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a. ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(B=\frac{2x+2}{\sqrt{x}}+\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{2x+2}{\sqrt{x}}+\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}=\frac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
b. Ta có \(B-5=\frac{2x+2\sqrt{x}+2}{\sqrt{x}}-5=\frac{2x-3\sqrt{x}+2}{\sqrt{x}}=\frac{2\left(x-2.\sqrt{x}.\frac{3}{4}+\frac{9}{16}\right)-\frac{9}{8}+2}{\sqrt{x}}\)
\(=\frac{2\left(\sqrt{x}-\frac{3}{4}\right)^2+\frac{7}{8}}{\sqrt{x}}\)
Ta thấy \(\hept{\begin{cases}2\left(\sqrt{x}-\frac{3}{4}\right)^2+\frac{7}{8}>0\\\sqrt{x}>0\forall x>0\end{cases}\Rightarrow B-5>0\Rightarrow B>5}\)
Vậy \(B>5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(P=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b: \(P=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi x=1/4
![](https://rs.olm.vn/images/avt/0.png?1311)
mk làm luôn
a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)
=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
=\(\frac{\left(3x+3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}\)
=\(\frac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
=
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(A=\frac{\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)-\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}+1-3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-4\sqrt{x}+1-3\sqrt{x}-1}{\left(3\sqrt{x}\right)^2-1}-\frac{8\sqrt{x}}{9x-1}}{1-1-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-7\sqrt{x}}{9x-1}-\frac{8\sqrt{x}}{9x-1}}{-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{3x-7\sqrt{x}-8\sqrt{x}}{9x-1}\left(\frac{-\left(3\sqrt{x}+1\right)}{3}\right)\)
\(A=\frac{3x-15\sqrt{x}}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{3\left(x-3\sqrt{x}\right)}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{\left(x-3\sqrt{x}\right)\left(-3\sqrt{x}-1\right)}{9x-1}\)
\(A=\frac{3x\sqrt{x}+8x+3\sqrt{x}}{9x-1}\)
\(A=\frac{3x\sqrt{x}}{9x-1}+\frac{8x}{9x-1}+\frac{3\sqrt{x}}{9x-1}\)
\(A=\frac{x\sqrt{x}}{x-\frac{1}{3}}+\frac{8x}{9x-1}+\frac{\sqrt{x}}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)}{x-\frac{1}{3}}+\frac{\frac{8}{3}x}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)+\frac{8}{3}x}{x-\frac{1}{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cái này là toán lớp 9 chứ.
a)
ĐKXĐ : \(x\ne\pm4\)
\(A=\left(\frac{x-\sqrt{x}+7}{x-4}+\frac{\sqrt{x}+2}{x-4}\right):\left(\frac{\left(\sqrt{x}+2\right)^2}{x-4}-\frac{\left(\sqrt{x}-2\right)^2}{x-4}-\frac{2\sqrt{x}}{x-4}\right)\)
\(=\left(\frac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}\right):\left(\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{x-4}\right)\)
\(=\frac{x+9}{x-4}\cdot\frac{x-4}{6\sqrt{x}}=\frac{x+9}{6\sqrt{x}}\)
b)
Ta có
\(x+9-6\sqrt{x}=\left(\sqrt{x}-3\right)^2\ge0\)
\(\Rightarrow x+9\ge6\sqrt{x}\)
\(\Rightarrow\frac{x+9}{6\sqrt{x}}\ge1\)
\(\Leftrightarrow A\ge1\)
\(\Leftrightarrow\frac{1}{A}\le1\)
\(\Rightarrow A\ge\frac{1}{A}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1
a, Với \(x=9\)thì \(A=\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{3}{\sqrt{x}}+1=\frac{3}{3}+1=2\)
b, Để \(A=\frac{5}{2}\)thì \(\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{3}{\sqrt{x}}+1=\frac{5}{2}< =>\frac{3}{\sqrt{x}}=\frac{3}{2}< =>x=4\)
Bài 2
a, \(B=\frac{\sqrt{x}-2}{\sqrt{x}}+\frac{4\sqrt{x}+2}{x+\sqrt{x}}\left(đk:x>0\right)\)
\(=1-\frac{2}{\sqrt{x}}+\frac{4\sqrt{x}+2}{x+\sqrt{x}}=\frac{x+5\sqrt{x}+2}{x+\sqrt{x}}-\frac{2}{\sqrt{x}}\)
\(=\frac{x\sqrt{x}+5x+2\sqrt{x}-2x-2\sqrt{x}}{x\sqrt{x}+x}=\frac{x\sqrt{x}+3x}{x\sqrt{x}+x}\)
\(=1+\frac{2x}{x\left(\sqrt{x}+1\right)}=1+\frac{2}{\sqrt{x}+1}=\frac{\sqrt{x}+3}{\sqrt{x}+1}\)
\(A=\frac{3+\sqrt{x}}{\sqrt{x}}\)Thay x = 9 ta có :
\(VT=\frac{3+\sqrt{9}}{\sqrt{9}}=\frac{3+3}{3}=2\)
Bài ra ta có : \(A=\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{5}{2}\)
\(\Leftrightarrow\frac{3}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}}=\frac{5}{2}\Leftrightarrow\frac{3}{\sqrt{x}}+1=\frac{5}{2}\)
\(\Leftrightarrow\frac{3}{\sqrt{x}}=\frac{3}{2}\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ \(\sqrt{\frac{m}{1-2x+x^2}}\cdot\sqrt{\frac{4m-8mx+4mx^2}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}}\cdot\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}}\cdot\sqrt{\frac{4m\left(1-x\right)^2}{81}}\)
\(=\sqrt{\frac{m}{\left(1-x\right)^2}\cdot\frac{4m\left(1-x\right)^2}{81}}\)
\(=\sqrt{\frac{4m^2}{81}}=\sqrt{\frac{\left(2m\right)^2}{9^2}}=\frac{2\left|m\right|}{9}\)
3/\(\frac{a+b}{b^2}\sqrt{\frac{a^2b^4}{a^2+2ab+b^2}}\)
\(=\frac{a+b}{b^2}\sqrt{\frac{\left(ab^2\right)^2}{\left(a+b\right)^2}}\)
\(=\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{\left|a+b\right|}\)
TH1: \(\Rightarrow\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{-\left(a+b\right)}=-\left|a\right|\)
TH2: \(\Rightarrow\frac{a+b}{b^2}\cdot\frac{\left|a\right|b^2}{a+b}=\left|a\right|\)
2/\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\frac{\sqrt{a}-a}{1-\sqrt{a}}\right)\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1}\cdot\frac{1-\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{\left(1-a\sqrt{a}+\sqrt{a}-a\right)\cdot\left(1-\sqrt{a}\right)}{\left(1-a\right)^2}\)
\(=\frac{1-a\sqrt{a}+\sqrt{a}-a-\sqrt{a}+a^2-a+a\sqrt{a}}{\left(1-a\right)^2}\)
\(=\frac{a^2-2a+1}{\left(1-a\right)^2}\)
\(=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}=\frac{-\left(1-a\right)^2}{\left(1-a\right)^2}=-1\)