=(a+2b)2−2⋅(a+2b)⋅3c+(3c)2 =a2+4ab+4b2−6ac−12bc+9c2故答案为: a2+4ab+4b2−6ac−12bc+9c2 可把前两项看做一个整体,利用完全平方公式展开,再计算即可. 本题考查了多项式乘以多项式,可以用一个多项式里的每一项乘以另一个多项式里的每一项,也可以用完全平方公式或平方差公...
解:原式=⎡⎢⎣⎤⎥⎦⎛ ⎛⎜ ⎜⎝⎞⎟⎠a-2b-3c2=⎛ ⎛⎜ ⎜⎝⎞⎟⎠a-2b2-2⎛ ⎛⎜ ⎜⎝⎞⎟⎠a-2b·3c+3c2=a2-4ab+4b2-6ac+12bc+3c2 故答案为: a2-4ab+4b2-6ac+12bc+3c2 先化成=[(a-2b)-c]²,根据完全平方公式展开得出(...
2A代表耐压100伏,332就是容量为3300皮法(33乘以10的二次幂等于3300)。J表示精度正负5%,耐压等级:1A是10伏、1B是12.5伏、1C是16伏、1D是20伏、1E是25伏、1F是31.5伏;2A是100伏、2B是125伏、2C是160伏、2D是200伏、2E是250伏、2F是315伏;3A是1000伏、3B 是1250伏、3C是1600伏…...
=[(a+2b)-3c]^2 =(a+2b)^2-6c(a+2b)+(3c)^2 =a^2+4b^2+9c^2+4ab-6ac-12bc
2b). Previously, it has been hypothesized that a lack of hypomethylation in pediatric tumors might stem from the generally higher methylation levels in cells from younger patients9; however, adolescent and adult T-ALL samples in our cohort also lacked a significant reduction of methylation in PMDs...
decreased modified Ashcroft scores (a semi-quantitative grading for the severity of lung fibrosis) and lung fibrotic area relative to vehicle-treated mice. Additionally, INS018_055- and nintedanib-treated animals showed a marked reduction in α-SMA- and collagen I-positive regions (Fig.3c). ...
=(a-2b+3c+a+2b-3c)(a-2b+3c-a-2b+3c)=4a(3c-2b)[ab(3-b)-2a(b-2分之b的平方)](-3a的平方b的3次方)=(3ab-ab^2-2ab+ab^2)(-3a^2b^3)=ab(-3a^2b^3)=-3a^3b^4-2的100次方×0.5的100次方×(-1)的2005次方÷(-1)的-5次方=-2^100*(1/2)^100*(-1)/(-1)=-1[(x+...
1、a+b+c+d=I(只喜欢1者+只喜欢2者+3者都喜欢+3者都不喜欢=总集)2、a+2b+3c=A+B+C(三个集合相加时,喜欢1者的部分加了1次,2者的部分加了2次,喜欢3者的部分加了3次)3、b+3c=X+Y+Z(题目中的固定表达方式为喜欢A和B的有X人、喜欢A和C的有Y人,喜欢B和C的有Z人)...
a+2b=3c,abc分别是:a=3c-2b b=(3c-a)/2 c=(a+2b)/3
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