{y-1/2x+1=0 {2y+3(x-2/2)-10=10 {11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5二元一次方程 和 三元{y-1/2x+1=0 {2y+3(x-2/2)-10=10{11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/17 11:54:59
{y-1/2x+1=0{2y+3(x-2/2)-10=10{11x+3z=9{3x+2y+z=8{2x-6y+4z=5二元一次方程和三元{y-1/2x+1=0{2y+3(x-2/2)-10=10{11
{y-1/2x+1=0 {2y+3(x-2/2)-10=10 {11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5二元一次方程 和 三元{y-1/2x+1=0 {2y+3(x-2/2)-10=10{11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5
{y-1/2x+1=0 {2y+3(x-2/2)-10=10 {11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5
二元一次方程 和 三元
{y-1/2x+1=0 {2y+3(x-2/2)-10=10
{11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5
{y-1/2x+1=0 {2y+3(x-2/2)-10=10 {11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5二元一次方程 和 三元{y-1/2x+1=0 {2y+3(x-2/2)-10=10{11x+3z=9 {3x+2y+z=8 {2x-6y+4z=5
你这个是分两题的吗.
∵y-1/2x+1=0 ∴y=1且x≠-1/2
∵ 2y+3(x-2/2)-10=10 ∴x=14
同理下一题利用11x+3z=9将z用x带掉,即z=(9-11x)/3
再带入{3x+2y+z=8 {2x-6y+4z=5将他们看成二元一次方程 进行计算
若2/x-1/y=3,求[y/x-y/x-y(x-y/x-x+y)]/x-2y/x的值
(3x-y)^2+(3x+y)(3x-y),x=1,y=-2
已知2x-y=0,求分式【1+(3*y*y*y/x*x*x-y*y*y)】/【1+(2y/x-y)】的值注:x*x*x-y*y*y=(x-y)*(x*x+xy+y*y)
二元一次方程 :2(x+y)-(x-y)=3 (x+y)-2(x-y)=1
{4x-3y-10=0 3x-2y=0,{3x-4(x-y)=2 2x-3y=1,{2(x+y)-(x-y)=3 (x+y)-2(x-y)=1
(x-y)/(x+y)=3求( 3x-2y-1)/(x+y-5)
2(x+y)-3(x-y)=1 6(x+y)+(x-y)=51
{3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1
变量x,y满足约束条件,x+y>=3,x-y>=-1,2x-y
(x-y)^2+(x+y)(x-y) 其中 X =3 Y=-1
{x+y=1 ,xy=-6{x(2x-3)=0,y=x²-1{(3x+4y-3)(3x+4y+3)=0,3x+2y=5{(x-y+2)(x+y)=0,x²+y²=8{(x+y)((x+y-1)=0,(x-y)(x-y-1)=0
x+y+2(-x-y+1)=3(1-y-x)-4-(y+x-1),x+y=?
{(x+y)/2+(x-y)/3=1,(x+y)-5(x-y)=2.求x=?,y=?
设x大于1,y大于0,x^y+x^-y=2根号二,x^y-x^-y等于?
2x+y-2=0,x-3y-1
A={(x,y)|y=-x^2+mx-1},B={(x,y)|y=3-x,0A={(x,y)|y=-x^2+mx-1},B={(x,y)|y=3-x,0
2x(2x+3y)-y(x-3y),其中(2x-1)^2+|2-y|=0
若x,y满足|2x-y-3|+|3x+2y+1|=0,则x= y=