Giair phương trình:\(2x^2-5x+5=\sqrt{5x-1}\)
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1 cách ngu học
\(\left(2x+2\right)\sqrt{5x-6}=x^2+7x-6\)
\(\Leftrightarrow4.\left(x+1\right)^2.\left(5x-6\right)=\left(x^2+7x-6\right)^2\)
\(\Leftrightarrow20x^3-24x^2+40x^2-48x+20x-24=\left(x^2+7x-6\right)^2\)
\(\Leftrightarrow20x^3+16x^2-28x-24=\left(x^2+7x-6\right)^2\)
\(\Leftrightarrow20x^3+16x^2-28x-24-\left(x^2+7x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
pt <=> (x^5-x^4)-(4x^4-4x^3)+(4x^2-4x)-(x-1) = 0
<=> (x-1).(x^4-4x^3+4x-1) = 0
<=> (x-1).[(x^4-x^3)-(3x^3-3x)+(x-1)] = 0
<=> (x-1).(x-1).(x^3-3x^2-3x+1) = 0
<=>(x-1)^2.[(x^3+x^2)-(4x^2+4x)+(x+1)] = 0
<=> (x-1)^2.(x+1).(x^2-4x+1) = 0
<=> x-1=0 hoặc x+1=0 hoặc x^2-4x+1=0
<=> x=1 hoặc x=-1 hoặc x=2+\(\sqrt{3}\)hoặc x = 2-\(\sqrt{3}\)
k mk nha
ĐKXĐ: \(x\ge\dfrac{1}{5}\)
\(\Leftrightarrow\sqrt{3x+5}-\sqrt{2x+6}+\sqrt{5x-1}-2=0\)
\(\Leftrightarrow\dfrac{x-1}{\sqrt{3x+5}+\sqrt{2x+6}}+\dfrac{5\left(x-1\right)}{\sqrt{5x-1}+2}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{3x+5}+\sqrt{2x+6}}+\dfrac{5}{\sqrt{5x-1}+2}\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
đk x>=1/5
\(2\left(x^2-3x+2\right)+\left(x+1\right)=\sqrt{5x-1}\)
\(\Leftrightarrow2\left(x^2-3x+2\right)+\frac{\left(x+1\right)^2-\left(\sqrt{5x-1}\right)}{x+1+\sqrt{5x-1}}=0\)
\(\Leftrightarrow2\left(x^2-3x+2\right)+\frac{x^2-3x+2}{x+1+\sqrt{5x-1}}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(2+\frac{1}{x+1+\sqrt{5x-1}}\right)=0\)
\(\Leftrightarrow x^2-3x+2=0\left(do.\frac{1}{x+1+\sqrt{5x-1}}+2>0.với.mọi.x\ge\frac{1}{5}\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Ta có:
\(\frac{x^2-5x+1}{2x+1}+2=\frac{x^2-5x+4x+1+2}{2x+1}\)
=\(\frac{x^2-x+3}{2x+1}=\frac{x^2-4x+1}{x+1}\)
=> (x2 - x +3)(x+1)=(x2 - 4x+1)(2x+1)
=>x3 +2x+3=2x3-7x2-2x+1
=>0=x3-7x2-4x-2
Đây là cách làm của mình :
\(\Leftrightarrow\frac{x^2-5x+1}{2x+1}+1+1=\frac{x^2-4x+1}{x+1}\)
\(\Leftrightarrow\frac{x^2-5x+1}{2x+1}+1=\frac{x^2-4x+1}{x+1}-1\)
\(\Leftrightarrow\frac{x^2-3x+2}{2x+1}=\frac{x^2-5x}{x+1}\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{2x+1}=\frac{x^2-5x}{x+1}\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=\left(2x+1\right)\left(x^2-5x\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-1\right)=\left(2x+1\right)\left(x^2-5x\right)\)
Bạn tự nhân phân phối vào nha :
\(\Leftrightarrow x^3-2x^2-x+2=2x^3-9x^2-5x\)
\(\Leftrightarrow x^3-7x^2-4x-2=0\)
Đến đây chỉ có nước bấm máy tính thôi chứ phân tích bình thường không ra được đâu
CASIO fx-570VN PLUS : Mode --> 5 --> 4 : giải pt bậc 3 một ẩn
Kết quả cho là x = 7.563793497...
\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)
\(ĐKXĐ:x\ne-1;x\ne-\frac{1}{2}\)
\(PT:\Leftrightarrow\frac{x^2-4x+1}{x+1}+1+\frac{x^2-5x+1}{2x+1}=0\)
\(\Leftrightarrow\frac{x^2-3x+2}{x+1}+\frac{x^2-3x+2}{2x+1}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(\frac{1}{x+1}+\frac{1}{2x+1}\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(3x+2\right)=0\)
\(x-1=0\Leftrightarrow x=1\)
\(x-2=0\Leftrightarrow x=2\)
\(3x+2=0\Leftrightarrow3x=-2\Leftrightarrow x=-\frac{2}{3}\)
\(\Rightarrow\hept{\begin{cases}x=1\\x=2\\x=-\frac{2}{3}\end{cases}}\)
\(\frac{x^2-4x+1}{x+1}+2=-\frac{x^2-5x+1}{2x+1}\)
\(\Leftrightarrow\left(x^2-4x+1\right)\left(x+1\right)+2\left(x+1\right)\left(2x+1\right)=-\left(x^2-5x+1\right)\left(x+1\right)\)
\(\Leftrightarrow2x^3-3x^2+4x+3=-x^3+4x^2+4x-1\)
\(\Leftrightarrow2x^3-3x^2+3+x^2-4x+1=0\)
\(\Leftrightarrow3x^2-7x^2+4=0\)
\(\Leftrightarrow\left(3x^2-4x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x^2+2x-6x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x\left(3x+2\right)-2\left(3x+2\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}3x+2=0\\x-2=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\x=2\\x=1\end{cases}}\)
vậy:...
bình phương 2 vế rồi giải phương trình bậc 4 là ra
2x^2 - 5x + 5 = \(\sqrt{5x-1}\)
<=> (2x^2 - 5x + 5)^2 = (\(\sqrt{5x-1}\))^2
<=> 4x^4 - 20x^3 + 45x^2 - 50x + 25 = 5x - 1
<=> 4x^4 - 20x^3 + 45x^2 - 50x + 25 - 5x + 1 = 0
<=> 4x^4 - 20x^3 + 45x^2 - 55x + 26 = 0
<=> (4x^3 - 16x^2 + 29x - 26)(x - 1) = 0
<=> (4x^2 - 8x + 13)(x - 2)(x - 1) = 0
mà 4x^2 - 8x + 16 # 0 nên:
(x - 2)(x - 1) =0
=> x = 2; x = 1