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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
\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)
a) \(ĐK:a\ne1;a\ne0\)
\(A=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}=\left[\frac{a^2-2a+1}{a^2+a+1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}\)\(=\left[\frac{a^3-3a^2+3a-1}{a^3-1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}=\frac{a^3-1}{a^3-1}.\frac{4a}{a^2+4}=\frac{4a}{a^2+4}\)
b) Ta có: \(a^2+4\ge4a\)(*)
Thật vậy: (*)\(\Leftrightarrow\left(a-2\right)^2\ge0\)
Khi đó \(\frac{4a}{a^2+4}\le1\)
Vậy MaxA = 1 khi x = 2
a) \(ĐKXĐ:\hept{\begin{cases}a\ne1\\a\ne0\end{cases}}\)
\(M=\left(\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right)\div\frac{a^3+4a}{4a^2}\)
\(\Leftrightarrow M=\left(\frac{\left(a-1\right)^2}{a^2+a+1}-\frac{1-2a^2+4a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\right):\frac{a^2+4}{4a}\)
\(\Leftrightarrow M=\frac{\left(a-1\right)^3-1+2a^2-4a+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a}{a^2+4}\)
\(\Leftrightarrow M=\frac{a^3-3a^2+3a-1-1+2a^2-4a+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a}{a^2+4}\)
\(\Leftrightarrow M=\frac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a^2}{a^2+4}\)
\(\Leftrightarrow M=\frac{4a^2}{a^2+4}\)
b) Ta có : \(\frac{4a^2}{a^2+4}=\frac{4\left(a^2+4\right)-16}{a^2+4}\)
\(=4-\frac{16}{a^2+4}\)
Để M đạt giá trị lớn nhất
\(\Leftrightarrow\frac{16}{a^2+4}\)min
\(\Leftrightarrow a^2+4\)max
\(\Leftrightarrow a\)max
Vậy để M đạt giá trị lớn nhất thì a phải đạ giá trị lớn nhất.
a) \(ĐKXĐ:\hept{\begin{cases}a\ne\pm2\\a\ne1\\a\ne0\end{cases}}\)
\(A=\left(\frac{4a}{2+a}+\frac{8a^2}{4-a^2}\right):\left(\frac{a-3}{a^2-2a}-\frac{2}{a}\right)\)
\(\Leftrightarrow A=\frac{8a-4a^2+8a^2}{\left(2-a\right)\left(2+a\right)}:\frac{a-3-2a+4}{a\left(a-2\right)}\)
\(\Leftrightarrow A=\frac{4a^2+8a}{\left(2-a\right)\left(2+a\right)}:\frac{-a+1}{a\left(a-2\right)}\)
\(\Leftrightarrow A=\frac{4a}{2-a}:\frac{-a+1}{a\left(a-2\right)}\)
\(\Leftrightarrow A=\frac{4a^2\left(a-2\right)}{\left(a-2\right)\left(a-1\right)}\)
\(\Leftrightarrow A=\frac{4a^2}{a-1}\)
b) Để A nhận giá trị nguyên
\(\Leftrightarrow\frac{4a^2}{a-1}\inℤ\)
\(\Leftrightarrow4a^2⋮a-1\)
\(\Leftrightarrow4\left(a^2-1\right)+4⋮a-1\)
\(\Leftrightarrow4\left(a-1\right)\left(a+1\right)+4⋮a-1\)
\(\Leftrightarrow4⋮a-1\)
\(\Leftrightarrow a-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow a\in\left\{0;2;-1;3;-3;5\right\}\)
Ta sẽ loại các giá trị ở đkxđ
Vậy để \(A\inℤ\Leftrightarrow a\in\left\{2;-1;3;-3;5\right\}\)
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
\(\left(\frac{2+a}{2-a}-\frac{4a^2}{a^2-4}+\frac{a-2}{a+2}\right)\div\frac{a-3}{2a-a^2}\)(ĐK: \(a\ne0,a\ne\pm2,a\ne3\))
\(=\frac{\left(2+a\right)\left(2+a\right)+4a^2+\left(a-2\right)\left(2-a\right)}{\left(2-a\right)\left(2+a\right)}\div\frac{a-3}{a\left(2-a\right)}\)
\(=\frac{4+4a+a^2+4a^2-\left(a^2-4a+4\right)}{\left(2-a\right)\left(2+a\right)}.\frac{a\left(2-a\right)}{a-3}\)
\(=\frac{4a^2+8a}{\left(2-a\right)\left(2+a\right)}.\frac{a\left(2-a\right)}{a-3}\)
\(=\frac{4a^2}{a-3}\)
\(\left(\frac{2+a}{2-a}-\frac{4a^2}{a^2-4}+\frac{a-2}{a+2}\right)\div\frac{a-3}{2a-a^2}\)
\(=\left(\frac{2+a}{2-a}+\frac{4a^2}{4-a^2}+\frac{a-2}{2+a}\right)\div\frac{a-3}{2a-a^2}\)
\(=\left(\frac{\left(2+a\right)^2}{\left(2-a\right)\left(2+a\right)}+\frac{4a^2}{\left(2-a\right)\left(2+a\right)}+\frac{\left(a-2\right)\left(2-a\right)}{\left(2-a\right)\left(2+a\right)}\right)\div\frac{a-3}{2a-a^2}\)
\(=\left(\frac{4+4a+a^2+4a^2-a^2+4a-4}{\left(2-a\right)\left(2+a\right)}\right)\div\frac{a-3}{2a-a^2}\)
\(=\frac{4a^2+8a}{\left(2-a\right)\left(2+a\right)}\div\frac{a-3}{2a-a^2}\)
\(=\frac{4a\left(a+2\right)}{\left(2-a\right)\left(2+a\right)}\div\frac{a-3}{2a-a^2}\)
\(=\frac{4a}{2-a}.\frac{2a-a^2}{a-3}=\frac{4a\left(2a-a^2\right)}{\left(2-a\right)\left(a-3\right)}=\frac{4a^2\left(2-a\right)}{\left(2-a\right)\left(a-3\right)}=\frac{4a^2}{a-3}\)