数学;请帮忙解一下2道方程 (1)(25+2h)(2h+2h)=888 (2) (100-2x)(90-x)=8448

作者&投稿:莫支 (若有异议请与网页底部的电邮联系)
解:(1) 25+2h)*4h=888.
8h^2+100h-888=0.
2h^2+25h-222=0.
h={(-25±√[25^2-4*2*(-222)]}/2*2.
=[-25√(625+1776)]/4.
=(-25±√2401)/4.
=(-25±49)/4.
∴ h1=(-25+49)/4=6;
h2=(-25-49)/4=-18.5.

(2) (100-2x)(90-x)=8448.
2x^2-280x+9000-8448=0.
x^2-140x+276=0.
x=[(140±√(140^2-4*276)]/2.
=(140±136)/2.
∴ x1=(140+136)/2=138;
x2=(140-136)/2=2.

1)原式化简得
8h^2+100h-888=0
根号b^2-4ac=196
根据x=[-b+-(根号b^2-4ac)]/2a
h1=96/16=6
h2=-296/16=-18.5
2)原式=x^2-140x+276=0
根号b^2-4ac=136
x1=138
X2=2

数学;请帮忙解一下2道方程 (1)(25+2h)(2h+2h)=888 (2) (100-2x)(90-x)=8448~

解:(1)(25+h)*(2h+2h)=888; (25+h)*4h=888; (25+h)*h=222; 25h+h^2=22;
h^2+25h+12.5^2=222+12.5^2; (h+12.5)^2=378.5;所以h+12.5=19.45或h+12.5=-19.45
即h=6.95或h=-31.95
(2) (100-2x)(90-x)=8448; (50-x)(90-x)=4224;4500-140x+x^2=4224; x^2-140x+276=0;
x^2-140x+4900=4900-276;(x-70)^2=4624;所以x-70=68或x-70=-68
即x=138或x=2

H2,2代表氢分子是有两个氢原子组成的

2H,这个只能说是单独的两个氢原子(一般方程式里面不会出现这个,到是会有出现氢离子的情况)