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Mình giải câu a thấy số xấu và câu b không thỏa dạng nên mình sửa đề lại nha. Hi vọng đúng với đề gốc của bạn.
a. Thay x=25 vào A ta được: A=\(\frac{7}{\sqrt{25}+8}=\frac{7}{13}\)
b. B=\(\frac{\sqrt{x}}{\sqrt{x-3}}+\frac{2\sqrt{x}-24}{x-9}\)
\(\Leftrightarrow B=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\frac{2\sqrt{x}-24}{x-9}\)
\(\Leftrightarrow B=\frac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow B=\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)(\sqrt{x}+3)}\)
\(\Leftrightarrow B=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow B=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)
c. ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne-8\\x\ne-3\end{matrix}\right.\)
P=A.B=\(\frac{7}{\sqrt{x}+8}.\frac{\sqrt{x}+8}{\sqrt{x}+3}=\frac{7}{\sqrt{x}+3}\)
Để P nguyên thì \(\sqrt{x}+3\:\in\) Ư(7) \(\Leftrightarrow\sqrt{x}+3\in\){\(\pm1;\pm7\)}
\(\sqrt{x}+3\: =1\) \(\Leftrightarrow\)\(\sqrt{x}=-2\:\left(KTM\right)\)
\(\sqrt{x}+3=-1\text{}\Leftrightarrow\sqrt{x}=-4\) (KTM)
\(\sqrt{x}+3=7\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\) (TM)
\(\sqrt{x}+3=-7\Leftrightarrow\sqrt{x}=-10\:\left(KTM\right)\)
Vậy...
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
a) Thay x=4 zô là đc . ra kết quả \(\frac{7}{6}\)là dúng
b) \(B=\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\)
\(=\frac{3x+3\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\)
\(=>P=A.B=\frac{3\sqrt{x}+1}{x+\sqrt{x}}.\frac{3\left(x+\sqrt{x}\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}=\frac{3}{3\sqrt{x}-1}\)
c) xét \(\frac{1}{P}=\frac{3\sqrt{x}-1}{3}\)
do \(\sqrt{x}\ge0=>3\sqrt{x}-1\ge-1\)\(=>\frac{3\sqrt{x}-1}{3}\ge-\frac{1}{3}\)
\(=>\frac{1}{P}\ge-\frac{1}{3}\)
dấu = xảy ra khi x=0
zậy ..
mk làm luôn.
a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
=\(\frac{3x+3\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(\frac{3.\left(x+\sqrt{x}\right).\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
mk làm phần rút gọn xong mk bận nên bn tự làm câu b nha ^^
Khi x=25
=> A=\(\frac{7}{\sqrt{25+8}}=\frac{7}{\sqrt{\text{3}\text{3}}}\)=\(\frac{7\sqrt{33}}{33}\)
b) B= \(\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}+\frac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
B= \(\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\)
B= \(\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)