K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
23 tháng 1 2020
1) 2x4 - 9x3 + 14x2 - 9x + 2 = 0
<=> (2x4 - 4x3) - (5x3 - 10x2) + (4x2 - 8x) - (x - 2) = 0
<=> 2x3(x - 2) - 5x2(x - 2) + 4x(x - 2) - (x - 2) = 0
<=> (2x3 - 5x2 + 4x - 1)(x - 2) = 0
<=> [(2x3 - 2x2) - (3x2 - 3x) + (x - 1)](x - 2) = 0
<=> [2x2(x - 1) - 3x(x - 1) + (x - 1)](x - 2) = 0
<=> (2x2 - 2x - x + 1)(x - 1)(x - 2) = 0
<=> (2x - 1)(x - 1)2(x - 2) = 0
<=> 2x - 1=0
hoặc x - 1 = 0
hoặc x - 2 = 0
<=> x = 1/2
hoặc x = 1
hoặc x = 2
Vậy S = {1/2; 1; 2}
NT
2
HY
18 tháng 2 2020
\(b.6x^4+25x^3+12x^2-25x+6=0\\\Leftrightarrow 6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\\\Leftrightarrow 6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\\\Leftrightarrow \left(6x^3+13x^2-14x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(6x^3+18x^2-5x^2-15x+x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left[6x^2\left(x+3\right)-5x\left(x+3\right)+\left(x+3\right)\right]\left(x+2\right)=0\\ \Leftrightarrow\left(6x^2-5x+1\right)\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(6x^2-3x-2x+1\right)\left(x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left[3x\left(2x-1\right)-\left(2x-1\right)\right]\left(x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(2x-1\right)\left(x+3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\2x-1=0\\x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=\frac{1}{2}\\x=-3\\x=-2\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{\frac{1}{3};\frac{1}{2};-3;-2\right\}\)
HY
18 tháng 2 2020
\(2x^4-9x^3+14x^2-9x+2=0\\\Leftrightarrow 2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\\\Leftrightarrow 2x^3\left(x-1\right)-7x^2\left(x-1\right)+7x\left(x-1\right)-2\left(x-1\right)=0\\\Leftrightarrow \left(2x^3-7x^2+7x-2\right)\left(x-1\right)=0\\\Leftrightarrow \left[2\left(x^3-1\right)-7x\left(x-1\right)\right]\left(x-1\right)=0\\\Leftrightarrow \left(x-1\right)^2\left[2\left(x^2+x+1\right)-7x\right]=0\\\Leftrightarrow \left(2x^2+2x+2-7x\right)\left(x-1\right)^2=0\\\Leftrightarrow \left(2x^2-5x+2\right)\left(x-1\right)^2=0\\\Leftrightarrow \left(2x^2-x-4x+2\right)\left(x-1\right)^2=0\\\Leftrightarrow \left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left(x-1\right)^2=0\\\Leftrightarrow \left(x-2\right)\left(2x-1\right)\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-1=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\2x=1\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{2;\frac{1}{2};1\right\}\)
16 tháng 5 2018
\(\dfrac{2}{2x-6}+\dfrac{2}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\) ( x # 3 ; x # -1)
⇔ \(\dfrac{2}{2\left(x-3\right)}+\dfrac{2}{2\left(x+1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
⇔ \(\dfrac{x+1}{\left(x-3\right)\left(x+1\right)}+\dfrac{x-3}{\left(x+1\right)\left(x-3\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
⇔ x + 1 + x - 3 - 2x = 0
⇔ - 2 = 0 ( vô lý )
Vậy , phương trình vô nghiệm
19 tháng 12 2019
a, 2x(x-3)-2x2=12
⇔2x2-6x-2x2=12
⇔-6x=12
⇔x=-2
b,(x-2)2 -x(x+3)= 25
⇔(x-2)2 -25-x(x+3)=0
⇔[(x-2)2-52]-x(x+3)=0
⇔(x+3)(x-7)-x(x+3)=0
⇔(x+3)(x-7-x)=0
⇔(x+3)(-7)=0
⇔x+3=0
⇔x=-3
c, 2x (x-3) +4(3-x)=0
⇔ 2x (x-3) -4(x-3)=0
⇔(x-3)(2x-4)=0
⇒\(\left[{}\begin{matrix}x-3=0\\2x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
d,x2 -9x -10= 0
⇔x2 -10x+x-10=0
⇔x(x-10) + (x-10)=0
⇔(x-10)(x+1)=0
\(\Rightarrow\left[{}\begin{matrix}x-10=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-1\end{matrix}\right.\)
2 tháng 3 2020
\(\left(3x-5\right)\left(-2x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\-2x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=5\\-2x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-7}{2}\end{cases}}}\)
A
2 tháng 3 2020
\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)
\(\Leftrightarrow9x^2-1=2x-3+6x^2-9x\)
\(\Leftrightarrow9x^2-1=-7x-3+6x^2\)
\(\Leftrightarrow9x^2-1+7x+3-6x^2=0\)
\(\Leftrightarrow3x^2+2+7x=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}\)
18 tháng 6 2022
a: \(\Leftrightarrow\dfrac{x+5}{2x-1}+\dfrac{2x-1}{x+5}-2=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+5\right)+\left(2x-1\right)^2-2\left(2x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2+10x+25+4x^2-4x+1-2\left(2x^2+10x-x-5\right)=0\)
\(\Leftrightarrow5x^2+6x+26-4x^2-18x+10=0\)
\(\Leftrightarrow x^2-12x+36=0\)
=>x=6
b: \(\dfrac{9x-27}{2x-7}-\dfrac{8x-28}{x-3}=0\)
\(\Leftrightarrow9\left(x-3\right)^2-4\left(2x-7\right)^2=0\)
\(\Leftrightarrow\left(3x-9\right)^2-\left(4x-14\right)^2=0\)
\(\Leftrightarrow\left(3x-9-4x+14\right)\left(3x-9+4x-14\right)=0\)
\(\Leftrightarrow\left(5-x\right)\left(7x-23\right)=0\)
hay \(x\in\left\{5;\dfrac{23}{7}\right\}\)
NT
3
T
10 tháng 7 2019
Em thử nhá, có khi biến đổi sai sót cũng không chừng :D
a) \(\left(2x-1\right)^3=\left(2x\right)^3\)
\(\Leftrightarrow\left(2x-1-2x\right)\left[\left(2x-1\right)^2+2x\left(2x-1\right)+\left(2x\right)^2\right]=0\)
\(\Leftrightarrow\left[\left(2x-1\right)^2+2x\left(2x-1\right)+\left(2x\right)^2\right]=0\) (rút gọn rồi chia hai vế cho -1)
\(\Leftrightarrow4x^2-4x+1+4x^2-2x+4x^2=0\)
\(\Leftrightarrow12x^2-6x+1=0\Leftrightarrow\text{vô nghiệm}\)
b) PT \(\Leftrightarrow x^3+x^2+3x+\frac{1}{9}=0\)
Đặt \(x=y-\frac{1}{3}\) suy ra:
\(\left(y-\frac{1}{3}\right)^3+\left(y-\frac{1}{3}\right)^2+3\left(y-\frac{1}{3}\right)+\frac{1}{9}=0\)
Rút gọn lại ta được: \(\frac{27y^3+72y-22}{27}=0\)
\(\Leftrightarrow27y^3+72y-22=0\)
Tới đây nghiệm xấu vẫn là xấu :(
10 tháng 7 2019
a) \(\left(2x-1\right)^3=8x^3\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(2x\right)^3\)
\(\Leftrightarrow2x-1=2x\)
\(\Leftrightarrow0=1\) ( vô lí )
\(\Rightarrow x\in\varnothing\)
Đúng không ta ???
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 4 2019
a/
\(9x^2+25y^2+1+30xy-6x-10y+4y^2-20y+25=0\)
\(\Leftrightarrow\left(3x+5y-1\right)^2+\left(2y-5\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y-1=0\\2y-5=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\frac{23}{6}\\y=\frac{5}{2}\end{matrix}\right.\)
b/
\(4x^2+4y^2+8xy+x^2-2x+1+y^2+2y+1=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
c/
\(y^2-2y+1+2=\frac{6}{x^2+2x+1+3}\)
\(\Leftrightarrow\left(y-1\right)^2+2=\frac{6}{\left(x+1\right)^2+3}\)
Ta có \(VT=\left(y-1\right)^2+2\ge2\)
\(\left(x+1\right)^2+3\ge3\Rightarrow VP=\frac{6}{\left(x+1\right)^2+3}\le\frac{6}{3}=2\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}y-1=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
d/
\(\frac{-9x^2+18x-9-8}{x^2-2x+1+2}=y^2+4y+4-4\)
\(\Leftrightarrow\frac{-9\left(x-1\right)^2-8}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow\frac{-9\left(x-1\right)^2-18+10}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow-9+\frac{10}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow\frac{10}{\left(x-1\right)^2+2}=\left(y+2\right)^2+5\)
Ta có \(\left(x-1\right)^2+2\ge2\Rightarrow\frac{10}{\left(x-1\right)^2+2}\le\frac{10}{2}=5\Rightarrow VT\le5\)
\(\left(y+2\right)^2+5\ge5\Rightarrow VP\ge5\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Ta có x2 - 1 = ( x + 1)(x - 1)
=> \(\frac{x^2-1}{x-1}\) = x + 1
=> đpcm
Ta có
x2 - 1 = (x + 1) (x - 1)
=>\(\frac{x^2-1}{x-1}\) = x+ 1