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22 tháng 7 2020
Bài 1:
a) Ta có: \(P=\frac{x}{x+2}+\frac{x+3}{x-2}+\frac{6-9x}{4-x^2}\)
\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x+3\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{6-9x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2-2x+x^2+5x+6-6+9x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x^2+12x}{\left(x-2\right)\left(x+2\right)}\)
b) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Để P=3 thì \(\frac{2x^2+12x}{\left(x-2\right)\left(x+2\right)}=3\)
\(\Leftrightarrow2x^2+12x=3\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow2x^2+12x=3\left(x^2-4\right)\)
\(\Leftrightarrow2x^2+12x=3x^2-12\)
\(\Leftrightarrow2x^2+12x-3x^2+12=0\)
\(\Leftrightarrow-x^2+12x+12=0\)
\(\Leftrightarrow x^2-12x-12=0\)
\(\Leftrightarrow x^2-12x+36-24=0\)
\(\Leftrightarrow\left(x-6\right)^2=24\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=\sqrt{24}\\x-6=-\sqrt{24}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6+2\sqrt{6}\left(nhận\right)\\x=6-2\sqrt{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: khi P=3 thì \(x\in\left\{6+2\sqrt{6};6-2\sqrt{6}\right\}\)
Bài 2:
a) Ta có: \(B=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}\)
\(=\frac{2a^2}{\left(a+1\right)\left(a-1\right)}+\frac{a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}-\frac{a\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}\)
\(=\frac{2a^2+a^2-a-a^2-a}{\left(a+1\right)\cdot\left(a-1\right)}=\frac{2a^2-2a}{\left(a+1\right)\left(a-1\right)}\)
\(=\frac{2a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}=\frac{2a}{a+1}\)
b) ĐKXĐ: \(a\notin\left\{1;-1\right\}\)
Để B là số nguyên thì \(2a⋮a+1\)
\(\Leftrightarrow2a+2-2⋮a+1\)
\(\Leftrightarrow-2⋮a+1\)
\(\Leftrightarrow a+1\inƯ\left(-2\right)\)
\(\Leftrightarrow a+1\in\left\{1;-1;2;-2\right\}\)
hay \(a\in\left\{0;-2;1;-3\right\}\)
mà \(a\notin\left\{1;-1\right\}\)
nên \(a\in\left\{0;-2;-3\right\}\)
Vậy: khi B có giá trị nguyên thì \(a\in\left\{0;-2;-3\right\}\)
Bài 3:
Ta có: \(Q=\frac{4}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)
\(=\frac{4\left(x-2\right)+2\left(x+2\right)+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8+2x+4+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x-2}\)
Bài 4:
a) Ta có: \(P=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right)\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}+3\right)\)
\(=\left(\frac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}+\frac{3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\right)\)
\(=\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}+2+3\sqrt{x}-6}{\sqrt{x}-2}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\cdot4\cdot\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}-2\right)^2}\)
\(=\frac{-16x+16\sqrt{x}}{\left(\sqrt{x}-2\right)^2}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Để P=-4 thì \(\frac{-16x+16\sqrt{x}}{\left(\sqrt{x}-2\right)^2}=-4\)
\(\Leftrightarrow-16x+16\sqrt{x}=-4\left(\sqrt{x}-2\right)^2\)
\(\Leftrightarrow-16x+16\sqrt{x}=-4\left(x-4\sqrt{x}+4\right)\)
\(\Leftrightarrow-16x+16\sqrt{x}=-4x+16\sqrt{x}-16\)
\(\Leftrightarrow-16x+16\sqrt{x}+4x-16\sqrt{x}+16=0\)
\(\Leftrightarrow-12x+16=0\)
\(\Leftrightarrow-12x=-16\)
hay \(x=\frac{4}{3}\)(nhận)
Vậy: Khi P=-4 thì \(x=\frac{4}{3}\)
25 tháng 7 2019
\(\text{Câu 1: Sửa đề}\)
\( a)M = \left( {1 - \dfrac{{4\sqrt x }}{{x - 1}} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{x - 2\sqrt x }}{{x - 1}}\\ M = \left[ {1 - \dfrac{{4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \left[ {1 + \dfrac{{ - 4\sqrt x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \left[ {1 + \dfrac{{ - 3\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) - 3\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \sqrt x \left( {\sqrt x - 3} \right).\dfrac{1}{{x - 2\sqrt x }}\\ M = \dfrac{{x - 3\sqrt x }}{{x - 2\sqrt x }} \)
\( b)M = \dfrac{1}{2} \Rightarrow \dfrac{{x - 3\sqrt x }}{{x - 2\sqrt x }} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( {x - 3\sqrt x } \right) = x - 2\sqrt x \\ \Leftrightarrow 2x - 6\sqrt x = x - 2\sqrt x \\ \Leftrightarrow - 4\sqrt x = - x\\ \Leftrightarrow 16x = {x^2}\\ \Leftrightarrow 16x - {x^2} = 0\\ \Leftrightarrow x\left( {16 - x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 16 - x = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 16 \end{array} \right. \)
25 tháng 7 2019
\(\text{Câu 2}:\)
\( a)\sqrt {49x - 98} - 14\sqrt {\dfrac{{x - 2}}{{49}}} = 3\sqrt {x - 2} + 8\left( {x \ge 2} \right)\\ \Leftrightarrow 7\sqrt {x - 2} - 3\sqrt {x - 2} = 8 + 14\sqrt {\dfrac{{x - 2}}{{49}}} \\ \Leftrightarrow 4\sqrt {x - 2} = 8 + 14\sqrt {\dfrac{{x - 2}}{{49}}} \\ \Leftrightarrow 4\sqrt {x - 2} = 8 + 14\dfrac{{\sqrt {x - 2} }}{7}\\ \Leftrightarrow 4\sqrt {x - 2} = 8 + 2\sqrt {x - 2} \\ \Leftrightarrow 4\sqrt {x - 2} - 2\sqrt {x - 2} = 8\\ \Leftrightarrow 2\sqrt {x - 2} = 8\\ \Leftrightarrow \sqrt {x - 2} = 4\\ \Leftrightarrow x - 2 = 16\\ \Leftrightarrow x = 16 + 2 = 18 \text{(thỏa mãn điều kiện)} \)
N
18 tháng 3 2019
1.\(x=7+4\sqrt{3}\)
\(=\left(\sqrt{3}+2\right)^2\)
Thay x=\(\left(2+\sqrt{3}\right)^2\), ta có:
\(A=\frac{3+\sqrt{3}}{4+\sqrt{3}}\)
2. \(B=\frac{\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(B=\frac{-3}{2-\sqrt{x}}\left(đpcm\right)\)
3. \(\frac{B}{A}=\frac{\frac{-3}{2-\sqrt{x}}}{\frac{\sqrt{x}+1}{\sqrt{x}+2}}=\frac{-3}{2-\sqrt{x}}.\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(\frac{B}{A}< -1\Rightarrow\frac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}< -1\)
\(\Leftrightarrow\frac{3\sqrt{x}+6+x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{x-2\sqrt{x}+4}{x-\sqrt{x}-2}< 0\)
\(\Rightarrow x-\sqrt{x}-2< 0\)(Vì \(x-2\sqrt{x}+4>0\))
\(\Leftrightarrow-1< x< 2\)
Xét x=0 thì a=0 và M = 0
Xét x khác 0 thì a khác 0
\(M=\frac{x^2}{x^4+x^2+1}=\frac{x}{x^2-x+1}.\frac{x}{x^2+x+1}\) (1)
\(\Rightarrow\frac{x^2+x+1}{x}=\frac{x^2-x+a}{x}+\frac{2x}{x}\)
\(=\frac{1}{a}+2=\frac{1+2a}{a}\) (2)
Từ (1) và (2) \(\Rightarrow M=a.\frac{a}{1+2a}=\frac{a^2}{1+2a}\)
\(\Rightarrow M=\frac{a^2}{1+2a}\)
+) Xét trường hợp x=0
\(\Rightarrow a=0\)
\(\Rightarrow\frac{a^2}{a+2a}=0\Rightarrow M=0\)
Vậy ...