a) \(\sqrt{x}>1\), b)\(\sqrt{x}< 3\), c) \(\sqrt[2]{x}=14\)
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24 tháng 7 2019
a) \(B=\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{7\sqrt{x}-9}{x-9}\)
\(B=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{x-9}-\frac{7\sqrt{x}-9}{x-9}\)
\(B=\frac{x+2\sqrt{x}-3-7\sqrt{x}+9}{x-9}\)
\(B=\frac{x-5\sqrt{x}+6}{x-9}\)
\(B=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
b) c) ?
24 tháng 7 2019
b mình làm đc rồi, nó ko liên quan gì đến a và c đâu
5 tháng 8 2019
a) \(A=\frac{-\sqrt{x}+2+4}{\sqrt{x}-2}=-1+\frac{4}{\sqrt{x}-2}\)
Để \(A\in Z\Leftrightarrow\sqrt{x}-2\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-2;0;1;3;4;6\right\}\)
Mà \(x\in Z;\sqrt{x}\ge0\Rightarrow x\in\left\{0;1;9;16;36\right\}\)
b)\(A=\frac{4\sqrt{x}-2+3}{2\sqrt{x}-1}=2+\frac{3}{2\sqrt{x}-1}\)
Để \(A\in Z\Leftrightarrow2\sqrt{x}-1\in\left\{-3;-1;1;3\right\}\)
\(\Leftrightarrow2\sqrt{x}\in\left\{-2;0;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-1;0;1;2\right\}\Leftrightarrow x\in\left\{0;1;4\right\}\)
19 tháng 7 2018
a)\(3-\sqrt{3}+\sqrt{15}-3\sqrt{5}=\sqrt{3}\left(\sqrt{3}-1\right)-\sqrt{15}\left(\sqrt{3}-1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{15}\right)=\sqrt{3}\left(\sqrt{3}-1\right)\left(1-\sqrt{5}\right)\)\(\)b)\(\sqrt{1-a}+\sqrt{1-a^2}=\sqrt{1-a}.1+\sqrt{1-a}.\sqrt{1+a}=\sqrt{1-a}\left(\sqrt{1+a}+1\right)\)
19 tháng 7 2018
c)\(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b+\sqrt{ab}\right)=\left(\sqrt{a}-\sqrt{b}\right)\left(a+2\sqrt{ab}+b\right)=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)^2\)
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Cô Hoàng Huyền
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12 tháng 7 2017
a) \(P=\frac{\left(x\sqrt{x}-3\right)-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-\left(x+4\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{x+8}{\sqrt{x}+1}\)
b) Ta có \(x=14-6\sqrt{5}=9-2.3.\sqrt{5}+5=\left(3-\sqrt{5}\right)^2\)
Vậy nên \(\sqrt{x}=3-\sqrt{5}\)
Suy ra \(P=\frac{\left(3-\sqrt{5}\right)^2+8}{3-\sqrt{5}+1}=\frac{58-2\sqrt{5}}{11}\)
c) \(P=\frac{x+8}{\sqrt{x}+1}=\frac{\left(x-1\right)+9}{\sqrt{x}+1}=\left(\sqrt{x}-1\right)+\frac{9}{\sqrt{x}+1}\)
\(=\left(\sqrt{x}+1\right)+\frac{9}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{9}{\sqrt{x}+1}}-2=4\)
minP = 4 khi \(\sqrt{x}+1=\frac{9}{\sqrt{x}+1}\Rightarrow\sqrt{x}+1=3\Rightarrow x=4.\)
\(a,\sqrt{x}>1\Leftrightarrow\sqrt{x}>\sqrt{1}\Leftrightarrow x>1.\)
\(b,\sqrt{x}< 3\Rightarrow\sqrt{x}< \sqrt{9}\Rightarrow x< 9\)
\(c,\sqrt{x}=14\Rightarrow\sqrt{x}=\sqrt{196}\Leftrightarrow x=196\)