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\(B=\left(\frac{2x}{x-3}-\frac{x-1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\left(ĐK:x\ne\pm3\right)\)
\(=\frac{2x\left(x+3\right)-\left(x-1\right)\left(x-3\right)-x^2-1}{x^2-9}:\frac{x+3-x+1}{x+3}\)
\(=\frac{2x^2+6x-x^2+3x+x-3-x^2-1}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}\)
\(=\frac{10x-4}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}=\frac{10x-4}{4\left(x-3\right)}\)
\(B=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)
\(=\left[\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+3-x+1}{x+3}\right)\)
\(=\left(\frac{2x^2+6x-x^2+3x-x+3-x^2-1}{\left(x+3\right)\left(x-3\right)}\right):\frac{4}{x+3}\)
\(=\frac{8x-1}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{4}\)\(=\frac{8x-1}{4\left(x-3\right)}\)
a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.
Thay x=-2 và B ta có :
\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)
b) Rút gọn :
\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)
\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
Xấu nhỉ ??
\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
a. A=\(1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left(\frac{x+1+x+1-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right).\frac{x\left(x^2-x+1\right)}{x^2\left(x-2\right)}\)
\(=1+\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b.\(\left|x-\frac{3}{4}\right|=\frac{5}{4}\Rightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)
Với \(x=2\Rightarrow A=\frac{2-1}{2+1}=\frac{1}{3}\)
Với \(x=-\frac{1}{2}\Rightarrow A=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
a xác định khi và chỉ khi x^2 -1 khác 0 suy ra x^2 khác 1 suy ra x khác 1
\(\frac{x^2-9}{x^2+2x+1}\)khác 0 suy ra x^2-9 khác 0 suy ra x^2 khác 9 suy ra x khác 3
1-x khác 0 suy ra x khác 1
vậy xác định khi x khác 1 và 3
b A = \(\frac{x+3}{x^2-1}\cdot\frac{x^2+2x+1}{x^2-9}-\frac{x}{1-x}\)
= \(\frac{\left(x+3\right)\cdot\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)}-\frac{x}{1-x}\)
= \(\frac{x+1}{\left(x-1\right)\left(x-3\right)}+\frac{x}{x-1}\)
= \(\frac{x+1+x\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{x+1+x^2-3x}{\left(x-1\right)\left(x-3\right)}=\frac{x^2-2x+1}{\left(x-1\right)\left(x-3\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x-3\right)}=\frac{x-1}{x-3}\)
\(A=\frac{x^3+2x^2+x}{x^3+x}=\frac{x^3+3x}{x^2+x}=\frac{x^2+3}{x^2+1}\)
\(B=\frac{x^2-9}{3-x}=\frac{x^2-9}{-\left(3-x\right)}=\frac{\left(x-3\right)\left(x+3\right)}{x+3}=x-3\)
Em mới lớp 7 nên rút gọn bừa ạ !!!
\(A=\frac{x^3+2x^2+x}{x^3+x}=\frac{\left(x^3+x\right)+2x^2}{x^3+x}=\frac{x^3+x}{x^3+x}+\frac{2x^2}{x^3+x}=\frac{2x^2}{x^3+x}\)\(=\frac{x^2.x.x}{x.\left(x^2+1\right)}=\frac{x^2.x}{x^2+1}\)
Lm thử sức thôi ạ !!!