Thu gon
\(\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{2}{\sqrt{x}-2}\right)\left(x^2-4\right)\) (x>=0, x# 4)
Những câu hỏi liên quan
rut gon bieu thuc A=\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\) tìm giá trị của x để A>0
\(A=\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)(DK : \(x\ge0;x\ne4\))
\(=\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{x-4+10-x}{\sqrt{x}+2}\)
\(=\frac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{6}=\frac{1}{2-\sqrt{x}}\)
Để A > 0 thì \(2-\sqrt{x}>0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Vậy để A < 0 thì x < 4
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Bảo Ngọc kết luận hơi sai một chút nhé. Để A > 0 thì x < 4 nhé :)
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\(B=\left(\frac{\left(x-2\right)}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\) rut gon
=\(\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\)
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Rut gon
a)\(\frac{\left(\sqrt{x}+1\right).\left(x-\sqrt{xy}\right).\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right).\left(\sqrt{x^3}+x\right)}\)
b) \(\frac{\left(2-\sqrt{x}\right)^2-\left(\sqrt{x}+3\right)}{1+2.\sqrt{x}}\)
Lời giải:
a)
\(=\frac{(\sqrt{x}+1)\sqrt{x}(\sqrt{x}-\sqrt{y}))\sqrt{x}+\sqrt{y})}{(x-y)x(\sqrt{x}+1)}=\frac{(\sqrt{x}+1)\sqrt{x}(x-y)}{(x-y)x\sqrt{x}+1)}=\frac{1}{\sqrt{x}}\)
b)
\(=\frac{(2-\sqrt{x}-\sqrt{x}-3)(2-\sqrt{x}+\sqrt{x}+3)}{1+2\sqrt{x}}=\frac{(-1-2\sqrt{x}).5}{2\sqrt{x}+1}=\frac{-5(2\sqrt{x}+1)}{2\sqrt{x}+1}=-5\)
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Rut gon
a)\(\frac{\left(\sqrt{x}+1\right).\left(x-\sqrt{xy}\right).\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right).\left(\sqrt{x^3}+x\right)}\)
b) \(\frac{\left(2-\sqrt{x}\right)^2-\left(\sqrt{x}+3\right)}{1+2.\sqrt{x}}\)
\(a,\frac{\left(\sqrt{x}+1\right)\cdot\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\sqrt{x}\left(x+1\right)}\)\(=\frac{\left(\sqrt{x}+1\right)\sqrt{x}\left(x-y\right)}{\left(x-y\right)\sqrt{x} \left(x+1\right)}\)\(=\frac{\sqrt{x}+1}{x+1}\)
\(b,\frac{\left(2-\sqrt{x}\right)^2-\sqrt{x}-3}{1+2\sqrt{x}}=\frac{4+x-4\sqrt{x}-\sqrt{x}-3}{1+2\sqrt{x}}=\frac{1+x-5\sqrt{x}}{1+2\sqrt{x}}\)
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rut gon bieu thuc: \(\frac{\sqrt{\sqrt{\frac{x-1}{x+1}}+\sqrt{\frac{x+1}{x-1}}-2}\left(2x+\sqrt{x^2+1}\right)}{\sqrt{\left(x+1\right)^3}-\sqrt{\left(x-1\right)^2}}\)
Thu gọn biểu thức :
\(\left(\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{\sqrt{x}+2}{\sqrt{x}-2}\right).\left(\sqrt{x}-\frac{4}{\sqrt{x}}\right)\) với x >0 và x # 4
\(\left(\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{\sqrt{x}+2}{\sqrt{x}-2}\right).\left(\sqrt{x}-\frac{4}{\sqrt{x}}\right)\)
\(=\left(\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right).\left(\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}}-\frac{4}{\sqrt{x}}\right)\)
\(=\frac{x-4\sqrt{x}+4-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}\right)^2-4}{\sqrt{x}}\)
\(=\frac{-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}}=-8\)
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Cho A = \(\left(\frac{x-\sqrt{x}+7}{x-4}+\frac{1}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{x-4}\right)\)
a) Rut gon A
b) So sanh A voi \(\frac{1}{A}\)
Pleft(frac{sqrt{x}+2}{sqrt{x}+1}-frac{x-sqrt{x}-3}{x-sqrt{x}-2}right):left(frac{x-sqrt{x}}{x-sqrt{x}-2}+frac{2}{sqrt{x}-2}right)frac{left(sqrt{x}+2right)left(sqrt{x}-2right)-x+sqrt{x}+3}{left(sqrt{x}+1right)left(sqrt{x}-2right)}:frac{x-sqrt{x}+2left(sqrt{x}+1right)}{left(sqrt{x}+1right)left(sqrt{x}-2right)}frac{x-4-x+3+sqrt{x}}{left(sqrt{x}+1right)left(sqrt{x}-2right)}.frac{left(sqrt{x}+1right)left(sqrt{x}-2right)}{x-sqrt{x}+2sqrt{x}+2}frac{sqrt{x}-1}{x+sqrt{x}+2}
Đọc tiếp
\(P=\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-3}{x-\sqrt{x}-2}\right):\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{2}{\sqrt{x}-2}\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x+\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-\sqrt{x}+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-x+3+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}+2\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+2}\)
#)Hỏi j đi bn, bn ph hỏi cái j chứ làm lun rùi còn để cộng đồng ngắm ak ???
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Bó cả tay lẫn chân !!! Bất lực như gặp cực hình !
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Chắc là bạn ấy hỏi bạn ấy làm có đúng ko ha gì đó ?
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Xem thêm câu trả lời
RÚT GON:1.frac{xsqrt{x}-3}{x-2sqrt{x}-3}-frac{2left(sqrt{x}-3right)}{sqrt{x}+1}+frac{sqrt{x}+3}{3-sqrt{x}}2.frac{left(x+sqrt{x}+1right)^2+1}{left(x+1right)^2}-frac{left(x-sqrt{x}-1right)^2-1}{left(1-xright)^2}3.frac{3x+sqrt{9x}-3}{3+sqrt{x}-2}-frac{sqrt{x}+1}{sqrt{x}+2}+frac{sqrt{x}-2}{1-sqrt{x}}
Đọc tiếp
RÚT GON:
\(1.\)\(\frac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(2.\)\(\frac{\left(x+\sqrt{x}+1\right)^2+1}{\left(x+1\right)^2}-\frac{\left(x-\sqrt{x}-1\right)^2-1}{\left(1-x\right)^2}\)
\(3.\)\(\frac{3x+\sqrt{9x}-3}{3+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{1-\sqrt{x}}\)