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đk : x khác -2 ; 2 ; -1
\(P=\dfrac{x+2-4}{x^2-4}\left(\dfrac{x+2}{x+1}\right)=\dfrac{1}{x+2}.\dfrac{x+2}{x+1}=\dfrac{1}{x+1}\)
\(P=\left(\dfrac{1}{x-2}-\dfrac{4}{x^2-4}\right).\left(1+\dfrac{1}{x+1}\right)\)
\(\Rightarrow P=\left(\dfrac{x+2}{\text{}\left(x-2\right)\left(x+2\right)}-\dfrac{4}{\left(x-2\right)\left(x+2\right)}\right).\left(\dfrac{x+1}{x+1}+\dfrac{1}{x+1}\right)\)
\(\Rightarrow P=\dfrac{x+2-4}{\text{}\left(x-2\right)\left(x+2\right)}.\dfrac{x+1+1}{x+1}\)
\(\Rightarrow P=\dfrac{x-2}{\text{}\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{x+1}\)
\(\Rightarrow P=\dfrac{1}{x+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đk : xkhác -1 ; 1 ; 1/2
\(A=\dfrac{3-3x+x+1-8}{\left(1-x\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}=\dfrac{-2x-4}{\left(1-x\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)
\(=\dfrac{2x+4}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}=\dfrac{2x+4}{1-2x}\)
\(=\left(\dfrac{3}{x+1}-\dfrac{1}{x-1}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x^2-1}{1-2x}\)
\(=\dfrac{3x-3-x-1+8}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{-2x+1}\)
\(=\dfrac{2x+4}{-2x+1}=\dfrac{-2x-4}{2x-1}\)
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\(P=\left(\dfrac{2x}{x+3}+\dfrac{x}{x-3}-\dfrac{3x^2+3}{x^2-9}\right):\left(\dfrac{2x-2}{x-3}-1\right)\)
\(\Rightarrow P=\dfrac{2x\left(x-3\right)+x\left(x+3\right)-\left(3x^2+3\right)}{\left(x+3\right)\left(x-3\right)}:\dfrac{2x-2-\left(x-3\right)}{x-3}\)
\(\Rightarrow P=\dfrac{2x^2-6x+x^2+3x-3x^2-3}{\left(x+3\right)\left(x-3\right)}:\dfrac{2x-2-x+3}{x-3}\)
\(\Rightarrow P=\dfrac{-3x-3}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+1}{x-3}\)
\(\Rightarrow P=\dfrac{-3\left(x+1\right)}{\left(x+3\right)\left(x-3\right)}.\dfrac{x-3}{x+1}\)
\(\Rightarrow P=\dfrac{-3}{x+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)
\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)
\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))
\(\Leftrightarrow x=-36\).
Vậy nghiệm của pt là x = -36.
2) x(x+1)(x+2)(x+3)= 24
⇔ x.(x+3) . (x+2).(x+1) = 24
⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24
Đặt \(x^2\)+ 3x = b
⇒ b . (b+2)= 24
Hay: \(b^2\) +2b = 24
⇔\(b^2\) + 2b + 1 = 25
⇔\(\left(b+1\right)^2\)= 25
+ Xét b+1 = 5 ⇒ b=4 ⇒ \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0
⇒(x-1)(x+4)=0⇒x=1 và x=-4
+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0
⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))2 = - \(\dfrac{15}{4}\) Hay ( \(x^2\) +\(\dfrac{3}{2}\) )2= -\(\dfrac{15}{4}\) (vô lí)
⇒x= 1 và x= 4
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a) Ta có: \(\dfrac{AE}{AB}=\dfrac{2}{5}\)
\(\dfrac{AF}{AC}=\dfrac{4}{10}=\dfrac{2}{5}\)
Do đó: \(\dfrac{AE}{AB}=\dfrac{AF}{AC}\)\(\left(=\dfrac{2}{5}\right)\)
Xét ΔAEF và ΔABC có
\(\dfrac{AE}{AB}=\dfrac{AF}{AC}\)(cmt)
\(\widehat{A}\) chung
Do đó: ΔAEF\(\sim\)ΔABC(c-g-c)
Suy ra: \(\dfrac{AE}{AB}=\dfrac{EF}{BC}\)(Các cặp cạnh tương ứng tỉ lệ)
\(\Leftrightarrow\dfrac{2}{5}=\dfrac{EF}{12}\)
hay EF=4,8(cm)
Vậy: EF=4,8cm
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x3 _ x2 _ 4x - 4 = 0
x mũ 2(x+1)- 4(x+1)=0
(x mũ 2 - 4) (x+1)=0
(x+2) (x-2) (x+1) =0
suy ra (x+2)=0
(x-2)=0
(x+1)=0
vậy x=-2
x=2
x= -1
good luck!
Sửa đề : \(x^3-x^2-4x+4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow x=\pm2;1\)
Tách nhỏ ra bạn
Bài 7:
\(P=\dfrac{2x^2-6x+x^2+3x-3x^2-3}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x-2-x+3}{x-3}\)
\(=\dfrac{-3\left(x+1\right)}{\left(x+3\right)}\cdot\dfrac{1}{x+1}=\dfrac{-3}{x+3}\)
Bài 8:
\(A=\dfrac{x^2+8x+8-x^2-4x-4}{x\left(x+2\right)}:\dfrac{x^2-x-3+x+2}{x\left(x+2\right)}\)
\(=\dfrac{4x+4}{x^2-1}=\dfrac{4}{x-1}\)