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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) ĐK: \(x\ne\left\{0;\pm7;49\right\}\)
b) \(\left(\frac{x}{x^2-49}-\frac{x-7}{x^2-7x}\right):\frac{2x-7}{x^2+7x}-\frac{x}{x-7}\)
Xét \(\frac{x}{x^2-49}-\frac{x-7}{x^2-7x}\)
= \(\frac{x^2}{x\left(x-7\right)\left(x+7\right)}-\frac{\left(x-7\right)\left(x+7\right)}{x\left(x-7\right)\left(x+7\right)}\)
=\(\frac{x^2-\left(x-7\right)\left(x+7\right)}{x\left(x-7\right)\left(x+7\right)}\)
=\(\frac{x^2-\left(x^2-49\right)}{x.\left(x-7\right)\left(x+7\right)}\)
=\(\frac{49}{x\left(x+7\right)\left(x-7\right)}\) (ĐK: x\(\ne\) { 0; 7;-7;49}
sau đó: chia trước trừ sau
Đoạn sau dễ chắc bạn tự làm được
Làm bài tốt
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 \(x=-\frac{1}{2}\Rightarrow A=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
Huỳnh Thoại m ghi thế bố t cx chả hỉu k it lm ns luôn đi lại còn bày đặt giỏi đã ngu còn tỏ ra ngu hơn
a) \(P=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)
\(P=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}\)
\(P=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)
\(P=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x\left(x+1\right)-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x-1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x+1\right)}\)
b) \(Q=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x^2+4x-5}{2\left(x+5\right)}\)
a/ \(\frac{7x-14y}{x^2-4y^2}=\frac{7\left(x-2y\right)}{x^2-\left(2y\right)^2}=\frac{7\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{7}{x+2y}.\)
b/ \(\frac{1-\frac{2y}{x}+\frac{y^2}{x^2}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{x^2-2xy+y^2}{x^2}}{\frac{y-x}{xy}}=\frac{\left(x-y\right)^2}{x^2}.\frac{xy}{-\left(x-y\right)}=-\frac{y\left(x-y\right)}{x}=\frac{y\left(y-x\right)}{x}\)
ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne5\end{cases}}\)
\(A=\frac{x-5}{\left(x-2\right)\left(x-5\right)}+\frac{x^2-x-2}{\left(x-2\right)\left(x-5\right)}-\frac{2\left(x-2\right)^2}{\left(x-2\right)\left(x-5\right)}\)
\(=\frac{x-5+x^2-x-2-2x^2+8x-8}{\left(x-2\right)\left(x-5\right)}\)
\(=\frac{-x^2+8x-15}{\left(x-2\right)\left(x-5\right)}=\frac{\left(3-x\right)\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=\frac{3-x}{x-2}\)
\(A=\frac{1}{x-2}+\frac{x^2-x-2}{x^2-7x+10}-\frac{2x-4}{x-5}\)ĐK : \(x\ne2;5\)
\(=\frac{x-5+x^2-x-2-\left(2x-4\right)\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\frac{x^2-7-2x^2+8x-8}{\left(x-2\right)\left(x-5\right)}\)
\(=\frac{-x^2+8x-15}{\left(x-2\right)\left(x-5\right)}=\frac{-\left(x-5\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5-x}{x-2}\)