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Lời giải:
a) \(x=\frac{23(5-\sqrt{2})}{5+\sqrt{2}}=\frac{23(5-\sqrt{2})^2}{(5+\sqrt{2})(5-\sqrt{2})}=\frac{23(5-\sqrt{2})^2}{5^2-2}=(5-\sqrt{2})^2\)
\(\Rightarrow x=5-\sqrt{2}\)
Do đó: \(B=\frac{5-\sqrt{2}+2}{5-\sqrt{2}-5}=\frac{7-\sqrt{2}}{-\sqrt{2}}=\frac{\sqrt{2}-7}{\sqrt{2}}\)
b)
\(A=\frac{x+3\sqrt{x}}{x-25}+\frac{1}{\sqrt{x}+5}=\frac{x+3\sqrt{x}}{(\sqrt{x}-5)(\sqrt{x}+5)}+\frac{\sqrt{x}-5}{(\sqrt{x}-5)(\sqrt{x}+5)}\)
\(=\frac{x+4\sqrt{x}-5}{(\sqrt{x}-5)(\sqrt{x}+5)}=\frac{(\sqrt{x}-1)(\sqrt{x}+5)}{(\sqrt{x}-5)(\sqrt{x}+5)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-5}\)
Ta có: \(\frac{A}{B}=\frac{\sqrt{x}-1}{\sqrt{x}-5}:\frac{\sqrt{x}+2}{\sqrt{x}-5}=\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{4}{7}\)
\(\Rightarrow 7(\sqrt{x}-1)=4(\sqrt{x}+2)\)
\(\Rightarrow \sqrt{x}=5\Rightarrow x=25\)
c)
\(\frac{A}{B}=\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\)
Vì \(\sqrt{x}\geq 0\Rightarrow \sqrt{x}+2\geq 2\Rightarrow \frac{3}{\sqrt{x}+2}\leq \frac{3}{2}\)
\(\Rightarrow \frac{A}{B}=1-\frac{3}{\sqrt{x}+2}\geq 1-\frac{3}{2}=\frac{-1}{2}\)
Vậy \(P_{\min}=\frac{-1}{2}\Leftrightarrow x=0\)
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
a: \(M=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Khi a=9/25 thì \(M=\dfrac{\dfrac{3}{5}-4}{\dfrac{3}{5}-2}=\dfrac{-17}{5}:\dfrac{-7}{5}=\dfrac{17}{7}\)
c: Để |M|=1/6 thì M=1/6 hoặc M=-1/6
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=\dfrac{1}{6}\\\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6\sqrt{a}-24=\sqrt{a}-2\\6\sqrt{a}-24=-\sqrt{a}+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5\sqrt{a}=22\\7\sqrt{a}=26\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=\left(\dfrac{22}{5}\right)^2\\a=\left(\dfrac{26}{7}\right)^2\end{matrix}\right.\)
a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
a. \(A=\dfrac{x+2\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}ĐK\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x+2}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b. \(x=25\Leftrightarrow\sqrt{x}=5\Leftrightarrow A=\dfrac{5}{5-2}=\dfrac{5}{3}\)
c.\(A=-\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{1}{3}\Leftrightarrow3\sqrt{x}=\sqrt{x}-2\Leftrightarrow4\sqrt{x}-2=0\Leftrightarrow2\left(\sqrt{x}-1\right)=0\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(TM\right)\)Chúc bạn học tốt!
a) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}.\dfrac{196}{9}}=\sqrt{\dfrac{25}{81}}.\sqrt{\dfrac{16}{49}}.\sqrt{\dfrac{196}{9}}=\dfrac{5}{9}.\dfrac{4}{7}.\dfrac{14}{3}=\dfrac{40}{27}\)
b) \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{25}.2\dfrac{34}{81}}=\sqrt{\dfrac{49}{16}.\dfrac{64}{25}.\dfrac{196}{81}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{64}{25}}.\sqrt{\dfrac{196}{81}}=\dfrac{7}{4}.\dfrac{8}{5}.\dfrac{14}{9}=\dfrac{196}{45}\)
c) \(\dfrac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.49}{81}}=\dfrac{\sqrt{64}.\sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=\dfrac{56}{9}\)
d) \(\sqrt{21,6}.\sqrt{810}.\sqrt{11^2-5^2}=\sqrt{21,6.810.\left(11^2-5^2\right)}=\sqrt{216.81.\left(11+5\right)\left(11-5\right)}=\sqrt{36^2.9^2.4^2}=36.9.4=1296\)
a) \(\sqrt{\left(2x-1\right)^2}=3\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\dfrac{1}{3}\sqrt{15x}\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-\dfrac{1}{3}\sqrt{15x}=2\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{15x}=2\)
\(\Leftrightarrow\sqrt{15x}=6\)
\(\Leftrightarrow15x=6^2\Leftrightarrow15x=36\)
\(\Rightarrow x=\dfrac{5}{12}\)
a/ khi x = 9 thì A = \(\dfrac{\sqrt{9}+2}{\sqrt{9}-5}=\dfrac{5}{-2}=-\dfrac{5}{2}\)
b/ B = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{1}{\sqrt{x}-5}\left(đpcm\right)\)
c/ \(A=B\cdot\left|x-4\right|\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-5}=\dfrac{1}{\sqrt{x}-5}\cdot\left|x-4\right|\)
\(\Leftrightarrow\left|x-4\right|=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\sqrt{x}+2\)
Vì: \(\sqrt{x}+2>0\)=> đk: x > 4
\(\left|x-4\right|=\sqrt{x}+2\)
\(\Leftrightarrow x-4=\sqrt{x}+2\)
\(\Leftrightarrow x-\sqrt{x}-6=0\)
\(\Leftrightarrow\left(x-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{25}{4}=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-\dfrac{1}{2}=\dfrac{5}{2}\\\sqrt{x}-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-2\left(loai\right)\end{matrix}\right.\)
\(\sqrt{x}=3\Leftrightarrow x=9\left(TM\right)\)
Vậy x = 9 thì A = B.|x - 4|
bài 2:
a: \(\dfrac{25}{5-2\sqrt{3}}=\dfrac{125+10\sqrt{3}}{13}\)
b: \(\dfrac{8}{\sqrt{5}+2}=8\sqrt{5}-32\)
c: \(\dfrac{6}{2\sqrt{3}-\sqrt{7}}=\dfrac{12\sqrt{3}+6\sqrt{7}}{5}\)
d: \(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}=\dfrac{\sqrt{6}}{2}\)
Với `x ne 25` có:
`P=5/7<=>[-15x+25]/[x-25]=5/7`
`<=>7(-15x+25)=5(x-25)`
`<=>-105x+175=5x-125`
`<=>110x=300`
`<=>x=30/11` (t/m)