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Ví dụ cho bạn một bài, còn lại tương tự.
a)Ta có: \(3x^4-5x^3+8x^2-5x+3\)
\(=3x^2\left(x-\frac{5}{6}\right)^2+\frac{71}{12}\left(x-\frac{30}{71}\right)^2+\frac{138}{71}>0\)
Vậy phương trình vô nghiệm.
Bài 1 :
a) (3a+4b)3+(3a-4b)3-48a2b2
=27a3+108a2b+144ab2+64b3+27a3-108a2b+144ab2-64b3-48a2b2
=54a3+288ab2-48a2b2
=2a(27a2+144b2-24ab)
b) (5x+2y)(5x-2y)+(2x-y)3+(2x+y)3
=25x2-4y2+8x3-12x2y+6xy2-y3+8x3+12x2y+6xy2+y3
=16x3+25x2-y2+12xy2
=x2(16x+25)-y2(1-12x)
Bài 2 :
\(x^2-8x+7=0\)
\(\Leftrightarrow x^2-x-7x+7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
b)\(x^3-4x^2+3x=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{3}\\x=1\end{cases}}\)
c)Nếu đề đổi thành =1 thì có vẻ hợp lí hơn
d)\(\left(3x-1\right)^3-3\left(3x+2\right)^2+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-3\left(9x^2+12x+4\right)+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-27x^2-36x-12+13=0\)
\(\Leftrightarrow27x^3-54x^2-27x=0\)
\(\Leftrightarrow27x\left(x^2-2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}27x=0\\x^2-2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\-\left(x^2+2x+1\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\-\left(x+1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
#H
1) \(x^4-6x^3-x^2+54x-72=0\)
\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
Tự làm nốt...
2) \(x^4-5x^2+4=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
Tự làm nốt...
\(x^4-2x^3-6x^2+8x+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
...
\(2x^4-13x^3+20x^2-3x-2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)
Bí
a) Gần giống cho nó giống luôn.
cần thêm (-x^3+2x^2-x) là giống
\(\left(x-1\right)^4+x^3-2x^2+x=\left(x-1\right)^4+x\left(x^2-2x+1\right)=\left(x-1\right)^4+x\left(x-1\right)^2\)
\(\left(x-1\right)^2\left[\left(x-1\right)^2+x\right]\)
\(\left[\begin{matrix}x-1=0\Rightarrow x=0\\\left(x-1\right)^2+x=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\end{matrix}\right.\)
Nghiệm duy nhất: x=1
1, a,\(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\)
Từ đó suy ra \(x=-\dfrac{5}{2}\) hoặc \(x=3\)
b, \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\left(x-2\right)\left(3x-1\right)=0\)
Từ đó suy ra \(x=2\) hoặc \(x=\dfrac{1}{3}\)
c, \(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\)
Áp dụng hằng đẳng thức hiệu hai bình phương để suy ra:
\(\Leftrightarrow\left(3x+7\right)\left(x+3\right)=0\)
Từ đó suy ra \(x=-\dfrac{7}{3}\) hoặc \(x=-3\)
d, \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-4x+4-x+2=0\)
\(\Leftrightarrow\left(x-2\right)^2-\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
Từ đó suy ra \(x=2\) hoặc \(x=3\)
e, \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\)
Từ đó suy ra \(x=0\) hoặc \(x=\dfrac{1}{2}\) hoặc \(x=-3\)
CHÚC BẠN HỌC GIỎI.................
a/ \(x\left(x^2-2x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\pm\sqrt{3}\\\end{matrix}\right.\)
b/ \(\Leftrightarrow2x^3-4x^2+6x-x^2+2x-3=0\)
\(\Leftrightarrow2x\left(x^2-2x+3\right)-\left(x^2-2x+3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-2x+3\right)=0\)
c/ \(\Leftrightarrow3x^3-15x^2+9x+x^2-5x+3=0\)
\(\Leftrightarrow3x\left(x^2-5x+3\right)+\left(x^2-5x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x^2-5x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=\frac{5\pm\sqrt{13}}{2}\end{matrix}\right.\)
d/ \(x\left(x^2+6x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\pm\sqrt{14}\end{matrix}\right.\)
a/ \(x^4-4x^2+4+x^2-6x+9=0\)
\(\Leftrightarrow\left(x^2-2\right)^2+\left(x-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=2\\x=3\end{matrix}\right.\) \(\Rightarrow\) pt vô nghiệm
b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(3x^2+\frac{3}{x^2}-5x-\frac{5}{x}+8=0\)
\(\Leftrightarrow3\left(x+\frac{1}{x}\right)^2-5\left(x+\frac{1}{x}\right)+2=0\)
\(\Leftrightarrow\left(x+\frac{1}{x}-1\right)\left(3x+\frac{3}{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{x}-1=0\\3x+\frac{3}{x}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\3x^2-2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{1}{x}\right)^2+\frac{3}{4}=0\\3\left(x-\frac{1}{x}\right)^2+\frac{8}{3}=0\end{matrix}\right.\)
Cả 2 pt đều vô nghiệm nên pt đã cho vô nghiệm
\(3\left(x^2+\frac{1}{x}\right)^2-3\left(x+\frac{1}{x}\right)-2\left(x+\frac{1}{x}\right)+2=0\)
\(\Leftrightarrow3\left(x+\frac{1}{x}\right)\left(x+\frac{1}{x}-1\right)-2\left(x+\frac{1}{x}-1\right)=0\)
\(\Leftrightarrow\left(3x+\frac{3}{x}-2\right)\left(x+\frac{1}{x}-1\right)=0\)
Mà thực chất ko phải vậy đâu, đặt \(x+\frac{1}{x}=t\Rightarrow3t^2-5t+2=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=\frac{3}{2}\end{matrix}\right.\) (casio) rồi thay ngược lại :D