Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) $(x-3)^2-(x+2)(x-2)=-5$
$\Rightarrow x^2-2\cdot x\cdot3+3^2-(x^2-2^2)=-5$
$\Rightarrow x^2-6x+9-(x^2-4)=-5$
$\Rightarrow x^2-6x+9-x^2+4=-5$
$\Rightarrow-6x+13=-5$
$\Rightarrow-6x=-18$
$\Rightarrow x=3$
b) $x^3-2x^2-4x+8=0$
$\Rightarrow(x^3-2x^2)-(4x-8)=0$
$\Rightarrow x^2(x-2)-4(x-2)=0$
$\Rightarrow (x^2-4)(x-2)=0$
$\Rightarrow (x^2-2^2)(x-2)=0$
$\Rightarrow (x-2)(x+2)(x-2)=0$
$\Rightarrow (x-2)^2(x+2)=0$
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
$\text{#}Toru$
\(a,\Leftrightarrow\dfrac{3x^3+6x^2-3x-5x^2-10x+5}{x^2+2x-1}=10\\ \Leftrightarrow\dfrac{3x\left(x^2+2x-1\right)-5\left(x^2+2x-1\right)}{x^2+2x-1}=10\\ \Leftrightarrow3x-5=10\Leftrightarrow3x=15\Leftrightarrow x=5\\ b,\Leftrightarrow\left(x^4+2x^2-4x^2-8\right):\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-4\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x^2+2\right)=0\Leftrightarrow x=-2\left(x^2+2>0\right)\\ c,\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-4\right)^2}=0\Leftrightarrow\dfrac{x}{x-4}=0\Leftrightarrow x=0\)
c) (4x - 8)[x + (-3)] = 0
=> 4(x - 2)(x - 3) = 0
=> (x - 2)(x - 3) = 0
=> x - 2 = 0 hoặc x - 3 = 0
+) x - 2 = 0 => x = 2
+) x - 3 = 0 => x = 3
Vậy x \(\in\){2;3}
11(x - 6) = 4x + 11
=> 11x - 66 = 4x + 11
=> 11x - 4x = 11 + 66
=> 7x = 77
=> x = 77/7
=> x = 11
a,x.(x+7)=0
suy ra x=o hoặc x+7=0
vs x+7=0
x=0+7
x=7
vậy x=0 hoặc x=7
b(2+2x)(7-x)=0
suy ra 2+2x=0 hoặc 7-x=0
vs2+2x=0 vs7-x=0
2x =0-2 x=0+7
2x =(-2) x=7
x=(-2);2
x=-1
vậy x=-1 hoặc x=7
d(x^2-9)(3x+15)=0
suy ra x^2-9=0 hoặc 3x+15=0
vsx^2-9=0 vs 3x+15=0
x^2 =0+9 3x =0-15
x^2 =9 3x =-15
x^2 =3^2 x=(-15):3
suy ra x=3 hoặc x=-3 x=-5
vậy x=3 x=-3 hoặc x=-5
e,(4x-8)(x^2+1)=0
suy ra4x-8=0 hoặc x^2+1=0
vs 4x-8=0 vs x^2+1=0
4x =0+8 x^2 =0-1
4x =8 x^2 =-1
x =8:4 x^2 =-1^2 hoặc 1^2
x =2 suy ra x=-1 hoặc x=1
vậy x=2, x=-1 hoặc x=1
Tìm x
a) \(\left(x+1\right)\left(x+2\right)-x^2-x=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2-x\right)=0\)
\(\Leftrightarrow2\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
b) \(2x^2+5x-3=0\)
\(\Leftrightarrow2x^2+6x-x-3=0\)
\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-3\end{array}\right.\)
a) x2 -2x + 4x - 8 = 0
x.(x-2) + 4.(x-2) = 0
(x+4).(x-2) = 0
=> x + 4 = 0 => x = -4
x-2= 0 => x = 2
KL:,..
b) 2.(x-3) - (3-x) = 0
2.(x-3) + (x-3) = 0
(x-3).(2+1 ) = 0
(x-3).3 = 0
=> x - 3 = 0 => x =3
KL:...
a) x2 -2x + 4x - 8 = 0
x.(x-2) + 4.(x-2) = 0
(x+4).(x-2) = 0
=> x + 4 = 0 => x = -4
x-2= 0 => x = 2
KL:,..
b) 2.(x-3) - (3-x) = 0
2.(x-3) + (x-3) = 0
(x-3).(2+1 ) = 0
(x-3).3 = 0
=> x - 3 = 0 => x =3
KL:...
#