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18 tháng 8 2018
Câu 4 :
\(x^2+y^2-2\left(x-y-1\right)=0\)
\(\Leftrightarrow x^2+y^2-2x+2y+2=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+1\right)^2=0\)
Do \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Vậy \(x=1;y=-1\)
6 tháng 8 2019
1a) -3x2(2x3 - 2x + 1/3) = -6x5 + 6x3 - x2
b) (x4 + 2x3 - 2/3).(-3x4) = -3x8 - 6x7 + 2x4
c) (x + 3)(x - 4) = x2 - 4x + 3x - 12 = x2 - x - 12
d)(x - 4)(x2 + 4x + 16) = (x - 4)(x2 + 4x + 42) = x3 - 64
e) 4(x - 1/2)(x + 1/2)(4x2 + 1) =4(x2 - 1/4)(4x2 + 1) = 4(4x4 + x2 - x2 - 1/4) = 4(4x4 - 1/4) = 16x4 - 1
B2. a) (2 - x)(x2 + 2x + 4) + x(x - 3)(x + 4) - x2 + 24 = 0
=> 8 - x3 + x(x2 + 4x - 3x - 12) - x2 + 24 = 0
=> 8 - x3 + x3 + x2 - 12x - x2 + 24 = 0
=> -12x + 32 = 0
=> -12x = -32
=> x = -32 : (-12) = 8/3
b) (x/2 + 3)(5 - 6x) + (12x - 2)(x/4 + 3) = 0
=> 5x/2 - 3x2 + 15 - 18x + 3x2 + 36x - x/2 - 6 = 0
=> 20x + 9 = 0
=> 20x = -9
=> x = -9/20
CM
4 tháng 12 2018
a) \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
b) \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Rightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(3x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\3x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\3x=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
d) \(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
e) \(2x^3+6x^2=x^2+3x\)
\(\Rightarrow2x^3+6x^2-x^2-3x=0\)
\(\Rightarrow2x^3+5x^2-3x=0\)
\(\Rightarrow x\left(2x^2+5x-3\right)=0\)
\(\Rightarrow2x^2+5x-3=0\)
\(\Rightarrow2x^2-6x+x-3=0\)
\(\Rightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
f) \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)-2x^2\)
\(\Rightarrow\left(x^2-1\right)\left(x+2\right)-\left(x^3-8\right)-2x^2=0\)
\(\Rightarrow x^3+2x^2-x+2-x^3+8-2x^2=0\)
\(\Rightarrow-x+10=0\)
\(\Rightarrow x=10\)
15 tháng 2 2020
a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42
10 tháng 12 2021
\(a,\Rightarrow 2x^2-10x-3x-2x^2=26\\ \Rightarrow -13x=26\\ \Rightarrow x=-2\\ b, \Rightarrow -2x^2+3x+3-3x-3+2x^2-x=18\\ \Rightarrow -x=18\Rightarrow x=-18\)
29 tháng 11 2022
1: \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
2: \(=8x^2+12x-10x-15-4\left(2x^2-x+4x-2\right)+10x+7\)
\(=8x^2+12x-8-8x^2+4x-16x+8\)
\(=0\)
3: \(=\left(6x+1-6x+1\right)^2=4\)
5: \(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)
\(=3x^3+6x-3x^3+3x=9x\)
Bài làm
x4 + x3 + 6x2 = -5( x + 1 )
<=> x4 + x3 + 6x2 = -5x - 5
<=> x4 + x3 + 6x2 + 5x + 5 = 0
<=> x4 + x3 + x2 + 5x2 + 5x + 5 = 0
<=> ( x4 + x3 + x2 ) + ( 5x2 + 5x + 5 ) = 0
<=> x2( x2 + x + 1 ) + 5( x2 + x + 1 ) = 0
<=> ( x2 + 5 )( x2 + x + 1 ) = 0
Mà x2 + x + 1 > 0
=> x2 + 5 = 0
<=> x2 = -5 ( vô lí )
=> phương trình trên vô nghiệm
Học dốt :)) bài bn lm tốt nhưng lần sau ko cần phải phân tích rồi ghép tích đâu , cách đấy hơi loằng ngoằng nhưng nếu muốn độ chính xác cao thì bn cx nên kham khảo bài bn ấy !
\(x^4+x^3-6x^2=-5\left(x+1\right)\)
\(x^4+x^3-6x^2=-5x-5\)
\(x^4+x^3-6x^2+5x+5=0\)
=> vô nghiệm