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ĐK: \(x\ge-7\)
PT \(\Leftrightarrow\left(\sqrt[3]{x-8}-\left(x-8\right)\right)+\left[\sqrt{x+7}-4\right]+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\frac{-\left(x-9\right)\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}+\frac{x-9}{\sqrt{x+7}+4}+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left[x^2+x+2+\frac{1}{\sqrt{x+7}+4}-\frac{\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}\right]=0\)
\(\Leftrightarrow x=9\)
P/s:em chả biết đánh giá cái ngoặc to thế nào nữa:((((
a) Đkxđ : \(\left\{{}\begin{matrix}a\ge0\\a\ne9\end{matrix}\right.\)
A = \(\left(\frac{\sqrt{a}+3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\frac{\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right)\left(1-\frac{3}{\sqrt{a}}\right)\)
= \(\frac{2\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\frac{\sqrt{a}-3}{\sqrt{a}}\)
= \(\frac{2}{\sqrt{a}+3}\)
b) Để A > \(\frac{1}{2}\)
<=> \(\frac{2}{\sqrt{a}+3}>\frac{1}{2}\Leftrightarrow\frac{2}{\sqrt{a}+3}-\frac{1}{2}>0\)
<=> \(4-\sqrt{a}-3>0\Leftrightarrow1-\sqrt{a}>0\Leftrightarrow a< 1\)
Vậy để A >1/2 thì a <1
Mới đc câu a ak, thog cảm nha, trih độ mih thấp lắm:
\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}\)
=\(\frac{a+\sqrt{ab}-\sqrt{ab}+b}{a-b}-\frac{2b}{a-b}\)
=\(\frac{a+b-2b}{a-b}=\frac{a-b}{a-b}=1\)
2.
a)
\(\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(2-\frac{2\sqrt{a}-a}{\sqrt{a}-2}\right)\\ =\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(2+\frac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\right)\\ =\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)\\ =2^2-\left(\sqrt{a}\right)^2\\ =4-a\)
b)
\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}}\right):\frac{\sqrt{x}+1}{x}\\ =\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\frac{x}{\sqrt{x}+1}\\ =\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\cdot\frac{x}{\sqrt{x}+1}\\ =\frac{x-1}{\sqrt{x}}\cdot\frac{x}{\sqrt{x}+1}\\ =\sqrt{x}\left(\sqrt{x}-1\right)\\ =x-\sqrt{x}\)
c)
\(\left(\frac{1-x\sqrt{x}}{1-x}+\sqrt{x}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\\ =\left(\frac{1-\sqrt{x^3}}{1-x}+\sqrt{x}\right)\cdot\frac{\left(1-\sqrt{x}\right)^2}{\left(1-x\right)^2}\\ =\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\sqrt{x}\right)\cdot\frac{\left(1-\sqrt{x}\right)^2}{\left[\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\right]^2}\\ =\left(\frac{1+\sqrt{x}+x+\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\right)\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\\ =\frac{2x+2\sqrt{x}+1}{1+\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{2x+2\sqrt{x}+1}{\left(1+\sqrt{x}\right)^3}\)
1. (Ko viết lại đề nha :v)
a)
\(A=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\\ =\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\\ =\left(\frac{x+2\sqrt{x}-\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\\ =\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-1\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2}{x-1}\)
b) Để A đạt giá trị nguyên thì \(2⋮x-1\Leftrightarrow x-1\inƯ\left(2\right)\)
\(\Leftrightarrow x-1\in\left\{-1;1;-2;2\right\}\\ \Leftrightarrow x\in\left\{0;2;-1;3\right\}\)
Vậy......