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a)
2a^2+6a=2a(a+3) khác 0=> a khác 0 và a khác -3
a^2-9=(a-3)(a+3) khác 0=> a khác -3 và a khác 3
tỏng hợp a \(\ne\) {-3,0,3}
b)\(B=\frac{\left(a+3\right)^2}{2a\left(a+3\right)}\cdot\frac{\left(a^2-9\right)-6a+18}{\left(a-3\right)\left(a+3\right)}=\frac{\left(a+3\right)^2.\left(a-3\right)^2}{2a.\left(a-3\right)\left(a+3\right)^2}=\frac{a-3}{2a}\)
c)B=0\(\frac{\left(a-3\right)}{2a}=0=>a=3\Rightarrow\left(loai\right)\) kết luận ko có giá trị nào a ;B =0
d)\(B=1\Rightarrow\left(a-3\right)=2a\Rightarrow a=-3\left(loai\right)\)không có giá trị nào của a cho B=1
a) B xác định
\(\Leftrightarrow\begin{cases}2a^2+6a\ne0\\a^2-9\ne0\end{cases}\Leftrightarrow\begin{cases}2a\left(a+3\right)\ne0\\\left(a+3\right)\left(a-3\right)\ne0\end{cases}\Leftrightarrow\begin{cases}a\ne0\\a\ne-3\\a\ne3\end{cases}\)
Vậy để B xác định thì \(a\ne0\) và \(a\ne\pm3\)
b) \(B=\frac{\left(a+3\right)^2}{2a^2+6a}\cdot\left(1-\frac{6a-18}{a^2-9}\right)\)
\(=\frac{\left(a+3\right)^2}{2a\left(a+3\right)}\cdot\frac{\left(a+3\right)\left(a-9\right)}{\left(a+3\right)\left(a-3\right)}\)
\(=\frac{a+3}{2a}\cdot\frac{a-9}{a+3}\)
\(=\frac{a-9}{2a}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}2a^2+6a\ne0\\a^2-9\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2a\left(a+3\right)\ne0\\\left(a-3\right)\left(a+3\right)\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2a\ne0\\a-3\ne0\\a+3\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a\ne0\\a\ne3\\a\ne-3\end{matrix}\right.\)
b) \(B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\left(1-\dfrac{6a-18}{a^2-9}\right)\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\left(\dfrac{a^2-9}{a^2-9}-\dfrac{6a-18}{a^2-9}\right)\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{\left(a^2-9\right)-\left(6a-18\right)}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{a^2-9-6a+18}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{a^2-6a+9}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{\left(a-3\right)^2}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{\left(a-3\right)^2}{\left(a-3\right)\left(a+3\right)}\)
\(\Leftrightarrow B=\dfrac{a+3}{2a}.\dfrac{a-3}{a+3}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)\left(a-3\right)}{2a\left(a+3\right)}\)
\(\Leftrightarrow B=\dfrac{a-3}{2a}\)
a) \(A=\left(\frac{2}{2a-b}+\frac{6b}{b^2-4a^2}-\frac{4}{2a+b}\right):\left(a+\frac{4a^2+b^2}{4a^2-b^2}\right)\)
\(=\left(\frac{2}{2a-b}+\frac{6b}{\left(b-2a\right)\left(b+2a\right)}-\frac{4}{2a+b}\right):\left(a+\frac{4a^2+b^2}{4a^2-b^2}\right)\)
\(=\left(\frac{-2\left(b+2a\right)}{\left(b-2a\right)\left(b+2a\right)}+\frac{6b}{\left(b-2a\right)\left(b+2a\right)}-\frac{4\left(b-2a\right)}{\left(2a+b\right)\left(b-2a\right)}\right):\left(\frac{a\left(4a^2-b^2\right)}{4a^2-b^2}+\frac{4a^2+b^2}{4a^2-b^2}\right)\)
\(=\frac{-2b-4a+6b-4b+8a}{\left(b-2a\right)\left(b+2a\right)}:\frac{4a^3-ab^2+4a^2+b^2}{4a^2-b^2}\)
\(=\frac{4a}{\left(b-2a\right)\left(b+2a\right)}.\frac{\left(2a-b\right)\left(2a+b\right)}{4a^3-ab^2+4a^2+b^2}\)
\(=\frac{-4a}{\left(2a-b\right)\left(b+2a\right)}.\frac{\left(2a-b\right)\left(2a+b\right)}{4a^3-ab^2+4a^2+b^2}\)
\(=.\frac{-4a}{4a^3-ab^2+4a^2+b^2}\)
b) ĐKXĐ: \(\hept{\begin{cases}2a\ne b\\2a\ne-b\end{cases}}\)
Ta thấy \(a=\frac{1}{3};b=2\)thỏa mãn điều kiện \(\hept{\begin{cases}2a\ne b\\2a\ne-b\end{cases}}\)nên thay vào A ta được:
bạn thay vào tự tính nhé mà cái phần rút gọn bạn vừa làm vừa check giùm bài mik nhé =)) sợ sai
a, \(\frac{a\left(b+1\right)-b-1}{b\left(a-1\right)+a-1}=\frac{a\left(b+1\right)-\left(b+1\right)}{b\left(a-1\right)+\left(a-1\right)}=\frac{\left(b+1\right)\left(a-1\right)}{\left(b+1\right)\left(a-1\right)}=1\)
b, \(\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}=\frac{2a\left(b+1\right)-\left(b+1\right)}{3b\left(2a-1\right)+3\left(2a-1\right)}=\frac{\left(b+1\right)\left(2a-1\right)}{\left(2a-1\right)\left(b+1\right)3}=\frac{1}{3}\)
Sửa lại đề bài: 1 / 2a- b
( MÁY MK KO ĐÁNH ĐC PHÂN SỐ MONG BN THÔNG CẢM)
mới lm đc nhé bn!
a) ĐKXĐ: bn tự lm nhé !
bn biến đổi: 2a3-b+2a-a2b = (2a-b) + ( 2a3-a2b) = (2a-b) + a2(2a-b) = (2a-b)(a2+1)
rồi bn nhân 1 / 2a+b với a2+1 rồi trừ 2 phân thức với nhau sẽ ra 0 => A=0
b. Sử dụng các hằng đẳng thức
\(a^3+b^3+c^2-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=3\left(a^2+b^2+c^2-ab-bc-ca\right)\)
và \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
nên \(A=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{1}{2}.\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Do (a - b) + (b - c) + (c - a) = 0 nên áp dụng hđt \(X^2+Y^2+Z^2=-2\left(XY+YZ+ZX\right)\)khi X + Y + Z = 0, ta có:
\(A=-2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right).\)
Bài 1 :
\(b,ax^2+3ax+9=a^2\)
\(\Leftrightarrow a^2x+3ax+9-a^2=0\)
\(\Leftrightarrow ax\left(a+3\right)+\left(a+3\right)\left(3-a\right)=0\)
\(\Leftrightarrow\left(a+3\right)\left(ax+3-a\right)=0\)
Vì \(a\ne3\Rightarrow\left(a+3\right)\ne0\Rightarrow ax+3-a=0\)
\(\Leftrightarrow ax=a-3\)
Vì \(a\ne0\Rightarrow x=\frac{a-3}{a}\)
1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
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