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\(\text{a) ĐKXĐ: }a\ne1\)
\(\text{b) }M=\frac{a^2+1+a}{a^2+1}:\left[\frac{1}{a-1}-\frac{2a}{a^2\left(a-1\right)+\left(a-1\right)}\right]\)
\(M=\frac{a^2+a+1}{a^2+1}:\left[\frac{1}{a-1}-\frac{2a}{\left(a-1\right)\left(a^2+1\right)}\right]\)
\(M=\frac{a^2+a+1}{a^2+1}:\frac{a^2+1-2a}{\left(a-1\right)\left(a^2+1\right)}\)
\(M=\frac{a^2+a+1}{a^2+1}.\frac{\left(a-1\right)\left(a^2+1\right)}{\left(a-1\right)^2}\)
\(M=\frac{a^2+a+1}{a-1}\)
a/ Điều kiện xác định \(\hept{\begin{cases}a^2+a\ne0\\a^2-a\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}a\ne0\\a\ne1\\a\ne-1\end{cases}}}\)
b/ \(M=\frac{a^2-1}{2016+2015a^2}\left(\frac{2015a-2016}{a+a^2}+\frac{2016+2015a}{a^2-a}\right)\)
\(=\frac{\left(a-1\right)\left(a+1\right)}{2016+2015a^2}\left(\frac{2015a-2016}{a\left(a+1\right)}+\frac{2016+2015a}{a\left(a-1\right)}\right)\)
\(=\frac{\left(a-1\right)\left(a+1\right)}{2016+2015a^2}\left(\frac{2015a-2016}{a\left(a+1\right)}+\frac{2016+2015a}{a\left(a-1\right)}\right)\)
\(=\frac{\left(a-1\right)\left(a+1\right)}{2016+2015a^2}.\frac{2\left(2015a^2+2016\right)}{a\left(a+1\right)\left(a-1\right)}\)
\(=\frac{2}{a}=\frac{2}{2016}=\frac{1}{1008}\)
điều kiện dễ mà,mẫu phải khác 0=>điều kiện pài này là x khác 1
a) \(ĐKXĐ:\hept{\begin{cases}a\ne-3\\a\ne\pm2\end{cases}}\)
\(M=\frac{2a-a^2}{a+3}\left(\frac{a-2}{a+2}-\frac{a+2}{a-2}+\frac{4a^2}{4-a^2}\right)\)
\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{\left(a-2\right)^2-\left(a+2\right)^2-4a^2}{\left(a-2\right)\left(a+2\right)}\)
\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{a^2-4a+4-a^2-4a-4-4a^2}{\left(a-2\right)\left(a+2\right)}\)
\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{-4a^2-8a}{\left(a-2\right)\left(a+2\right)}\)
\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{-4a\left(a+2\right)}{\left(a-2\right)\left(a+2\right)}\)
\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{-4a}{a-2}\)
\(\Leftrightarrow M=\frac{4a^2\left(a-2\right)}{\left(a+3\right)\left(a-2\right)}\)
\(\Leftrightarrow M=\frac{4a^2}{a+3}\)
b) Để M = 1
\(\Leftrightarrow\frac{4a^2}{a+3}=1\)
\(\Leftrightarrow4a^2=a+3\)
\(\Leftrightarrow4a^2-a-3=0\)
\(\Leftrightarrow\left(4a+3\right)\left(a-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4a+3=0\\a-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a=-\frac{3}{4}\left(tm\right)\\a=1\left(tm\right)\end{cases}}\)
Vậy để \(M=1\Leftrightarrow a\in\left\{-\frac{3}{4};1\right\}\)
c) Để M > 0
\(\Leftrightarrow\frac{4a^2}{a+3}>0\)
\(\Leftrightarrow a+3>0\)(Vì 4a2 > 0, loại trường hợp = 0)
\(\Leftrightarrow a>-3\)
Vậy để \(M>0\Leftrightarrow a>-3\)
Để M < 0
\(\Leftrightarrow\frac{4a^2}{a+3}< 0\)
\(\Leftrightarrow a+3< 0\)(Vì 4a2 > 0, loại trường hợp = 0)
\(\Leftrightarrow a< -3\)
Vậy để \(M< 0\Leftrightarrow a< -3\)
Câu 1 :
a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)
\(\Leftrightarrow2x^2+8x+6=0\)
\(\Leftrightarrow x^2+4x+4-1=0\)
\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)
Vậy : \(x=-3\) thì P = 1.
Ta có :
\(A=\frac{a^2+2a}{2a+10}+\frac{a-5}{a}+\frac{50-5a}{2a\left(a+5\right)}\)
\(A=\frac{a^2+2a}{2\left(a+5\right)}+\frac{a-5}{a}+\frac{50-5a}{2a\left(a+5\right)}\)
a) Giá trị của biểu thức A xác định
\(\Leftrightarrow\hept{\begin{cases}a+5\ne0\\a\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}a\ne-5\\a\ne0\end{cases}}}\)
Vậy để giá trị của biểu thức A xác định \(\Leftrightarrow\hept{\begin{cases}a\ne-5\\a\ne0\end{cases}}\)
ĐKXĐ : \(\hept{\begin{cases}a\ne-5\\a\ne0\end{cases}}\)
b) Ta có :
\(A=\frac{a^2+2a}{2\left(a+5\right)}+\frac{a-5}{a}+\frac{50-5a}{2a\left(a+5\right)}\)
\(A=\frac{a\left(a^2+2a\right)+2\left(a+5\right)\left(a-5\right)+50-5a}{2a\left(a+5\right)}\)
\(A=\frac{a^3+2a^2+2\left(a^2-25\right)+50-5a}{2a\left(a+5\right)}\)
\(A=\frac{a^3+4a^2-50+50-5a}{2a\left(a+5\right)}\)
\(A=\frac{a\left(a^2+4a-5\right)}{2a\left(a+5\right)}\)
\(A=\frac{a^2+5a-a-5}{2\left(a+5\right)}\)
\(A=\frac{\left(a+5\right)\left(a-1\right)}{2\left(a+5\right)}=\frac{a-1}{2}\)
c) Thay a = -1 ( Thỏa mãn ĐKXĐ ) vào biểu thức A ta có :
\(A=\frac{-1-1}{2}=-1\)
Vậy tại a = -1 thì giá trị của biểu thức A là - 1
d) Cho A = 0 , ta có :
\(\frac{a-1}{2}=0\)
\(\Leftrightarrow a-1=0\Leftrightarrow a=1\)( Thỏa mãn ĐKXĐ )
Vậy a = 1 thì giá trị của biểu thức A = 0 .
\(a.ĐKXĐ:\)\(2a+10\ne0\) \(a\ne-5\)
\(a\ne0\) \(\Leftrightarrow\)\(a\ne0\) \(\Leftrightarrow\)\(\hept{\begin{cases}a\ne0\\a\ne-5\end{cases}}\)
\(2a\left(a+5\right)\ne0\) \(\hept{\begin{cases}a\ne0\\a\ne-5\end{cases}}\)
\(b.A=\frac{a\left(a+2\right)}{2\left(a+5\right)}+\frac{a-5}{a}+\frac{5\left(10-a\right)}{2a\left(a+5\right)}\)
\(=\frac{a\left(a+2\right)a}{2a\left(a+5\right)}+\frac{\left(a-5\right)2\left(a+5\right)}{2a\left(a+5\right)}+\frac{5\left(10-a\right)}{2a\left(a+5\right)}\)
\(=\frac{a^3+2a^2+\left(2a-10\right)\left(a+5\right)+5\left(10-a\right)}{2a\left(a+5\right)}\)
\(=\frac{a^3+2a^2+2a^2+10a-10a-50+50-5a}{2a\left(a+5\right)}\)
\(=\frac{a^3+4a^2-5a}{2a\left(a+5\right)}\)
\(=\frac{a\left(a^2+4a-5\right)}{2a\left(a+5\right)}\)
\(=\frac{a\left(a-1\right)\left(a+5\right)}{2a\left(a+5\right)}\)
\(=\frac{a-1}{2}\)với \(x\ne0\)và \(x\ne-5\)
\(c.\)Thay \(a=-1\left(t/mđk\right)\Leftrightarrow\frac{a-1}{2}\Rightarrow\frac{-1-1}{2}\)
\(=-1\left(t/mđk\right)\)
\(d.A=0\Leftrightarrow A=\frac{a-1}{2}=0\)
\(\Rightarrow a-1=2.0\)
\(\Rightarrow a-1=0\)
\(\Rightarrow a=1\left(t/mđk\right)\)
\(M=\left(\frac{1}{a^2-a}+\frac{1}{a-1}\right):\frac{a+1}{a^2-2a+1}\)
\(M=\left(\frac{1}{a\left(a-1\right)}+\frac{1}{a-1}\right):\frac{a+1}{\left(a-1\right)^2}\)ĐKXĐ : a khác 0, a khác 1
\(M=\frac{1+a}{a\left(a-1\right)}.\frac{\left(a-1\right)^2}{a+1}\)
\(M=\frac{a-1}{a}\)
\(M=\left(\frac{1}{a^2-a}+\frac{1}{a-1}\right):\frac{a+1}{a^2-2a+1}\)DK : \(x\ne0;\pm1\)
\(=\left(\frac{1}{a\left(a-1\right)}+\frac{1}{a-1}\right):\frac{a+1}{\left(a-1\right)^2}=\left(\frac{1}{a\left(a-1\right)}+\frac{a}{a\left(a-1\right)}\right):\frac{a+1}{\left(a-1\right)^2}\)
\(=\frac{a+1}{a\left(a-1\right)}.\frac{\left(a-1\right)^2}{a+1}=\frac{a-1}{a}\)