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3 tháng 8 2020
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có :
\(A=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{x-1}:\frac{1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+1}\)
\(=1\)
Vậy...
b/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có :
\(B=\left(\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}+6\right)\left(\frac{x\sqrt{x}-1}{x+\sqrt{x}+1}-3\right)\)
\(=\left(\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}+6\right)\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-3\right)\)
\(=\left(\sqrt{x}-2+6\right)\left(\sqrt{x}-1-3\right)\)
\(=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)\)
\(=x-16\)
Vậy..
c/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có :
\(C=\frac{2\sqrt{x}}{x-1}+\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}-x}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-2}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{2}{\sqrt{x}}\)
Vậy..
30 tháng 7 2020
mình sẽ xóa câu này mong bạn gửi lại câu hỏi khác để rõ ràng cho các bạn khác tham khảo nha
8 tháng 10 2020
đk: \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có:
\(P=\left(\frac{1}{1-\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\div\left(\frac{2x+\sqrt{x}-1}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(P=\frac{\sqrt{x}-1+\sqrt{x}}{\left(1-\sqrt{x}\right)\sqrt{x}}\div\frac{\left(2x+\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)+\left(2x+\sqrt{x}-1\right)\left(\sqrt{x}-x\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\)
\(P=\frac{2\sqrt{x}-1}{\left(1-\sqrt{x}\right)\sqrt{x}}\cdot\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{2x+\sqrt{x}-1}\)
\(P=\frac{\left(2\sqrt{x}-1\right)\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(1+\sqrt{x}\right)\sqrt{x}}\)
\(P=\frac{x-\sqrt{x}+1}{\sqrt{x}}=\frac{x\sqrt{x}-x+\sqrt{x}}{x}\)
PH
tính \(\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{\sqrt{x^3}-x}{1-\sqrt{x}}\)
1
19 tháng 8 2019
\(\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{\sqrt{x^3}-x}{1-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x}-\sqrt{x-1}}-\left(-x\right)\)
\(=\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x}-\sqrt{x-1}}+x\)
\(=\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x}-\sqrt{x-1}}+\frac{x}{1}\)
\(=\frac{\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}-\frac{\sqrt{x}+\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}\)\(+\frac{x}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}\)
\(=\frac{\sqrt{x}-\sqrt{x-1}-\left(\sqrt{x}+\sqrt{x-1}\right)+x}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}\)
\(=\frac{\sqrt{x}-\sqrt{x-1}-\left(\sqrt{x}+\sqrt{x-1}\right)+x}{1}\)
\(=\frac{x-2\sqrt{x-1}}{1}\)
\(=x-2\sqrt{x-1}\)
WH
4 tháng 8 2020
+ bắt bẻ : đkxđ x ≥ 0 vì có \(\sqrt{x}\)
+ giải thích : \(\frac{1}{\sqrt{x+3}+\sqrt{x+2}}=\frac{\sqrt{x+3}-\sqrt{x+2}}{\left(\sqrt{x+3}+\sqrt{x+2}\right)\left(\sqrt{x+3}-\sqrt{x+2}\right)}=\frac{\sqrt{x+3}-\sqrt{x+2}}{\sqrt{x+3}^2-\sqrt{x+2}^2}=\frac{\sqrt{x+3}-\sqrt{x+2}}{x+3-x-2}=\frac{\sqrt{x+3}-\sqrt{x+2}}{1}=\sqrt{x+3}-\sqrt{x+2}\)
tương tự vs mấy cái còn lại !!
~ bn làm vậy ai hiểu cho nỗi !! ~
với x >= 0 ; x khác 1
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}=\frac{x-\sqrt{x}}{x\sqrt{x}-1}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
Answer:
Với \(x\ne1;x\ge0\) có:
\(\frac{x+2}{x\sqrt{x}-1}\)\(+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)