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học lớp 9 chưa mà đòi đăng ? :))
a) Ta có : \(A=\frac{x+5\sqrt{x}}{x-25}=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}}{\sqrt{x}-5}\)
Để A nhận giá trị = 0 thì \(\sqrt{x}=0\)<=> x = 0 ( tmđk )
Vậy với x = 0 thì A = 0
b) \(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}\)
c) P = B : A = \(\frac{\frac{\sqrt{x}}{\sqrt{x}+3}}{\frac{\sqrt{x}}{\sqrt{x}-5}}=\frac{\sqrt{x}}{\sqrt{x}+3}\div\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{\sqrt{x}}{\sqrt{x}+3}\times\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
Xét hiệu P - 1 ta có :
\(\frac{\sqrt{x}-5}{\sqrt{x}+3}-1=\frac{\sqrt{x}-5}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}=\frac{-8}{\sqrt{x}+3}\)
Vì \(\hept{\begin{cases}-8< 0\\\sqrt{x}+3>0\end{cases}}\Rightarrow\frac{-8}{\sqrt{x}+3}< 0\)hay P - 1 < 0
=> P < 1
a) \(A=0\Rightarrow\frac{x+5\sqrt{x}}{x-25}=0\Rightarrow x+5\sqrt{x}=0\Leftrightarrow x=0\)(thỏa mãn).
b) \(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\)
\(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+3}\)
c) \(P=B\div A=\frac{\sqrt{x}}{\sqrt{x}+3}\div\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}=1-\frac{8}{\sqrt{x}+3}< 1\)
\(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}-\frac{5\sqrt{x}-2}{x-4}\)
\(Q=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(Q=\frac{x-3\sqrt{x}-2-5\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(Q=\frac{x-8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\)
ủa sao không thấy gọn ta
\(Q=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
b.\(Q< 1\)
\(\Leftrightarrow x-\sqrt{x}-2< x-5\sqrt{x}+6\)
\(\Leftrightarrow4\sqrt{x}-8< 0\)
\(\Leftrightarrow0\le x< 4\)
Vay de Q<1 thi \(0\le0< 4\)
\(A=\frac{1}{\sqrt{5}-2}-\sqrt{9-4\sqrt{5}}=\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\sqrt{5-4\sqrt{5}+4}\)
\(A=\frac{\sqrt{5}+2}{5-4}-\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}+2-\sqrt{5}+2=4\)
\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{x-\sqrt{x}}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b) Với x >0 và x khác 1 (1)
Ta có: \(\frac{1}{6}A>B\) <=> \(\frac{\sqrt{x}-1}{\sqrt{x}}< 4\cdot\frac{1}{6}\)
<=> \(\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{2}{3}\) <=> \(3\sqrt{x}-3< 2\sqrt{x}\) <=> \(\sqrt{x}< 3\) <=> x < 9 (2)
Từ (1) và (2) => 0 < x < 9 và x khác 1
Khi \(x=1,44\): \(A=\frac{1,44+7}{\sqrt{1,44}}=\frac{8,44}{1,2}=\frac{211}{30}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-1}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\)(ĐK: \(x\ge0,x\ne9\))
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2x-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(S=\frac{1}{B}+A=\frac{\sqrt{x}-3}{\sqrt{x}}+\frac{x+7}{\sqrt{x}}=\frac{x+\sqrt{x}+4}{\sqrt{x}}=\sqrt{x}+\frac{4}{\sqrt{x}}+1\)
\(\ge2\sqrt{\sqrt{x}.\frac{4}{\sqrt{x}}}+1=5\)
Dấu \(=\)khi \(\sqrt{x}=\frac{4}{\sqrt{x}}\Leftrightarrow x=4\)(thỏa mãn)
các bạn giúp mk làm câu c thôi ạ.
c) ta rút gọn được B \(=\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{\sqrt{x}-5+5}{\sqrt{x}-5}=1+\frac{5}{\sqrt{x}-5}\)
để B nhỏ nhất thì \(\sqrt{x}-5\) lớn nhất và \(\left(\sqrt{x}-5\right)\in U\left(5\right)=\left\{1;5\right\}\)
suy ra \(\sqrt{x}-5=5\Leftrightarrow x=100\left(tm\right)\)
vậy min B=2 \(\Leftrightarrow x=100\)