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\(a,=x^3+2x^2-x-2-x^3+8=2x^2-x+6\\ b,=\dfrac{\left(x-2y\right)\left(x+2y\right)}{2\left(x-2y\right)}=\dfrac{x+2y}{2}\\ c,=\dfrac{4xy+x^2-2xy+y^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{2x}{x+y}-\dfrac{y}{x-y}\\ =\dfrac{x\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)^2}-\dfrac{y}{x-y}=\dfrac{x}{x-y}-\dfrac{y}{x-y}=1\)
ĐKXĐ:\(x\ne\pm2\)
\(\dfrac{x+1}{x-2}-\dfrac{5}{2+x}=\dfrac{12}{x^2-4}+1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{12}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{x^2+3x+2-5x+10-12-x^2+4}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Rightarrow-2x+4=0\\ \Leftrightarrow x=2\left(ktm\right)\)
\(\Rightarrow x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow-2x+12=8\Leftrightarrow x=2\left(ktm\right)\)
pt vô nghiệm
A, Để biểu thức A có nghĩa thì
\(2x+10\ne0\Rightarrow x\ne-5\)
\(x\ne0\)
\(2x^2+10x\ne0\Rightarrow x\ne0;x\ne-5\)
Vậy điều kiện của x là \(x\ne0;x\ne-5\)
b) Ta có:
\(A=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x^2+10x}\)
\(A=\frac{x\cdot\left(x^2+2x\right)+\left(x-5\right)\cdot\left(2x+10\right)+50-5x}{2x^2+10x}\)
\(A=\frac{x^3+4x^2-5x}{2x^2+10x}\)
\(A=\frac{x\cdot\left(x-1\right)\cdot\left(x+5\right)}{2x\cdot\left(x+5\right)}\)
\(A=\frac{x-1}{2}\)
Để A=1 thì ta có\(\frac{x-2}{2}=1\Leftrightarrow x-2=2\Leftrightarrow x=4\)
Vậy x=4 thì A=1
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
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
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 : \(x^3-x^2-4x+4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow x=\pm2;1\)
