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Bài 1:
Thay x=9 vào biểu thức \(A=\frac{2\sqrt{x}+1}{\sqrt{x}+2}\), ta được:
\(\frac{2\cdot\sqrt{9}+1}{\sqrt{9}+2}=\frac{2\cdot3+1}{3+2}=\frac{7}{5}\)
Vậy: \(\frac{7}{5}\) là giá trị của biểu thức \(A=\frac{2\sqrt{x}+1}{\sqrt{x}+2}\) tại x=9
Bài 2:
a) Ta có: \(B=\left(\frac{x+14\sqrt{x}-5}{x-25}+\frac{\sqrt{x}}{\sqrt{x}+5}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(=\left(\frac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right)\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(=\frac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(=\frac{2x+10\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\cdot\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}-1}{\sqrt{x}+2}\)
Sửa đề :
a) \(A=\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{4}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-5}{x-\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{x-\sqrt{x}+4\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-4-x+\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}\)
b) \(A=4\)
\(\Leftrightarrow\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}=4\)
\(\Leftrightarrow x+3\sqrt{x}+4=4\sqrt{x}+4\)
\(\Leftrightarrow x-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy \(A=4\Leftrightarrow x\in\left\{0;1\right\}\)
a) x = 16 (tm) => A = \(\frac{\sqrt{16}-2}{\sqrt{16}+1}=\frac{4-2}{4+1}=\frac{2}{5}\)
b) B = \(\left(\frac{1}{\sqrt{x}+5}-\frac{x+2\sqrt{x}-5}{25-x}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)
B = \(\frac{\sqrt{x}-5+x+2\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)
B = \(\frac{x+3\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
B = \(\frac{x+5\sqrt{x}-2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
B = \(\frac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
c) P = \(\frac{B}{A}=\frac{\sqrt{x}-2}{\sqrt{x}+2}:\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
=> \(P\left(\sqrt{x}+2\right)\ge x+6\sqrt{x}-13\)
<=> \(\frac{\sqrt{x}+1}{\sqrt{x}+2}.\left(\sqrt{x}+2\right)-x-6\sqrt{x}+13\ge0\)
<=> \(-x-6\sqrt{x}+13+\sqrt{x}+1\ge0\)
<=> \(-x-5\sqrt{x}+14\ge0\)
<=> \(x+5\sqrt{x}-14\le0\)
<=> \(x+7\sqrt{x}-2\sqrt{x}-14\le0\)
<=> \(\left(\sqrt{x}+7\right)\left(\sqrt{x}-2\right)\le0\)
Do \(\sqrt{x}+7>0\) với mọi x => \(\sqrt{x}-2\le0\)
<=> \(\sqrt{x}\le2\) <=> \(x\le4\)
Kết hợp với Đk: x \(\ge\)0; x \(\ne\)4; x \(\ne\)25
và x thuộc Z => x = {0; 1; 2; 3}
d) M = \(3P\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\) <=>M = \(3\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\)
M = \(\frac{3\sqrt{x}+3}{x+\sqrt{x}+4}=\frac{x+\sqrt{x}+4-x+2\sqrt{x}-1}{\left(x+\sqrt{x}+\frac{1}{4}\right)+\frac{15}{4}}=1-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}}\le1\)(Do \(\left(\sqrt{x}-1\right)^2\ge0\) và \(\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}>0\))
Dấu "=" xảy ra <=> \(\sqrt{x}-1=0\) <=> \(x=1\)
Vậy MaxM = 1 khi x = 1
\(B=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(B=\frac{x-\sqrt{x}+3\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)
b/ \(C=\left(\frac{\sqrt{x}-1}{\sqrt{x}-5}.\frac{\sqrt{x}+6}{\sqrt{x}-1}\right).\frac{\sqrt{x}-5}{\sqrt{x}}\)
\(C=\frac{\sqrt{x}+6}{\sqrt{x}-5}.\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}+6}{\sqrt{x}}=1+\frac{6}{\sqrt{x}}\)
Cai này thì so sánh \(\frac{6}{\sqrt{x}}\) vs 2
Nếu0< x<9\(\Rightarrow\frac{6}{\sqrt{x}}< 2\)
Nếu x=9\(\Rightarrow\frac{6}{\sqrt{x}}=2\)
Nếu x>9\(\Rightarrow\frac{6}{\sqrt{x}}>2\)
bài tập nâng cao thì 3=1+2
Mà vế kia cx có 1 thì so sánh 2 cái còn lại chứ!
a) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b) \(\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}+1}\cdot\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}+1}\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
a)\(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x+1}\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b)\(\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}\cdot\frac{\sqrt{\left(\sqrt{y}-1\right)^{2^2}}}{\sqrt{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
