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\(a,x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(b,x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow\left(x-2\right)x\left(x^2+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=0\\x^2+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\\left[{}\begin{matrix}x^2=10\\x^2=-10\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\x=\sqrt{10}\\x=-\sqrt{10}\end{matrix}\right.\)\(c,\left(2x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow4x^2-4x+1=x^2+6x+9\)
\(\Leftrightarrow4x^2-4x+1-x^2-6x-9=0\)
\(\Leftrightarrow3x^2-10x-8=0\)
\(\Leftrightarrow3x^2-12x+2x-8=0\)
\(\Leftrightarrow3x\left(x-4\right)+2\left(x-4\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Phần d tương tự
Câu a :
\(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-4^2\right)=0\)
\(\Leftrightarrow x\left[\left(x+4\right)\left(x-4\right)\right]=0\)
\(\Rightarrow\) \(x=0\)
\(\Rightarrow\) \(x+4=0\Rightarrow x=-4\)
\(\Rightarrow x-4=0\Rightarrow x=4\)
Câu b :
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)\) \(=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Rightarrow x=0\)
\(\left(x-2\right)=0\Rightarrow x=2\)
\(x^2+10=0\) \(\Rightarrow\) x ( loại )
=>x=0 hoặc x=2 hoặc x= - căn 10
x3-5x2+x-5=0
=> x2.(x-5)+(x-5)=0
=> (x-5).(x2+1)=0
=> x-5=0 hoặc x2+1=0
=> x=5 hoặc x2=-1 (vô lí)
Vậy x=5.
x4-2x3+10x2-20x=0
=> x3.(x-2)+10x(x-2)=0
=> (x-2).(x3+10x)=0
=> x.(x-2).(x2+10)=0
=> x=0 hoặc x-2=0 hoặc x2+10=0
=> x=0 hoặc x=2 hoặc x2=-10 (vô lí)
Vậy x=0 hoặc x=2.
Phân tích đa thức thành nhân tử:
a) \(3a^2-3ab+9b-9a=3a\left(a-b\right)+9\left(b-a\right)=3\left(a-b\right)\left(a-3\right)\)
b) \(2xm^3-2m=2m\left(xm^2-1\right)\)
c) \(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
Tìm x:
a) \(8x^2+10x+3=0\)
\(\Leftrightarrow8x^2+12x-2x-3=0\Leftrightarrow4x\left(2x+3\right)-\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{1}{4}\end{array}\right.\)
b) \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)
1)
a) \(x^3-5x^2+x-5=0\Rightarrow x^2.\left(x-5\right)+\left(x-5\right)\)
\(\Rightarrow\left(x^2+1\right).\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(sai\right)\\x=5\end{cases}}\)\(KL:x=5\)
b) \(x^4-2x^3+10x^2-20x=0\Rightarrow x^3.\left(x-2\right)+10x\left(x-2\right)\)
\(\Rightarrow\left(x-2\right).\left(x^3+10x\right)\Rightarrow\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x\left(x^2+10\right)=0\Rightarrow x=0\end{cases}}\)
Vì nếu x2 + 10 = 0 => x2 = -10 ( sai )
Vậy...
Giải:
a) \(125-x^2\)
\(=\left(5\sqrt{5}\right)^2-x^2\)
\(=\left(5\sqrt{5}-x\right)\left(5\sqrt{5}+x\right)\)
Vậy ...
b) \(x^4-2x^3-10x^2+20x=0\)
\(\Leftrightarrow\left(x^4-2x^3\right)-\left(10x^2-20x\right)=0\)
\(\Leftrightarrow x^3\left(x-2\right)-10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2-10\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-\sqrt{10}\right)\left(x+\sqrt{10}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-\sqrt{10}=0\\x+\sqrt{10}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\sqrt{10}\\x=-\sqrt{10}\end{matrix}\right.\)
Vậy ...
(Câu b mình làm không ra nên sửa đề, nếu đề đúng thì mình xin lỗi vì không làm được)
a)
\(=x^2\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
b)
\(=a\left(a-b\right)+a-b\)
\(=\left(a+1\right)\left(a-b\right)\)
c)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left(x+1-y\right)\left(x+1+y\right)\)
d)
\(=x^3\left(x-2\right)+10x\left(x-2\right)\)
\(=x\left(x^2+10\right)\left(x-2\right)\)
e)
\(=x\left(x^2+2x+1\right)\)
\(=x\left(x+1\right)^2\)
f)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right)\left(x+y\right)\)
a,2x3+3x2+2x+3
=(2x3+2x)+(3x2+3)
=2x(x2+1)+3(x2+1)
=(x2+1)(2x+3)
b,a2-ab+a-b
=(a2-ab)+(a-b)
=a(a-b)+(a-b)
=(a-b)(a+1)
c,2x2+4x+2-2y2
=2(x2+2x+1-y2)
=2[(x2+2x+1)-y2 ]
=2[(x+1)2-y2 ]
=2(x+1-y)(x+1+y)
d,x4-2x3+10x2-20x
=(x4-2x3)+(10x2-20x)
=x3(x-2)+10x(x-2)
=(x-2)(x3+10x)
=(x-2)[x(x2+10)]
e,x3+2x2+x
=x(x2+2x+1)
=x(x+1)2
f,xy+y2-x-y
=(xy+y2)-(x-y)
=y(x+y)-(x+y)
=(x+y)(y-1)
