早教吧作业答案频道 -->数学-->
JAvascript一个三元方程3次方程functionequation(x,y,z){vara1,a2,a3,b1,b2,b3,c1,c2,c3,a,b,c,x,y,za1=parseFloat(prompt("a1","1"))a2=parseFloat(prompt("a2","1"))a3=parseFloat(prompt("a3","1"))b1=parseFloat(prompt("b1","1"))b2=parseFloat(prompt
题目详情
JAvascript 一个三元方程
3次方程
function equation(x,y,z){
var a1,a2,a3,b1,b2,b3,c1,c2,c3,a,b,c,x,y,z
a1=parseFloat(prompt("a1","1"))
a2=parseFloat(prompt("a2","1"))
a3=parseFloat(prompt("a3","1"))
b1=parseFloat(prompt("b1","1"))
b2=parseFloat(prompt("b2","1"))
b3=parseFloat(prompt("b3","1"))
c1=parseFloat(prompt("c1","1"))
c2=parseFloat(prompt("c2","1"))
c3=parseFloat(prompt("c3","1"))
a=parseFloat(prompt("a","1"))
b=parseFloat(prompt("b","1"))
c=parseFloat(prompt("c","1"))
if (a1*x+a2*y+a3*z==a && b1*x+b2*y+b3*z==b && c1*x+c2*y+c3*z==c)
document.write("
"+"x,y,z的值分别为"+x+y+z)
else
document.write(" ")
}
3次方程
function equation(x,y,z){
var a1,a2,a3,b1,b2,b3,c1,c2,c3,a,b,c,x,y,z
a1=parseFloat(prompt("a1","1"))
a2=parseFloat(prompt("a2","1"))
a3=parseFloat(prompt("a3","1"))
b1=parseFloat(prompt("b1","1"))
b2=parseFloat(prompt("b2","1"))
b3=parseFloat(prompt("b3","1"))
c1=parseFloat(prompt("c1","1"))
c2=parseFloat(prompt("c2","1"))
c3=parseFloat(prompt("c3","1"))
a=parseFloat(prompt("a","1"))
b=parseFloat(prompt("b","1"))
c=parseFloat(prompt("c","1"))
if (a1*x+a2*y+a3*z==a && b1*x+b2*y+b3*z==b && c1*x+c2*y+c3*z==c)
document.write("
"+"x,y,z的值分别为"+x+y+z)
else
document.write(" ")
}
▼优质解答
答案和解析
没明白你是什么意思?
看了JAvascript一个三元方...的网友还看了以下:
已知函数f(x)=ax+b,当x属于[a1,b1]时,f(x)的值域为[a2,b2],当x属于[a 2020-06-02 …
已知函数f(x)=ax+b,当x属于[a1,b1]时,f(x)的值域为[a2,b2],当x属于[a 2020-06-02 …
若x不等于y,且x,A1,A2,A3,y与x,b1,b2,b3,b4,y各成等差数列,则(b2减b 2020-07-09 …
由等式x^4+a1x^3+a2x^2+a3x+a4=(x+1)^4+b1(x+1)^3+b2(x+ 2020-07-14 …
解向量方程2向量x-3(向量x-2向量a)=向量0方程组5向量x+2向量y=向量a3向量x-向量y 2020-07-17 …
如图,直线l:y=x+1交y轴于点A1,在x轴正方向上取点B1,使OB1=OA1;过点B1作A2B 2020-07-19 …
线性代数问题1.已知排列135……(2n-1)(2n)(2n-2)……42,该排列的逆序数为()2 2020-08-03 …
线性相关问题3、(选择题)设向量组a1,a2,a3向量无关,向量B1可由a1,a2,a3线性表示,而 2020-10-30 …
JAvascript一个三元方程3次方程functionequation(x,y,z){vara1, 2020-10-31 …
(急)高一数学题由等式x*4+a1x*3+a2x*2+a3x+a4=(x+1)*4+b1(x+1)* 2020-10-31 …