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1 cách ngu học
\(\left(2x+2\right)\sqrt{5x-6}=x^2+7x-6\)
\(\Leftrightarrow4.\left(x+1\right)^2.\left(5x-6\right)=\left(x^2+7x-6\right)^2\)
\(\Leftrightarrow20x^3-24x^2+40x^2-48x+20x-24=\left(x^2+7x-6\right)^2\)
\(\Leftrightarrow20x^3+16x^2-28x-24=\left(x^2+7x-6\right)^2\)
\(\Leftrightarrow20x^3+16x^2-28x-24-\left(x^2+7x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
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...
\(ĐKXĐ:x\ne-1;x\ne-\frac{1}{2}\)
\(PT:\Leftrightarrow\frac{x^2-4x+1}{x+1}+1+\frac{x^2-5x+1}{2x+1}=0\)
\(\Leftrightarrow\frac{x^2-3x+2}{x+1}+\frac{x^2-3x+2}{2x+1}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(\frac{1}{x+1}+\frac{1}{2x+1}\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(3x+2\right)=0\)
\(x-1=0\Leftrightarrow x=1\)
\(x-2=0\Leftrightarrow x=2\)
\(3x+2=0\Leftrightarrow3x=-2\Leftrightarrow x=-\frac{2}{3}\)
\(\Rightarrow\hept{\begin{cases}x=1\\x=2\\x=-\frac{2}{3}\end{cases}}\)
\(\frac{x^2-4x+1}{x+1}+2=-\frac{x^2-5x+1}{2x+1}\)
\(\Leftrightarrow\left(x^2-4x+1\right)\left(x+1\right)+2\left(x+1\right)\left(2x+1\right)=-\left(x^2-5x+1\right)\left(x+1\right)\)
\(\Leftrightarrow2x^3-3x^2+4x+3=-x^3+4x^2+4x-1\)
\(\Leftrightarrow2x^3-3x^2+3+x^2-4x+1=0\)
\(\Leftrightarrow3x^2-7x^2+4=0\)
\(\Leftrightarrow\left(3x^2-4x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x^2+2x-6x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x\left(3x+2\right)-2\left(3x+2\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}3x+2=0\\x-2=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\x=2\\x=1\end{cases}}\)
vậy:...
Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+5x+12}=a>0\\\sqrt{2x^2+3x+2}=b>0\end{matrix}\right.\) \(\Rightarrow x+5=\dfrac{a^2-b^2}{2}\)
Phương trình trở thành:
\(a+b=\dfrac{a^2-b^2}{2}\)
\(\Leftrightarrow\left(a-b-2\right)\left(a+b\right)=0\)
\(\Leftrightarrow a-b-2=0\) (do \(a+b>0\))
\(\Leftrightarrow a=b+2\)
\(\Leftrightarrow\sqrt{2x^2+5x+12}=\sqrt{2x^2+3x+2}+2\)
\(\Leftrightarrow2x^2+5x+12=2x^2+3x+6+4\sqrt{2x^2+3x+2}\)
\(\Leftrightarrow x+3=2\sqrt{2x^2+3x+2}\) (\(x\ge-3\))
\(\Leftrightarrow x^2+6x+9=4\left(2x^2+3x+2\right)\)
\(\Leftrightarrow7x^2+6x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{7}\end{matrix}\right.\)
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
Mấy ý này bản chất ko khác nhau nhé, mình làm mẫu, bạn làm tương tự mấy ý kia nhé
a, \(\left|5x\right|=x+2\)
Với \(x\ge0\)thì \(5x=x+2\Leftrightarrow x=\dfrac{1}{2}\)
Với \(x< 0\)thì \(5x=-x-2\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\)
b, \(\left|7x-3\right|-2x+6=0\Leftrightarrow\left|7x-3\right|=2x-6\)
Với \(x\ge\dfrac{3}{7}\)thì \(7x-3=2x-6\Leftrightarrow5x=-3\Leftrightarrow x=-\dfrac{3}{5}\)( ktm )
Với \(x< \dfrac{3}{7}\)thì \(7x-3=-2x+6\Leftrightarrow9x=9\Leftrightarrow x=1\)( ktm )
Vậy phương trình vô nghiệm
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