Thu gọn đơn thức sau: 2ab-2bc.c+ab+\(\frac{1}{2}c^2b-4cb^2+2c.b.b\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NB
3
20 tháng 6 2022
a: \(=ab\cdot\dfrac{4}{3}a^2b^4\cdot7abc=\dfrac{28}{3}a^4b^6c\)
b: \(a^3b^3\cdot a^2b^2c=a^5b^5c\)
c: \(=\dfrac{2}{3}a^3b\cdot\dfrac{-1}{2}ab\cdot a^2b=\dfrac{-1}{3}a^6b^3\)
d: \(=-\dfrac{7}{3}a^3c^2\cdot\dfrac{1}{7}ac^2\cdot6abc=-2a^5bc^5\)
e: \(=\dfrac{-3}{2}\cdot\dfrac{1}{4}\cdot ab^2\cdot bca^2\cdot b=\dfrac{-3}{8}a^3b^4c\)
NT
0
19 tháng 2 2017
a. \(4ab.\frac{1}{3}ac-2aca-9a^2.\frac{1}{2}b+10a^2.\frac{1}{5}c+a^2b-a^2bc\)
\(=\left(4.\frac{1}{3}\right)\left(a.a\right).bc-2a^2c-\left(9.\frac{1}{2}\right)a^2b+\left(10.\frac{1}{5}\right)a^2c+a^2b-a^2bc\)
\(=\frac{4}{3}a^2bc-2a^2c-\frac{9}{2}a^2b+2a^2c+a^2b-a^2bc\)
\(=\left(\frac{4}{3}a^2bc-a^2bc\right)+\left(-2a^2c+2a^2c\right)+\left(-\frac{9}{2}a^2b+a^2b\right)\)
\(=\frac{1}{3}a^2bc+\left(-\frac{7}{2}a^2b\right)\)
b. \(2ab-2bc.c+ab+\frac{1}{2}c^2b-4cb^2+2bcb\)
\(=2ab-2bc^2+ab+\frac{1}{2}c^2b-4cb^2+2b^2c\)
\(=\left(2ab+ab\right)+\left(-2bc^2+\frac{1}{2}c^2b\right)+\left(-4cb^2+2b^2c\right)\)
\(=3ab+-\frac{3}{2}bc^2+-2b^2c\)
\(=b\left(3a-\frac{3}{2}c^2-2bc\right)\)
XO
9 tháng 2 2021
ĐKXĐ : \(\hept{\begin{cases}ab-2\ne0\\ab+2\ne0\\a^4b^4\ne0\end{cases}}\Rightarrow ab\ne\pm2;a\ne0;b\ne0\)
\(P=\left(\frac{1}{ab-2}+\frac{1}{ab+2}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\left(\frac{2ab}{a^2b^2-4}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\left(\frac{4a^3b^3}{a^4b^4-16}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\frac{8a^5b^5}{a^8b^8-16^2}.\frac{a^4b^4+16}{a^4b^4}=\frac{8a^5b^5\left(a^4b^4+16\right)}{\left(a^4b^4-16\right)\left(a^4b^4+16\right).a^4b^4}\)
\(=\frac{8ab}{a^4b^4-16}\)
b) Khi \(\frac{a^2+4}{b^2+9}=\frac{a^2}{9}\)
=> (a2 + 4).9 = a2(b2 + 9)
=> 9a2 + 36 = a2b2 + 9a2
=> a2b2 = 36
=> (ab)2 = 36
=> \(\orbr{\begin{cases}ab=6\left(tm\right)\\ab=-6\left(tm\right)\end{cases}}\)
Khi ab = 6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.6}{6^4-16}=\frac{48}{1280}=\frac{3}{80}\)
Khi ab = -6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.\left(-6\right)}{\left(-6\right)^4-16}=-\frac{3}{80}\)
NH
1
20 tháng 5 2020
\(5xy\left(-2x^2y\right)=-10x^3y^2\)
Bậc : 5
\(\left(\frac{5}{4}ab^3\right)\left(-2b^2c\right)^2=\left(\frac{5}{4}ab^3\right)\left(4b^4c^2\right)=5ab^7c^2\)
Bậc : 10