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\(A=\frac{x+3a}{2-x}+\frac{x-3a}{2+x}-\frac{2a}{4-x^2}+a\)
\(=\frac{\left(x+3a\right)\left(2+x\right)+\left(x-3a\right)\left(2-x\right)-2a}{\left(2-x\right)\left(2+x\right)}+a\)
\(=\frac{2x+x^2+6a+3ax+2x-x^2-6a+3ax-2a}{\left(2-x\right)\left(2+x\right)}+a\)
\(=\frac{6ax+4x-2a+a\left(4-x^2\right)}{\left(2-x\right)\left(2+x\right)}=\frac{6ax+4x-2a+4a-a\cdot x^2}{\left(2-x\right)\left(2+x\right)}=\frac{-a\cdot x^2+6\cdot ax+2a}{\left(2-x\right)\cdot\left(2+x\right)}\)
\(-a\cdot x^2+6a\cdot x+2a\)
\(=-a\cdot\frac{a^2}{\left(3a+2\right)^2}+6a\cdot\frac{a}{3a+2}+2a=\frac{-a^3}{\left(3a+2\right)^2}+\frac{6a^2\left(3a+2\right)+2a\left(3a+2\right)^2}{\left(3a+2\right)^2}\)
\(=\frac{-a^3+18a^3+12a^2+2a\left(9a^2+12a+4\right)}{\left(3a+2\right)^2}\)
\(=\frac{17a^3+12a^2+18a^3+24a^2+8a}{\left(3a+2\right)^2}=\frac{35a^3+36a^2+8a}{\left(3a+2\right)^2}\)
\(\left(2-x\right)\left(2+x\right)\)
\(=\left(2-\frac{a}{3a+2}\right)\left(2+\frac{a}{3a+2}\right)=\frac{6a+4-a}{\left(3a+2\right)}\cdot\frac{6a+4+a}{3a+2}\)
\(=\frac{\left(5a+4\right)\left(7a+4\right)}{\left(3a+2\right)^2}\)
\(A=\frac{-a\cdot x^2+6\cdot ax+2a}{\left(2-x\right)\cdot\left(2+x\right)}\)
\(=\frac{35a^3+36a^2+8a}{\left(3a+2\right)^2}:\frac{\left(5a+4\right)\left(7a+4\right)}{\left(3a+2\right)^2}=\frac{35a^3+36a^2+8a}{\left(5a+4\right)\left(7a+4\right)}\)
Rút gọn biểu thức
\(A=\frac{x+3a}{2-x}+\frac{x-3a}{2+x}-\frac{2a}{4-x^2}+a\)
Với \(x=\frac{a}{3a+2}\)
Rút gọn biểu thức:
\(A=\frac{x+3a}{2-x}+\frac{x-3a}{2+x}-\frac{2a}{4-x^2}+a\)
Với \(x=\frac{a}{3a+2}\)
https://hoc24.vn/hoi-dap/question/177629.html (câu 2)
https://hoc24.vn/hoi-dap/question/940816.html?pos=2486212 (câu 1)
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)