100(x+4)+10x+(x+2)=21[10x+(x+2)]
100x+400+10x+x+2=21[10x+x+2]
100x+10x+x+400+2=210x+21x+42
111x+402=231x+42
231x-111x=402-42
120x=360
x=360÷120
x=3
100(x+4)+10x+(x+2)=21[10x+(x+2)]
1个回答
相关问题
-
【 100x+10(x-2)+(26-2x) 】-【 10x+x 】=【 100(26-2x)+10(x-2)+x 】
-
判断函数f(x)=｛x^2+10x+21,x∈[-5,-2] ｛x^2-10x+21,x∈[2,5] 的奇偶性.
-
求证:多项式(x2-4﹚﹙x2-10x+21﹚+100的值一定是非负数
-
求证:多项式(x2-4)(x2-10x+21)+100的值一定是非负数
-
(2x-10)/21+(7x-4)/20=(2-x)/15+(8x+9)/4
-
解方程：4(x^2-7x+10)-(2x+3)(2x-1)=-21
-
接下列一元二次方程：如：3x^2-10x+3=0x=-(-10)±√(-10)^2-4x3x3/2X3x=10±√100
-
（x-10)^2+(√2x-x)^2=100
-
100+100*2%=x-10%x+（100-x）+（100-x）*5%
-
3y-1/2=1/3 x-8=2x+7 2x=5x-1 9-10x=9x-10 4/5x=5x-21