Giải pt: x^2-4x+4=x-2
\((x+2)(1-4x^2)=x^2+4x+4\)
Giải pt
\(\Leftrightarrow\left(x+2\right)\left(1-4x^2\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(1-4x^2-x-2\right)=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
\(\left(x+2\right)\left(1-4x^2\right)=x^2+4x+4\\ \Leftrightarrow\left(x+2\right)\left(1-4x^2\right)-\left(x+2\right)^2\\ \Leftrightarrow\left(x+2\right)\left(1-4x^2-x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(-4x^2-x-1\right)=0\\ \Leftrightarrow\left(x+2\right)\left(4x^2+x+1\right)=0\\ \Leftrightarrow x+2=0\left(vì.4x^2+x+1>0\right)\\ \Leftrightarrow x=-2\)
\(\dfrac{(x-2)^3}{4}=x^2+4x+4\)
Giải pt
\(\Leftrightarrow\dfrac{\left(x-2\right)^3}{4}=\left(x+2\right)^2\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^3}{4}=\dfrac{4\left(x+2\right)^2}{4}\)
\(\Leftrightarrow\left(x-2\right)^3=4\left(x+2\right)^2\)
\(\Leftrightarrow x^3-6x^2+12x-8=4\left(x^2+4x+4\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8=4x^2+16x+16\)
\(\Leftrightarrow x^3-10x^2-4x-24=0\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^3}{4}=\left(x+2\right)^2\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^3}{4}=\dfrac{4\left(x+2\right)^2}{4}\)
\(\Leftrightarrow\left(x-2\right)^3=4\left(x+2\right)^2\)
\(\Leftrightarrow x^3-6x^2+12x-8=4\left(x^2+4x+4\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8=4x^2+8x+8\)
\(\Leftrightarrow x^3-10x^2+4x=0\)
\(\Leftrightarrow x\left(x^2-10x+4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2-10x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\\left\{{}\begin{matrix}x=5+\sqrt{21}\\x=5-\sqrt{21}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{0;5+\sqrt{21};5-\sqrt{21}\right\}\)
giải pt
\(\sqrt{x^2}=x\)
\(\sqrt{x^2-4x+4}=x-2\)
a: Ta có: \(\sqrt{x^2}=x\)
\(\Leftrightarrow\left|x\right|=x\)
hay \(x\ge0\)
b: Ta có: \(\sqrt{x^2-4x+4}=x-2\)
\(\Leftrightarrow\left|x-2\right|=x-2\)
\(\Leftrightarrow x\ge2\)
\(\sqrt{x^2}=x\Leftrightarrow\left|x\right|=x\Leftrightarrow x\ge0\)
\(\sqrt{x^2-4x+4}=x-2\left(x\in R\right)\\ \Leftrightarrow\left|x-2\right|=x-2\\ \Leftrightarrow x-2\ge0\Leftrightarrow x\ge2\)
Giải pt sau:
\(\dfrac{x}{x^2+4x+4}+\dfrac{5x}{x^2+4}=-2\)
\(ĐKXĐ:x\ne-2\)
Ta thấy x=0 ko là nghiệm của phương trình. Do đó \(x\ne0\)
\(\Rightarrow\dfrac{1}{\dfrac{x^2+4x+4}{x}}+\dfrac{5}{\dfrac{x^2+4}{x}}=-2\) (chia cả tử và mẫu của 2 phân số vế trái cho x )
\(\Leftrightarrow\dfrac{1}{x+\dfrac{4}{x}+4}+\dfrac{5}{x+\dfrac{4}{x}}=-2\)
Đặt \(x+\dfrac{4}{x}=t\) (\(t\ne0,t\ne-4\))
\(pt\) trở thành: \(\dfrac{1}{t+4}+\dfrac{5}{t}=-2\) \(\Rightarrow t+5\left(t+4\right)=-2\left(t+4\right)t\Leftrightarrow t+5t+20=-2t^2-8t\Leftrightarrow2t^2+14t+20=0\Leftrightarrow t^2+7t+10=0\) \(\Leftrightarrow\left(t+2\right)\left(t+5\right)=0\Leftrightarrow\left[{}\begin{matrix}t=-2\left(1\right)\\t=-5\left(2\right)\end{matrix}\right.\)
Từ (1) \(\Rightarrow x+\dfrac{4}{x}=-2\Rightarrow x^2+4=-2x\Leftrightarrow x^2+2x+4=0\Leftrightarrow\left(x+1\right)^2+3=0\left(VL\right)\)
Từ (2) \(\Rightarrow x+\dfrac{4}{x}=-5\Rightarrow x^2+4=-5x\Leftrightarrow x^2+5x+4=0\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(TM\right)\\x=-4\left(TM\right)\end{matrix}\right.\) Vậy...
l x^2+1 l - (x^2-4x+4) =3x
Giải pt
\(|x^2+1|-(x^2-4x+4)=3x\\\Rightarrow x^2+1-x^2+4x-4=3x(\text{vì }x^2 + 1 > 0 \forall x )\\\Leftrightarrow 4x-3=3x\\\Leftrightarrow4x-3x=3\\\Leftrightarrow x=3\)
Vậy nghiệm của phương trình là \(x=3\).
Do \(x^2+1>0;\forall x\Rightarrow\left|x^2+1\right|=x^2+1\)
Phương trình trở thành:
\(x^2+1-\left(x^2-4x+4\right)=3x\)
\(\Leftrightarrow4x-3=3x\)
\(\Leftrightarrow x=3\)
Giải pt: x^4 - 4x^3 + 6x^2 - 4x - 15=0
Tham khảo:
Giải phương trình \(x^4-4x^3+6x^2-4x-15=0\) - Hoc24
\(\Leftrightarrow x^4-3x^3-x^3+3x^2+3x^2-9x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x^3-x^2+3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^3+x^2-2x^2-2x+5x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x^2-2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\\left(x-1\right)^2+4=0\left(vn\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Giải pt \(\sqrt{-x^2+4x-3}+\sqrt{-2x^2+8x+1}=x^3-4x^2+4x+4\)
giải pt :
a, \(x^2-4x-2=2\sqrt{x^3+1}\)
b, \(x^2-7x+1=4\sqrt{x^4+x^2+1}\)
c, \(3\sqrt{x^2+4x-5}+\sqrt{x-3}=\sqrt{11x^2+25+2}\)
Giải pt
(4x-3)^2-(2x+1)^2=0
3x-12-5x×(x-4)=0
(8x+2)×(x^2+5)×(x^2-4)=0
(4x - 3)2 - (2x + 1)2 = 0
\(\Leftrightarrow\) (4x - 3 - 2x - 1)(4x - 3 + 2x + 1) = 0
\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
3x - 12 - 5x(x - 4) = 0
\(\Leftrightarrow\) 3x - 12 - 5x2 + 20x = 0
\(\Leftrightarrow\) -5x2 + 23x - 12 = 0
\(\Leftrightarrow\) 5x2 - 23x + 12 = 0
\(\Leftrightarrow\) 5x2 - 20x - 3x + 12 = 0
\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0
\(\Leftrightarrow\) (x - 4)(5x - 3) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy ...
(8x + 2)(x2 + 5)(x2 - 4) = 0
\(\Leftrightarrow\) (8x + 2)(x2 + 5)(x - 2)(x + 2) = 0
Vì x2 \(\ge\) 0 \(\forall\) x nên x2 + 5 > 0 \(\forall\) x
\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=2\\x=-2\end{matrix}\right.\)
Vậy ...
Chúc bn học tốt!
a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3-2x-1\right)\left(4x-3+2x+1\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)
b) Ta có: \(3x-12-5x\left(x-4\right)=0\)
\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)
c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)
mà \(2>0\)
và \(x^2+5>0\forall x\)
nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-1\\x=2\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)