0
于是 2a-b=0 3a-c=0 3b-2c=0 解得 b=2a c=3a 于是 a:b:c=1:2:3
4) x+y=5 (1)
z²=xy+y-9 (2)
(1)式得到 x=5-y 代入(2)式
z² = (5-y)y+y-9
z² =5y-y²+y-9 右边移到左边 合并得到
y²-6y+9 +z² = 0
(y-3)²+z²= 0
于是 y=3,z=0 x=5-y=2 则x+2y+3z = 8
于是 x=1 y=-2 z=3,x-y-z = 0
(x-y-z)^2010 = 0
2) 原式左边 = m²n²+m²+n²+2mn +8mn+16 = m²n²+8mn+16 + m²+n²+2mn
= (mn+4)² + (m+n)² = 0
于是 mn=-4,m+n=0
3) 14(a²+b²+c²)=(a+2b+3c)²
14a² + 14b² + 14c² = a² +4b² +9c² +4ab + 6ac + 12bc
右边移到左边得到
13a² + 10b² + 5c² -4ab -6ac -12bc = 0
配方:
(4a² -4ab +b²) + (9a²-6ac+c²) + (9b² -12bc+4c²) = 0
(2a-b)² + (3a-c)² + (3b-2c)²= 0
于是 2a-b=0 3a-c=0 3b-2c=0 解得 b=2a c=3a 于是 a:b:c=1:2:3
4) x+y=5 (1)
z²=xy+y-9 (2)
(1)式得到 x=5-y 代入(2)式
z² = (5-y)y+y-9
z² =5y-y²+y-9 右边移到左边 合并得到
y²-6y+9 +z² = 0
(y-3)²+z²= 0
于是 y=3,z=0 x=5-y=2 则x+2y+3z = 8
代数式的恒等变形(1).已知x²+y²+z²-2x+4y-6z+14=0,求(x-y-z)
已知x²+y²+z²-2x+4y-6z+14=0,求代数式(x-y-z)的2013次方的值
代数式的恒等变形3已知(x+y+-z)/z=(x-y+z)/y=(-x+y+z)/x且xyz不等于0求分式(x+y)(y
已知x²+y²+z²-2x+4y-6z+14=0,求x+y-z/x+y+z的值
已知x,y,z满足x+y+2z=1,x²+y²+6z+1.5=0,求x,y,z的值
若x-y=6,xy=-8,求代数式(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)的值
已知x²+y²+z²+2x-4y+6z+14=0 求x+y+z的值
已知x的平方+y的平方+z的平方-2x+4y-6z+14=0,求代数式x+y+z的平方根
已知:(x+y-z)/z=(x-y+z)/y+(y+z-x)/x,且xyz≠0,求代数式[(x+y)(y+z)(x+z)
已知x、y、z满足方程组:x+y-z=6;y+z-x=2;z+x-y=0 求x、y、z的值
已知x²+y²-6x+2y+|z+3|+10=0,求x+z-y的值
已知x-y=2,y-z=2,x+z=14,求x²-z²的值