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2x5 - 7x4 + 5x3 + 5x2 - 7x + 2 = 0
<=> 2x5-4x4-3x4+6x3-x3+2x2+3x2-6x-x+2=0
<=> 2x4(x-2)-3x3(x-2)-x2(x-2)+3x(x-2)-(x-2)=0
<=>(x-2)(2x4-3x3-x2+3x-1)=0
<=>(x-2)(2x4-x3-2x3+x2-2x2+x+2x-1)=0
<=>(x-2)[x3(2x-1)-x2(2x-1)-x(2x-1)+2x-1]=0
<=>(x-2)(2x-1)(x3-x2-x+1)=0
<=>(x-2)(2x-1)[x2(x-1)-(x-1)]=0
<=>(x-2)(2x-1)(x-1)(x2-1)=0
<=>(x-2)(2x-1)(x-1)2(x+1)=0
=> x-2=0 => x=2
hoặc 2x-1=0=>x=1/2
hoặc x-1=0=>x=1
hoặc x+1=0=>x=-1
Vậy...
\(2x^5-7x^4+5x^3+5x^2-7x+2=0\)
\(\Leftrightarrow\left(2x^5-4x^4+2x^3\right)-\left(3x^4-6x^3+3x^2\right)-\left(3x^3-6x^2+3x\right)+\left(2x^2-4x+2\right)=0\)
\(\Leftrightarrow2x^3\left(x^2-2x+1\right)-3x^2\left(x^2-2x+1\right)-3x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(2x^3-3x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(2x^3+2x^2-5x^2-5x+2x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left[2x^2\left(x+1\right)-5x\left(x+1\right)+2\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\left(2x^2-5x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\left(2x^2-4x-x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\left[2x\left(x-2\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\)\(x-1=0\)
hoặc \(x+1=0\)
hoặc \(x-2=0\)
hoặc \(2x-1=0\)
\(\Leftrightarrow\)\(x=1\)
hoặc \(x=-1\)
hoặc \(x=2\)
hoặc \(x=\frac{1}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;-1;2;\frac{1}{2}\right\}\)
Bạn tự phân tích đa thức thành nhân tử nhé!
\(1.\)
\(2x^3+x+3=0\)
\(\Leftrightarrow\) \(\left(x+1\right)\left(2x^2-2x+3\right)=0\) \(\left(1\right)\)
Vì \(2x^2-2x+3=2\left(x^2-x+1\right)+1=2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}>0\) với mọi \(x\in R\)
nên từ \(\left(1\right)\) \(\Rightarrow\) \(x+1=0\) \(\Leftrightarrow\) \(x=-1\)
\(1;x^2+7x+10=0\Rightarrow x^2+2x+5x+10=0\Rightarrow x\left(x+2\right)+5\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x+5\right)=0\)
=> x + 2 = 0 hoặc x + 5 = 0
=> x = -2 hoặc x = - 5
2, x^4 - 5x^2 + 4 = 0
x^4 - 4x^2 - x^2 + 4 = 0
x^2 ( x^2 - 4) - ( x^2 - 4) = 0
( x^2 - 1)( x^2 - 4) = 0
( x - 1 )( x + 1)( x - 2)( x + 2) = 0
=> x= 1 hoặc x= -1 hoặc x = 2 hoặc x = - 2
Đúng cho mi8nhf mình giải tiếp cho
1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)
a) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
b) Ta có: \(-x^2+5x-6=0\)
\(\Leftrightarrow-\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow-\left(x^2-2x-3x+6\right)=0\)
\(\Leftrightarrow-\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow-\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow-\left[\left(x-2\right)\left(x-3\right)\right]=0\)
\(\Leftrightarrow-\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: x∈{2;3}
c) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
⇔(4x2-10x)-(2x-5)=0
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
d) Ta có: \(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow\left(2x^2+2x\right)+\left(3x+3\right)=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{-3}{2}\right\}\)
e) Ta có: \(x^3+2x^2-x-2=0\)
\(\Leftrightarrow\left(x^3+2x^2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;1;-1\right\}\)
g) Ta có: \(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow9x^2-6x+1-20x^2-20x-5+12x^2-3-x^2+2x-1=0\)
\(\Leftrightarrow-24x-8=0\)
\(\Leftrightarrow-8\left(3x+1\right)=0\)
⇔3x+1=0
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy: \(x=-\frac{1}{3}\)
h) \(2x^3-7x^2+7x-2=0\)
\(\Leftrightarrow2x^3-4x^2-3x^2+6x+x-2=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-2x-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy S = {2; 1; \(\frac{1}{2}\)}
i) \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\end{matrix}\right.\)
Vậy S = {1;-2}
A=x4_x3+3x3-3x2+8x2-8x+12x-12=0
A= x3.(x-1) +3x2.(x-1)+8x(x-1)+12(x-1)=0
A= (x-1)(x3+3x2+8x+12)=0
A=(x-1)(x3+2x2+x2+2x+6x+12)=0
A=(x-1)(x2(x+2)+x(x+2)+6(x+2))=0
A=(x-1)(x+2)(x2+x+6)=0 vi x2+x+6 >0
suy ra A=0 <=> x-1=0 hoac x+2=0 <=> x=1 hoac x=-2
vay S={-2;1} Hoc tot nha !
a)\(9x^2+5x+2=0\)
\(\Delta=5^2-4\cdot9\cdot2=-47< 0\)
Vô nghiệm
b)\(5x^2+4x-2=0\)
\(\Delta=4^2-4\cdot5\cdot\left(-2\right)=56\)
\(x_{1,2}=\frac{-4\pm\sqrt{56}}{10}\)
c)\(2x^3+7x^2+7x+2=0\)
\(\Rightarrow2x^3+6x^2+4x+x^2+3x+2=0\)
\(\Rightarrow2x\left(x^2+3x+2\right)+\left(x^2+3x+2\right)=0\)
\(\Rightarrow\left(x^2+3x+2\right)\left(2x+1\right)=0\)
\(\Rightarrow\left(x^2+2x+x+2\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[x\left(x+2\right)+\left(x+2\right)\right]\left(2x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(2x+1\right)=0\)
=>x=-1 hoặc x=-2 hoặc \(x=-\frac{1}{2}\)
1, \(x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+4x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+4x^2+4x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+3x^2+x^2+3x+x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2+x+1>0\right)\left(x+3\right)\left(x-2\right)=0\Leftrightarrow x=-3;x=2\)
2, \(2\left(x^3-1\right)-7x\left(x-1\right)=0\)
\(\Leftrightarrow2\left(x-1\right)\left(x^2+x+1\right)-7x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2-5x+2\right)=0\Leftrightarrow x=1;x=\dfrac{1}{2};x=2\)