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PA
4 tháng 3 2017
x4 - 3x3 + 4x2 - 3x + 1 = 0
<=> x4 - 2x2 + x2 - x3 + 2x2 - x + x2 - 2x + 1 = 0
<=> x2(x2 - 2x + 1) - x(x2 - 2x + 1) + (x2 - 2x + 1) = 0
<=> (x2 - 2x + 1)(x2 - x + 1) = 0
<=> (x - 1)2(x2 - x + 1) = 0
<=> x - 1 = 0 (vì x2 - x + 1 \(\ge\) 0,75 > 0)
<=> x = 1
Vậy tập nghiệm của pt là S = {1}
HH
25 tháng 6 2018
\(x^3+9x=0\)
<=> \(x\left(x^2+9\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\x^2+9=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x\in\varnothing\end{cases}}\)
<=> \(x=0\)
\(9x^2-4-2\left(3x-2\right)^2=0\)
<=> \(\left(9x^2-4\right)-2\left(3x-2\right)^2=0\)
<=> \(\left[\left(3x\right)^2-2^2\right]-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)
<=> \(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)
<=> \(\left(3x-2\right)\left(-3x+6\right)=0\)
<=> \(\left(3x-2\right)3\left(-x+2\right)=0\)
<=> \(3\left(3x-2\right)\left(2-x\right)=0\)
<=> \(\orbr{\begin{cases}3x-2=0\\2-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=2\\x=2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{2}{3}\\x=2\end{cases}}\)
\(\left(x^3-x^2\right)-4x+8x-4=0\)
<=> \(\left(x^3-x^2\right)+\left(4x-4\right)=0\)
<=> \(x^2\left(x-1\right)+4\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x^2+4\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x^2+4=0\end{cases}}\)
<=> \(x=1\)
\(\left(25x^2-10x\right):\left(-5x\right)-3\left(x-2\right)=4\)
<=> \(5x\left(5x-2\right)\left(-\frac{1}{5x}\right)-3\left(x-2\right)=4\)
<=> \(-\left(5x-2\right)-3\left(x-2\right)=4\)
<=> \(\left(5x-2\right)+3\left(x-2\right)=-4\)
<=> \(5x-2+3x-6=-4\)
<=> \(8x-8=-4\)
<=> \(8\left(x-1\right)=-4\)
<=> \(x-1=-\frac{1}{2}\)
<=> \(x=-\frac{3}{2}\)
VD
27 tháng 8 2017
\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
27 tháng 6 2017
a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....
b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........
c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......
d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......
3
22 tháng 2 2019
b) \(\frac{4x}{4x^2-8x+7}+\frac{5x}{4x^2-10x+7}=1\)
Giả sử x = 0 ta có :
\(0+0=1\)( vô lý )
=> \(x\ne0\)
Chia cả tử và mẫu của 2 phân thức cho x ta được :
\(\frac{4x:x}{\left(4x^2-8x+7\right):x}+\frac{5x:x}{\left(4x^2-10x+7\right):x}=1\)
\(\Leftrightarrow\frac{4}{4x-8+\frac{7}{x}}+\frac{5}{4x-10+\frac{7}{x}}=1\)
Đặt \(a=4x+\frac{7}{x}-9\)
\(\Leftrightarrow\frac{4}{a+1}+\frac{5}{a-1}=1\)
\(\Leftrightarrow\frac{4\left(a-1\right)+5\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}=\frac{a^2-1}{a^2-1}\)
\(\Rightarrow9a+1=a^2-1\)
\(\Leftrightarrow a^2-9a-2=0\)
Tự giải tiếp
22 tháng 2 2019
b) \(\frac{x^4+4}{x^2-2}=5x\)
\(\Leftrightarrow x^4+4=5x\left(x^2-2\right)\)
\(\Leftrightarrow x^4+4-5x^3+10x=0\)
\(\Leftrightarrow x^4-2x^3-3x^3+6x^2-6x^2+12x-2x+4=0\)
\(\Leftrightarrow x^3\left(x-2\right)-3x^2\left(x-2\right)-6x\left(x-2\right)-2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-3x^2-6x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+x^2-4x^2-4x-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+1\right)-4x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-4x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
\(x^2-4x-2=0\)
\(\Leftrightarrow x^2-4x+4-6=0\)
\(\Leftrightarrow\left(x-2\right)^2=\left(\pm\sqrt{6}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}+2\\x=-\sqrt{6}+2\end{cases}}\)
Vậy....
5 tháng 6 2022
a: \(=\dfrac{5\left(x^2+2xy+y^2\right)}{3\left(x^3+y^3\right)}\)
\(=\dfrac{5\left(x+y\right)^2}{3\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{5\left(x+y\right)}{3\left(x^2-xy+y^2\right)}\)
b: \(=\dfrac{x^2-4xy+4y^2-4}{2x\left(x-2y+2\right)}=\dfrac{\left(x-2y-2\right)\left(x-2y+2\right)}{2x\left(x-2y+2\right)}\)
\(=\dfrac{x-2y-2}{2x}\)
c: \(=\dfrac{2\left(x^2+5x+1\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(4x^4-10x^3+8x^2-5x-1=0\)
\(\left(x^4-x^3+2x^2\right)-\left(4x^3-4x^2+8x\right)+\left(2x^2-2x+4\right)=0\)
\(x^2\left(x^2-x+2\right)-4x\left(x^2-x+2\right)+2\left(x^2-x+2\right)=0\)
\(\left(x^2-x+2\right)\left(x^2-4x+2\right)=0\)
\(\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]\left(x^2-4x+2\right)=0\)
Vì \(\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]>0\)\(\Rightarrow x^2-4x+2=0\)
\(\Rightarrow\left(x-2\right)^2=2\)\(\Rightarrow x-2=\pm\sqrt{2}\)
\(\Rightarrow\orbr{\begin{cases}x=\sqrt{2}+2\\x=2-\sqrt{2}\end{cases}}\)