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Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
a, ĐKXĐ : x^2-9 khác 0 ; x-3 khác 0 ; x+3 khác 0 => x khác -3 và 3
A = x^2+3+2.(x-3)-(x+3)/(x-3).(x+3) = x^2+x-6/(x-3).(x+3) = (x-2).(x+3)/(x-3).(x+3) = x-2/x-3
b, Để A = 1/2 => x-2 = 2.(x-3) = 2x-6
=> x = 4 (tm ĐKXĐ)
k mk nha
\(M=\left(\frac{x^2+x+10}{x^2-9}-\frac{1}{x-3}\right):\frac{1}{x-3}\) \(ĐKXĐ:x\ne\pm3\)
\(M=\left[\frac{x^2+x+10}{\left(x-3\right)\left(x+3\right)}-\frac{x+3}{\left(x-3\right)\left(x+3\right)}\right]:\frac{1}{x-3}\)
\(M=\left[\frac{x^2+x+10-x-3}{\left(x-3\right)\left(x+3\right)}\right]:\frac{1}{x-3}\)
\(M=\frac{x^2+7}{\left(x-3\right)\left(x+3\right)}.\left(x-3\right)\)
\(M=\frac{x^2+7}{x+3}\)
vậy \(M=\frac{x^2+7}{x+3}\)
khi \(x=5\)thì \(M=\frac{5^2+7}{5+3}=\frac{25+7}{8}=\frac{32}{8}=4\)
vậy \(M=4\)khi \(x=5\)
a, \(A=\frac{x^2+3x-x+3-x^2+1}{x^2-9}\)\(.\frac{x+3}{2}\) \(\left(x\ne3;-3\right)\)
\(A=\frac{2x+4}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{2}\)\(=\frac{2\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{2}\)\(=\frac{x+2}{x-3}\)
b, để \(A\in Z\Rightarrow\hept{\begin{cases}x+2⋮x-3\\x-3⋮x-3\end{cases}}\)\(\Rightarrow x+2-x+3=5⋮x-3\)\(\leftrightarrow x+3\in\left(1;5;-1;-5\right)\)
\(\leftrightarrow x\in\left(-2;2;-4;-8\right)\)
\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\frac{2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)
b) Ta có : \(A=\frac{x+4}{x-3}=\frac{x-3+7}{x-3}=1+\frac{7}{x-3}\)
Để A đạt giá trị nguyên thì \(\frac{7}{x-3}\)đạt giá trị nguyên
=> 7 ⋮ x - 3
=> x - 3 ∈ Ư(7) = { ±1 ; ±7 }
| x-3 | 1 | -1 | 7 | -7 |
| x | 4 | 2 | 10 | -4 |
So với ĐKXĐ ta thấy x = 4 , x = 10 , x = -4 thỏa mãn
Vậy với x ∈ { ±4 ; 10 } thì A đạt giá trị nguyên
(....) dùng để nhìn được chữ số ở phân số cuối cùng thôi, ko dùng để làm gì.
( ác ) là từ ( các )
(gia strij) là từ ( giá trị )
a) ĐK: \(x\ne0,x\ne\pm3\)
\(A=\left(\frac{x-3}{x^2-9}+\frac{1}{x+3}\right)\div\frac{x}{x+3}\)
\(=\left(\frac{1}{x+3}+\frac{1}{x+3}\right)\div\frac{x}{x+3}\)
\(=\frac{2}{x+3}\times\frac{x+3}{x}=\frac{2}{x}\)
b) \(\left|A\right|=\left|\frac{2}{x}\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{x}=3\\\frac{2}{x}=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{2}{3}\end{cases}}\)(thỏa mãn)
a, ĐKXĐ :\(x\ne3;x\ne-3\)
b, \(P=\frac{3\cdot\left(x-3\right)}{\left(x-3\right)\cdot\left(x+3\right)}+\frac{x+3}{\left(x+3\right)\cdot\left(x-3\right)}+\frac{18}{\left(x+3\right)\cdot\left(x-3\right)}\)
\(=\frac{3x-9+x+3+18}{\left(x+3\right)\cdot\left(x-3\right)}\)\(=\frac{4x+12}{\left(x-3\right)\cdot\left(x+3\right)}\)
\(=\frac{4\cdot\left(x+3\right)}{\left(x+3\right)\cdot\left(x-3\right)}=\frac{4}{x-3}\)
c, Với P = 4 \(\Rightarrow\frac{4}{x-3}=4\Rightarrow4=4\cdot\left(x-3\right)\)\(\Rightarrow1=x-3\Rightarrow x=4\)
\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}\)
a,\(ĐKXĐ:x\ne0;x\ne3;x\ne1\)
\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}=\left(\frac{9}{x\left(x-3\right)}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3\left(1-x\right)}\)
\(=\left(\frac{9+\left(x-2\right)\left(x-3\right)-x.x}{x\left(x-3\right)}\right).\frac{x}{3\left(1-x\right)}=\frac{9+x^2-5x+6-x^2}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}\)
\(=\frac{-5x+15}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=\frac{-5\left(x-3\right)}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=-\frac{5}{3\left(1-x\right)}\)
b, \(x=\frac{1}{2}\)
\(\Rightarrow P=-\frac{5}{3\left(1-\frac{1}{2}\right)}=-\frac{5}{3.\frac{1}{2}}=-5:\frac{3}{2}=-\frac{10}{3}\)
c, Để \(P\in z\)thì \(3\left(1-x\right)\inƯ\left(5\right)=\left(-5;-1;1;5\right)\)
\(3\left(1-x\right)=-5\Rightarrow1-x=-\frac{5}{3}\Rightarrow x=\frac{8}{3}\)
\(3\left(1-x\right)=-1\Rightarrow1-x=-\frac{1}{3}\Rightarrow x=\frac{4}{3}\)
\(3\left(1-x\right)=1\Rightarrow1-x=\frac{1}{3}\Rightarrow x=\frac{2}{3}\)
\(3\left(1-x\right)=5\Rightarrow1-x=\frac{5}{3}\Rightarrow x=-\frac{2}{3}\)
a) ĐKXĐ : \(\hept{\begin{cases}x-3\ne0\\x^2-9\ne0\\x+3\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne3\\x\ne\pm3\\x\ne-3\end{cases}}\Rightarrow x\ne\pm3\)
b) A = \(\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}=\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x-3}\)
Khi x = 3 => Không thỏa mãn ĐKXĐ
=> Không tồn tại A khi x = 3
a, Điều kiện xác định là :
\(\hept{\begin{cases}x-3\ne0\\x^2-9\ne0\\x+3\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne3\\\left(x-3\right)\left(x+3\right)\ne\\x\ne-3\end{cases}}0\Rightarrow x\ne\pm3}\)
Vậy \(x\ne\pm3\)
b, \(A=\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}\)
\(=\frac{3}{x-3}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\)
\(=\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x-3}\)
Thay x = 3 ( ktm đkxđ )
Ko tồn tại x