rút gọn biểu thức \(\dfrac{1+3\sqrt{3x^3}}{1+\sqrt{3x}}-\sqrt{3x}\)
Những câu hỏi liên quan
Cho Q = \(\dfrac{3x+9\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
Rút gọn biểu thức Q
Rút gọn biểu thức P= \(\dfrac{3x+9\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
Tìm x để P= 3.
Sửa đề: 3x+căn 9x-3
\(P=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
P=3
=>căn x+1=3căn x-3
=>-2căn x=-4
=>x=4
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rút gọn biểu thức sau:
\(N=\dfrac{5\sqrt{x}+3x}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{1-\sqrt{x}}+\dfrac{7}{\sqrt{x}+3}\left(x\ge0,x\ne1\right)\)
\(N=\dfrac{5\sqrt{x}+3x}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{1-\sqrt{x}}+\dfrac{7}{\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}+3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{7\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5\sqrt{x}+3x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{3x+8\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{7\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{20\sqrt{x}+6x-10}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
Đề sai không vậy?
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Ta có: \(N=\dfrac{3x+5\sqrt{x}}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{7}{\sqrt{x}+3}\)
\(=\dfrac{3x+5\sqrt{x}+3x+9\sqrt{x}-\sqrt{x}-3+7\sqrt{x}-7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{6x+20\sqrt{x}-10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
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A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
Rút gọn biểu thức trên
\(x\ge0,x\ne9\)
\(A=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right]:\)
\(\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(A=\left[\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right].\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right).\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)
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Cho biểu thức P=\(\dfrac{3x+\sqrt{x}}{x+\sqrt{x}}+\dfrac{3\left(x-\sqrt{x}+1\right)}{x\sqrt{x}+1}\)
a) rút gọn P
b)chứng minh P<4
a: \(P=\dfrac{3\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}+1}=\dfrac{3\sqrt{x}+4}{\sqrt{x}+1}\)
b: \(P-4=\dfrac{3\sqrt{x}+4-4\sqrt{x}-4}{\sqrt{x}+1}=-\dfrac{\sqrt{x}}{\sqrt{x}+1}< 0\)
=>P<4
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(\(\dfrac{2\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{3x+3}{x-9}\)):(\(\dfrac{2\sqrt{x}-2}{\sqrt{x}+3}-1\))
a) Rút gọn biểu thức
b) Tìm x để Q<\(\dfrac{-1}{2}\)
c) Tìm min Q
\(a,=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}-3}{\sqrt{x}+3}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}\)
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a: \(=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
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4 tháng 9 2023 lúc 15:54
RÚT GỌN BIỂU THỨC:
16) \(A = \dfrac{\sqrt{x}}{\sqrt{x} + 3} + \dfrac{2\sqrt{x}}{\sqrt{x} - 3} - \dfrac{3x + 9}{x - 9}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
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\(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\) (ĐK: \(x\ge0;x\ne9\))
\(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{\left(\sqrt{x}\right)^2-3^2}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2x+6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{3}{\sqrt{x}+3}\)
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\(\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}-1\)
a, tìm ĐKXĐ và rút gọn biểu thức đã cho
b, Timf điều kiện của x để P<0
a) \(ĐK:x\ge0,x\ne1\)
\(=\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+4+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2x+4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
b) \(P=\dfrac{2\sqrt{x}}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)
Kết hợp với đk:
\(\Rightarrow0\le x< 1\)
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CHo biểu thức :
A = \(\left(\frac{6x-4}{3\sqrt{3x^3}-8}-\frac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\frac{1+3\sqrt{3x^3}}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
a) Rút gọn biểu thức A
b) Tìm các giá trị nguyên của x đẻ biểu thức A nhận giá trị nguyên