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MT
7 tháng 7 2016
\(A=\frac{\left(a+2\right)^2\left(5a-15a^2\right)}{\left(a-3\right)\left(4a-a^3\right)}=\frac{\left(a+2\right)^2.5a.\left(1-3a\right)}{\left(a-3\right).a.\left(2-a\right)\left(a+2\right)}\)
\(=\frac{\left(a+2\right).5.\left(1-3a\right)}{\left(a-3\right).\left(2-a\right)}\)
15 tháng 12 2018
Đề sai sửa luôn !
\(a,M=\left(\frac{21}{x^2-9}+\frac{4-x}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21-\left(4-x\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+3-1}{x+3}\right)\)
\(=\frac{21-4x-12+x^2+3x-x^2+3x+x-3}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)
\(=\frac{3x+6}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{3}{x-3}\)
\(b,x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Kết hợp ĐKXĐ => x = 2
Thay vào \(M=\frac{3}{2-3}=\frac{3}{-1}=-3\)
Vậy ...........................
KT
23 tháng 12 2017
a) B = \(\frac{\left(a+3\right)^2}{2a^2+6a}\). \(\left(1-\frac{6a-18}{a^2-9}\right)\)
= \(\frac{\left(a+3\right)^2}{2a\left(a+3\right)}\). \(\left(1-\frac{6\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}\right)\)
= \(\frac{a+3}{2a}\). \(\left(1-\frac{6}{a+3}\right)\)
= \(\frac{a+3}{2a}\). \(\frac{a+3-6}{a+3}\)
= \(\frac{a+3}{2a}\). \(\frac{a-3}{a+3}\)
= \(\frac{a-3}{2a}\)
b) B = \(\frac{a-3}{2a}\)= 1
\(\Leftrightarrow\)\(a-3=2a\)
\(\Leftrightarrow\)\(a=-3\)
Vậy khi B = 1 thì a = -3
11 tháng 12 2018
\(a,ĐKXĐ:\hept{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}\Leftrightarrow x\ne\pm1}\)
\(b,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}+\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
\(=\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}:\frac{x-1-x\left(x+1\right)+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{x-1-x^2-x+2}\)
\(=\frac{4x}{1-x^2}\)
\(c,A\ge0\Leftrightarrow\frac{4x}{1-x^2}\ge0\)
\(\Leftrightarrow\hept{\begin{cases}4x\ge0\\1-x^2\ge0\end{cases}\left(h\right)\hept{\begin{cases}4x\le0\\1-x^2\le0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2\le1\end{cases}\left(h\right)\hept{\begin{cases}x\le0\\x^2\ge1\end{cases}}}\)
\(\Leftrightarrow0\le x\le1\left(h\right)x\le-1\)
Vậy ///////
2 tháng 5 2019
Trả lời :
Cần j bạn ?
Hok_Tốt
#Thiên_Hy
________
12 tháng 11 2018
\(\frac{12x^3y^2}{18xy^5}=\frac{12x^3y^2:6xy}{18xy^5:6xy}=\frac{2x^2y}{3y^4}\),\(\frac{-21b^2y^2}{-28by}=\frac{-21b^2y^2:-7by}{-28by:-7by}=\frac{3by}{4}\)\(\frac{-49a^3}{14b^3}=\frac{-49a^3:7}{14b^3:7}=\frac{-7a^3}{2b^3}\)\(\frac{18ab}{27bc}=\frac{18ab:9b}{27bc:9b}=\frac{2a}{3c}\)
12 tháng 11 2018
a) \(\frac{18ab}{27bc}=\frac{2.9.ab}{3.9.bc}=\frac{2a}{3c}\)
b) \(\frac{-21b^2y^2}{-28by}=\frac{3.\left(-7\right).b.b.y.y}{4.\left(-7\right).b.y}=\frac{3by}{4}\)
c) \(\frac{-49a^3}{14b^3}=\frac{7.\left(-7\right).a^3}{7.2.b^3}=\frac{-7a^3}{2b^3}\)
d) \(\frac{12x^3y^2}{18xy^5}=\frac{2.6.x.x^2.y^2}{3.6.x.y^2.y^3}=\frac{2x^2}{3y^3}\)
\(\frac{a^3-125}{3a^2+15a+75}\)
\(=\frac{\left(a-5\right)\left(a^2+5a+25\right)}{3\left(a^2+5a+25\right)}\)
\(=\frac{a-5}{3}\)
Ủng hộ nhé~~~~!!! :3
\(\frac{a^3-125}{3a^2+15a+75}=\frac{\left(a-5\right).\left(a^2+5a+25\right)}{3.\left(a^2+5a+25\right)}=\frac{a-5}{3}\\ \)