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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 ..
Bài làm:
a) Tại x = 2 thì giá trị của B là:
\(B=-\frac{10}{2-4}=\frac{-10}{-2}=5\)
b) Ta có:
\(A=\frac{x+2}{x+5}+\frac{-5x-1}{x^2+6x+5}-\frac{1}{1+x}\)
\(A=\frac{x+2}{x+5}-\frac{5x+1}{\left(x+1\right)\left(x+5\right)}-\frac{1}{x+1}\)
\(A=\frac{\left(x+2\right)\left(x+1\right)-5x-1-\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}\)
\(A=\frac{x^2+3x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\)
\(A=\frac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}\)
\(A=\frac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}\)
\(A=\frac{x-4}{x+5}\)
c) Ta có: \(P=A.B=\frac{x-4}{x+5}\cdot\frac{-10}{x-4}=\frac{-10}{x+5}\)
Để \(-\frac{10}{x+5}\inℤ\Rightarrow\left(x+5\right)\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
=> \(x\in\left\{-15;-10;-7;-6;-4;-3;0;5\right\}\)
a) \(B=\frac{-10}{x-4}\)( ĐKXĐ : \(x\ne4\))
Tại x = 2 ( tmđk ) thì \(B=\frac{-10}{2-4}=\frac{-10}{-2}=5\)
b) \(A=\frac{x+2}{x+5}+\frac{-5x-1}{x^2+6x+5}-\frac{1}{1+x}\)
ĐKXĐ : \(x\ne-5,x\ne-1\)
\(A=\frac{x+2}{x+5}-\frac{5x+1}{\left(x+1\right)\left(x+5\right)}-\frac{1}{x+1}\)
\(A=\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}-\frac{5x+1}{\left(x+1\right)\left(x+5\right)}-\frac{1\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}\)
\(A=\frac{x^2+3x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\)
\(A=\frac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}\)
\(A=\frac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}=\frac{x-4}{x+5}\)
c) \(P=A\cdot B=\frac{x-4}{x+5}\cdot\frac{-10}{x-4}=\frac{-10}{x+5}\)( ĐKXĐ : \(x\ne-5\))
Để P nguyên => \(\frac{-10}{x+5}\)nguyên
=> -10 chia hết cho x + 5
=> x + 5 thuộc Ư(-10) = { ±1 ; ±2 ; ±5 ; ±10 }
| x+5 | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
| x | -4 | -6 | -3 | -7 | 0 | -10 | 5 | -15 |
Các giá trị của x đều tmđk
Vậy x = { -4 ; -6 ; -3 ; -7 ; 0 ; -10 ; 5 ; -15 }
a, Ta có : \(x=25\Rightarrow\sqrt{x}=\sqrt{25}=5\)
\(\Rightarrow Q=\frac{5-1}{5+1}=\frac{4}{6}=\frac{2}{3}\)
b, \(P=\frac{x\sqrt{x}-1}{x-\sqrt{x}}+\frac{x\sqrt{x}+1}{x+\sqrt{x}}-\frac{4}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{4}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1+x-\sqrt{x}+1-4}{\sqrt{x}}=\frac{2x-2}{\sqrt{x}}\)
c, Ta có : \(P.Q.\sqrt{x}< 8\)hay \(\frac{2x-2}{\sqrt{x}}.\sqrt{x}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)< 8\)
\(\Leftrightarrow\frac{2\left(x-1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< 8\Leftrightarrow2\left(\sqrt{x}-1\right)^2< 8\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2< 4\Leftrightarrow\sqrt{x}-1< 2\Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)