\(^{B=\frac{\sqrt{x}}{\sqrt{x}-5}+\frac{10\sqrt{x}}{25-x}-\frac{5}{\sqrt{x}+5}}\) \(\left(x\ge0;x\ne25\right)\)
a) Rút gọn B
b) tính GTNN của biểu thức B
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NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 9 2019
ĐKXĐ: ...
\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x+\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+5}-\frac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\frac{25-x+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)
\(=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}+5\right)}{-\left(\sqrt{x}+3\right)}=\frac{5}{\sqrt{x}+3}\)
b/ \(B=\frac{x+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)
\(\Rightarrow B\ge2\sqrt{\frac{\left(\sqrt{x}+3\right).25}{\sqrt{x}+3}}-6=4\)
\(B_{min}=4\) khi \(\left(\sqrt{x}+3\right)^2=25\Rightarrow x=4\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2019
Câu 1:
\(\frac{A}{B}\ge\frac{x}{4}+5\Leftrightarrow\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\ge\frac{x}{4}+5\)
\(\Rightarrow\sqrt{x}+4\ge\frac{x}{4}+5\Rightarrow x-4\sqrt{x}+4\le0\)
\(\Rightarrow\left(\sqrt{x}-2\right)^2\le0\Rightarrow\sqrt{x}-2=0\Rightarrow x=4\)
Câu 2:
Bạn coi lại đề, biểu thức B không hợp lý
7 tháng 7 2017
a. P=\(\frac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}:\frac{25-x-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}{\cdot\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}:\frac{-x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-5}{\sqrt{x}+5}.\frac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{5}{\sqrt{x}+3}\)
b. P=\(\frac{5}{\sqrt{x}+3}\)
P nguyên \(\Leftrightarrow\sqrt{x}+3\inƯ\left(5\right)\Rightarrow\sqrt{x}+3\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{2\right\}\)\(\Rightarrow x=4\)
Vậy x=4 thì P nguyên
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 6 2019
ĐKXĐ:...
\(\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}{\left(9-x\right)}\)
\(=\frac{5\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{5}{\sqrt{x}+3}\)
16 tháng 8 2021
b4 :
\(a,x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(b,x-5=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)
\(c,x+2\sqrt{xy}+y=\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(d,x-4\sqrt{x}\sqrt{y}+4y=\left(\sqrt{x}-2\sqrt{y}\right)^2\)
b5:
\(a,ĐK:x\ge1\)
\(\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}-\frac{4}{5}\sqrt{25\left(x-1\right)}=1\)
\(\Leftrightarrow3\sqrt{x-1}+2\sqrt{x-1}-4\sqrt{x-1}=1\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
\(b,ĐK:x\ge5\)
\(\frac{1}{3}\sqrt{9\left(x-5\right)}+\frac{1}{2}\sqrt{4\left(x-5\right)}-\frac{7}{5}\sqrt{25\left(x-5\right)}=2\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{x-5}-7\sqrt{x-5}=2\)
\(\Leftrightarrow-5\sqrt{x-5}=2\)
\(\Leftrightarrow\sqrt{x-5}=-\frac{2}{5}\left(voli\right)\)
\(c,ĐK:x>0\)
\(\sqrt{x}+\frac{9}{\sqrt{x}}=6\)
\(\Leftrightarrow x+9=6\sqrt{x}\)
\(\Leftrightarrow x-6\sqrt{x}+9=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)^2=0\)
\(\Leftrightarrow x=9\left(tm\right)\)
a)B=√x(√x+5)/x-25 -10√x/x-25 -5(√x-5)/x-25=(x+5√x-10√x-5√x+25)/x-25=(x-10√x+25)/x-25=(√x-5)^2/(√x+5)(√x-5)=√x-5/√x+5
a.
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\dfrac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\dfrac{5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
b.
\(B=\dfrac{\sqrt{x}+5-10}{\sqrt{x}+5}=1-\dfrac{10}{\sqrt{x}+5}\ge1-\dfrac{10}{0+5}=-1\)
\(B_{min}=-1\) khi \(x=0\)