Tìm x để \(\frac{8\sqrt{x}+8}{6\sqrt{x}+9}>\frac{8}{3}\)
Những câu hỏi liên quan
\(A=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1+\frac{3\sqrt{x}-1}{3\sqrt{x}+1}\right),\left(x>0,x\ne\frac{1}{9}\right)\)
1. Rút gọn A
2. Tìm x để \(P=\frac{5}{6}\)
\(A=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{6\sqrt{x}}{3\sqrt{x}+1}\)
\(A=\left[\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right].\frac{3\sqrt{x}+1}{6\sqrt{x}}\)
\(A=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{6\sqrt{x}}\)
\(A=\frac{3\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}.\frac{1}{6\sqrt{x}}\)
\(A=\frac{\sqrt{x}+1}{6\sqrt{x}-2}\)
\(A=\frac{5}{6}\Leftrightarrow\frac{\sqrt{x}+1}{6\sqrt{x}-2}=\frac{5}{6}\)
\(\Leftrightarrow6\sqrt{x}+6=30\sqrt{x}-10\)
\(\Leftrightarrow24\sqrt{x}=16\)
\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\Leftrightarrow x=\frac{4}{9}\)
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\(A=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]\div\frac{6\sqrt{x}}{3\sqrt{x}+1}\)
\(A=\left[\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]\times\frac{3\sqrt{x}+1}{6\sqrt{x}}\)
\(A=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}\times\frac{1}{6\sqrt{x}}\)
\(A=\frac{3\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\times\frac{1}{6\sqrt{x}}\)
\(A=\frac{\sqrt{x}+1}{6\sqrt{x}-2}\)
\(A=\frac{5}{6}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{6\sqrt{x}-2}=\frac{5}{6}\)
\(\Leftrightarrow6\sqrt{x}+6=30\sqrt{x}-10\)
\(\Leftrightarrow24\sqrt{x}=16\)
\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{4}{9}\)
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Tìm GTNN của A = \(x-\sqrt{xy}+\frac{7y}{12}+\frac{\sqrt{y}}{6}-\frac{8}{3}+\frac{9}{16\sqrt{x}}\)
A=(\(\frac{x+2\sqrt{x}+4}{x\sqrt{x}-8}+\frac{x+2\sqrt{x}+1}{x-1}\)):(3+\(\frac{1}{\sqrt{x}-2}+\frac{2}{\sqrt{x}-1}\))
Tìm x để A<1
\(N=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{4\sqrt{x}}{3}\)
a) Rút gọn N
b) Tìm x để N=\(\frac{8}{9}\)
c) Tìm x để \(\frac{1}{N}>\frac{3\sqrt{x}}{4}\)
1.Tìm điều kiện xác định( hay tìm x) để mỗi biểu thức sau có nghĩa:
A=\(\sqrt{\frac{x-9}{5-x}}\)
B=\(\frac{\sqrt{7x^2+8x-15}}{6-x}\)
C=\(\frac{4}{5}\sqrt{\frac{-x+8}{27x^3-64}}\)
Cho biểu thức : \(C=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}-1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1+\frac{3\sqrt{x}-1}{3\sqrt{x}+1}\right),x>0,x\ne\frac{1}{9}\)
Rút gọn CTìm \(x\) để \(P=\frac{5}{6}\)
Cho biểu thức:
P=\(\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{3x-8\sqrt{x}+27}{9-x}\)
a, Tìm ĐKXĐ của x để P có nghĩa
b, Rút gọn biểu thức P
c, Tìm giá trị x nguyên để P nhận giá trị nguyên
a) đk: \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
b) Ta có:
\(P=\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{3x-8\sqrt{x}+27}{9-x}\)
\(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)+2\sqrt{x}\cdot\left(\sqrt{x}-3\right)-3x+8\sqrt{x}-27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{x+5\sqrt{x}+6+2x-6\sqrt{x}-3x+8\sqrt{x}-27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{7\sqrt{x}-21}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{7\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{7}{\sqrt{x}+3}\)
c) Nếu x không là số chính phương => P vô tỉ (loại)
=> x là số chính phương khi đó để P nguyên thì:
\(\left(\sqrt{x}+3\right)\inƯ\left(7\right)\) , mà \(\sqrt{x}+3\ge3\left(\forall x\ge0\right)\)
\(\Rightarrow\sqrt{x}+3=7\Leftrightarrow\sqrt{x}=4\Rightarrow x=16\)
Vậy x = 16 thì P nguyên
Cho \(A=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
Với \(x\ge0,x\ne\frac{1}{9}\)
a, Rút gọn A
b,Tìm x để A<1
mk làm luôn
a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)
=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
=\(\frac{\left(3x+3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}\)
=\(\frac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
=
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Cho \(A=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
Với \(x\ge0,x\ne\frac{1}{9}\)
a, Rút gọn A
b,Tìm x để A<1
a/ \(A=\frac{\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)-\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}+1-3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-4\sqrt{x}+1-3\sqrt{x}-1}{\left(3\sqrt{x}\right)^2-1}-\frac{8\sqrt{x}}{9x-1}}{1-1-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-7\sqrt{x}}{9x-1}-\frac{8\sqrt{x}}{9x-1}}{-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{3x-7\sqrt{x}-8\sqrt{x}}{9x-1}\left(\frac{-\left(3\sqrt{x}+1\right)}{3}\right)\)
\(A=\frac{3x-15\sqrt{x}}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{3\left(x-3\sqrt{x}\right)}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{\left(x-3\sqrt{x}\right)\left(-3\sqrt{x}-1\right)}{9x-1}\)
\(A=\frac{3x\sqrt{x}+8x+3\sqrt{x}}{9x-1}\)
\(A=\frac{3x\sqrt{x}}{9x-1}+\frac{8x}{9x-1}+\frac{3\sqrt{x}}{9x-1}\)
\(A=\frac{x\sqrt{x}}{x-\frac{1}{3}}+\frac{8x}{9x-1}+\frac{\sqrt{x}}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)}{x-\frac{1}{3}}+\frac{\frac{8}{3}x}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)+\frac{8}{3}x}{x-\frac{1}{3}}\)
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