Bài 1 : cho biểu thức A = [(1-×^2)/(1-×)]/(1-×^2)/(1-×-×^2+×^3) với x khác -1 và 1. Nếu x; y là số thực thỏa mãn 3x^2+y^2+2x-2y= 1. Hãy tìm tất cả các giá trị nguyên dương của A
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
b: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(\dfrac{1}{x-3}-\dfrac{1}{x+3}+\dfrac{2x}{9-x^2}\)
\(=\dfrac{1}{x-3}-\dfrac{1}{x+3}-\dfrac{2x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x+3-x+3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=-\dfrac{2}{x+3}\)
c: ĐKXĐ: \(x\notin\left\{2;0\right\}\)
Sửa đề: \(\dfrac{x+1}{x-2}+\dfrac{4-5x}{x^3+4x}:\dfrac{x-2}{x^2+4}\)
\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x^2+4\right)}\cdot\dfrac{x^2+4}{x-2}\)
\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x-2\right)}\)
\(=\dfrac{x\left(x+1\right)+4-5x}{x\left(x-2\right)}=\dfrac{x^2+x-5x+4}{x\left(x-2\right)}\)
\(=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Phân tích đa thức thành nhân tử
a) 2( x + 1 ) - 3y( x + 1 ) = ( x + 1 )( 2 - 3y )
b) x2 - 5x + 4 = x2 - x - 4x + 4 = x( x - 1 ) - 4( x - 1 ) = ( x - 1 )( x - 4 )
Tìm x
a) x( x - 3 ) + 7x - 21 = 0
<=> x( x - 3 ) + 7( x - 3 ) = 0
<=> ( x - 3 )( x + 7 ) = 0
<=> x - 3 = 0 hoặc x + 7 = 0
<=> x = 3 hoặc x = -7
b) ( x - 2 )2 + x( 3 - x ) = 6
<=> x2 - 4x + 4 + 3x - x2 = 6
<=> -x + 4 = 6
<=> -x = 2
<=> x = -2
\(A=\frac{x-2}{x}\)và \(B=\frac{x}{x-2}-\frac{2x}{x^2-4}\)( x ≠ 0 ; x ≠ ±3 )
a) Tại x = 23 ( tmđk ) => \(A=\frac{23-2}{23}=\frac{21}{23}\)
b) P = A.B
\(=\frac{x-2}{x}\times\left(\frac{x}{x-2}-\frac{2x}{x^2-4}\right)\)
\(=\frac{x-2}{x}\times\left(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{x-2}{x}\times\frac{x^2+2x-2x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{1}{x}\times\frac{x^2}{x+2}=\frac{x}{x+2}\)
Để P = 4 => \(\frac{x}{x+2}=4\)
=> 4( x + 2 ) = x
=> 4x + 8 - x = 0
=> 3x + 8 = 0
=> x = -8/3 ( tmđk )
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\sqrt{x}-1\)
a) \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
Đk: \(x>0\) và \(x\ne1\)
\(\Rightarrow A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
\(=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
b) Thay \(x=3+2\sqrt{2}\) vào A ta được:
\(A=\sqrt{3+2\sqrt{2}}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1\)
\(=\sqrt{2}+1-1=\sqrt{2}\)
(Vì \(\sqrt{2}+1>0\Rightarrow\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\))