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1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3)
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27
<=> - 3x^2 - 18x - 22 = 0
<=> 3x^2 + 18x + 22 = 0
Nửa chu vi mảnh đất là:
120 : 2 = 60 (m)
Chiều dài hơn chiều rộng là:
5 + 5 = 10 (m)
Chiều rộng là:
( 60 - 10 ) : 2 = 25 (m)
Chiều dài là:
25 + 10 = 35 (m)
Diện tích là:
25 35 = 875 ( )
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\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8+3\left(x^2-6x+9\right)=x^3+6x^2+12x+8-x^3-9x+3x^2+27-6x^2+5\)
\(\Leftrightarrow-3x^2+3x-9+3\left(x^2-6x+9\right)=3x^2+3x+40\)
\(\Leftrightarrow-3x^2+3x-9+3x^2-18x+27-3x^2-3x-40=0\)
\(\Leftrightarrow-3x^2-18x-22=0\)
\(\Leftrightarrow3x^2+18x+22=0\)
\(\Delta=18^2-4\cdot3\cdot22=60\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-18-2\sqrt{15}}{6}=\dfrac{-9-\sqrt{15}}{3}\\x_2=\dfrac{-18+2\sqrt{15}}{6}=\dfrac{-9+\sqrt{15}}{3}\end{matrix}\right.\)
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a)2x.(x+3)-3.(x^2+1)=x+1-x.(x-2)
<=> 2x2 + 6x - 3x2 - 3 = x - 1 - x2 + 2x
<=> 2x2 + 6x - 3x2 - x + x2 - 2x = -1 +3
<=> 3x = 2
<=> x = 2/3
b)(x+2).(x-2)-(x-3).(x+5)=0
<=> x2 - 4 - x2 - 5x - 3x - 15 = 0
<=> -5x - 3x = 4 + 15
<=> -8x = 19
<=> x = -19/8
Phần c tương tự ạ
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b) (x-3)2-(x-3)(x+3)=0
<=>(x-3)[(x-3)-(x+3)]=0
<=>(x-3)[x-3-x-3]=0
<=>(x-3)*(-6)=0
<=>x=3
a) (x-2)(x+2)-7=(x-1)2
<=>x2-4-7=x2-2x+1
<=>x2-11-x2+2x-1=0
<=>(x2-x2)+2x-11-1=0
<=>2x-12=0
<=>2x=12
<=>x=6
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1) \(x\left(x-1\right)-2\left(1-x\right)=0\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow x^2+x=2\)
\(\Leftrightarrow x\left(x+1\right)=2\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
2) \(\left(x+1\right)^2=x+1\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)
3) \(\left(x-3\right)^3+\left(3-x\right)=0\)
\(\Leftrightarrow\left(x-3\right)^3+3=x\)
\(\Leftrightarrow\left(x-4\right)\left(x-3\right)\left(x-2\right)=0\)
4) \(x^3=x^5\)
\(\hept{\begin{cases}x=-1\\x=1\\x=0\end{cases}}\)
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Cho hoi dap de hoi chi khong duoc noi lung tung day la pham loi trong hoi dap
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(-3+x^2)(-3+x^2)(-3+x^2)(-3+x^2)(-3+x^2)=1
(-3+x2)5=1
-3+x2 = 1
x2 = 4
x = \(\pm2\)
Ta có: x + 3 = (x + 3)2
<=>x + 3 - (x + 3)2= 0
<=>(x + 3)(- 2 -x)=0
<=>x=-3 hoặc x=-2
x + 3 = (x + 3)2
<=> x + 3 = x2 + 6x + 9
<=> x + 3 - x2 - 6x - 9 = 0
<=> -5x - 6 - x2 = 0
<=> x2 + 5x + 6 = 0
<=> (x + 2)(x + 3) = 0
<=> x + 2 = 0 hoặc x + 3 = 0
<=> x = -2 x = -3
=> x = -2 hoặc x = -3