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Bài làm
a) \(Q=\left(\frac{1}{x-1}-\frac{1}{x}\right):\left(\frac{x+1}{x-2}-\frac{x+2}{x-1}\right)\)
\(Q=\left(\frac{x}{x\left(x-1\right)}-\frac{x-1}{x\left(x-1\right)}\right):\left(\frac{x^2-1}{\left(x-2\right)\left(x-1\right)}-\frac{x^2-4}{\left(x-2\right)\left(x-1\right)}\right)\)
\(Q=\left(\frac{x-x+1}{x\left(x-1\right)}\right):\left(\frac{x^2-1-x^2+4}{\left(x-2\right)\left(x-1\right)}\right)\)
\(Q=\frac{1}{x\left(x-1\right)}:\frac{3}{\left(x-2\right)\left(x-1\right)}\)
\(Q=\frac{1}{x\left(x-1\right)}.\frac{\left(x-2\right)\left(x-1\right)}{3}\)
\(Q=\frac{x-2}{3x}\)
ĐKXĐ: \(\frac{x-2}{3}\ge0\)
Vì \(\frac{x-2}{3}\ge0\). Mà 3 > 0
=> x - 2 > 0
<=> x > 2
Vậy x > 2 thì biểu thức Q có nghĩa.
b) \(C=\left(\frac{x+2}{x^2-x}+\frac{x-2}{x^2+x}\right).\frac{x^2-1}{x^2+2}\)
\(C=\left(\frac{x+2}{x\left(x-1\right)}+\frac{x-2}{x\left(x+1\right)}\right).\frac{x^2-1}{x^2+2}\)
\(C=\left(\frac{\left(x+2\right)\left(x+1\right)}{x\left(x^2-1\right)}+\frac{\left(x-2\right)\left(x-1\right)}{x\left(x^2-1\right)}\right).\frac{x^2-1}{x^2+2}\)
\(C=\left(\frac{x^2+2x+x+2+x^2-x-2x+2}{x\left(x^2-1\right)}\right).\frac{x^2-1}{x^2+2}\)
\(C=\frac{2x^2+4}{x\left(x^2-1\right)}.\frac{x^2-1}{x^2+2}\)
\(C=\frac{2\left(x^2+2\right)}{x\left(x^2-1\right)}.\frac{x^2-1}{x^2+2}\)
\(C=\frac{2}{x}\)
ĐKXĐ: \(\frac{2}{x}\ge0\)
Vì \(\frac{2}{x}\ge0\),
Mà 2 > 0
=> x > 0
Vậy x > 0 thì biểu thức C có nghĩa.
mọi ng ơi mk viết thiếu dấu ngoặc nha.thiếu ngoặc lownns nha. đóng ngoắc ở trước dấu chia
a) đk: \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
b) Ta có:
\(A=\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}}{x-1}\right)\div\frac{1}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}+1-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}+1\right)\)
\(A=\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}+1\right)\)
\(A=\frac{1}{\sqrt{x}-1}\)
c) Ta có; \(A=-\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{\sqrt{x}-1}=-\frac{1}{2}\)
\(\Leftrightarrow\sqrt{x}-1=-2\)
\(\Leftrightarrow\sqrt{x}=-1\) (vô lý)
Vậy không tồn tại x để A = -1/2
a/ Ta có
P = \(\frac{1+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) - \(\frac{2+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) - \(\frac{1+\sqrt{x}}{x+\sqrt{x}+1}\)
= \(\frac{-\sqrt{x}}{1+\sqrt{x}+x}\)
a) Đkxđ : \(\left\{{}\begin{matrix}x\ne0\\x\ne1\\x\ne2\end{matrix}\right.\)
\(Q=\left(\frac{1}{x-1}-\frac{1}{x}\right):\left(\frac{x+1}{x-2}-\frac{x+2}{x-1}\right)\)
\(Q=\frac{x-\left(x-1\right)}{x\left(x-1\right)}:\frac{\left(x+1\right)\left(x-1\right)-\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}\)
\(Q=\frac{1}{x\left(x-1\right)}:\frac{x^2-1-\left(x^2-4\right)}{\left(x-2\right)\left(x-1\right)}\)
\(Q=\frac{1}{x\left(x-1\right)}:\frac{3}{\left(x-2\right)\left(x-1\right)}=\frac{1}{x\left(x-1\right)}.\frac{\left(x-2\right)\left(x-1\right)}{3}=\frac{x-2}{3x}\)
b) Đkxđ : \(\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
\(C=\left(\frac{x+2}{x^2-x}+\frac{x-2}{x^2+x}\right).\frac{x^2-1}{x^2+2}\)
\(C=\left(\frac{x+2}{x\left(x-1\right)}+\frac{x-2}{x\left(x+1\right)}\right).\frac{x^2-1}{x^2+2}\)
\(C=\frac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}.\frac{x^2-1}{x^2+2}\)
\(C=\frac{x^2+3x+2+x^2-3x+2}{x\left(x^2-1\right)}.\frac{x^2-1}{x^2+2}=\frac{2x^2+4}{x\left(x^2+2\right)}=\frac{2}{x}\)