điền vào chỗ trống:
(x-2).(6x^2-5x+1)
=x.6x^2+x.(-5x)+x.1+(-2).6x^2+(-2)(-5x)+(-2).1
=6x^3-5x^2+.....-12x^2+......-2
=6x^3-.......11x-2
phân tích đa thức
x^4+6x^3+11x^2+6x+1
x^4+x^3+x^2+x+1
6x^4+5x^3-38x^2+5x+6
x^4+5x^3-12x^2+5x+1
a)\(x^4+6x^3+11x^2+6x+1\)
\(=x^4+9x^2+1+6x^3+6x+2x^2\)
\(=\left(x^2+3x+1\right)^2\)
\(x^4+5x^3-12x^2+5x+1\)
\(=\left(x^4-2x^3+x^2\right)+\left(7x^3-14x^2+7x\right)+\left(x^2-2x+1\right)\)
\(=x^2\left(x^2-2x+1\right)+7x\left(x^2-2x+1\right)+\left(x^2-2x+1\right)\)
\(=\left(x^2+7x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x^2+7x+1\right)\left(x-1\right)^2\)
phân tích đa thức
a)x^4+6x^3+11x^2+6x+1
b)x^4+x^3+x^2+x+1
c)6x^4+5x^3-38x^2+5x+6
d)x^4+5x^3-12x^2+5x+1
dễ mà bạn xin 20 phút làm ra giấy nhé :))
a) \(\left(x^4+6x^3+9x^2\right)+2x^2+6x+1\)
\(\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(\left(x^2+3x+1\right)^2\)
b) \(x^4+x^3+x^2+x+1\)
câu b, chúa sẽ c/m x ko tồn tại , và nó là 1 đa thức bất khả Q . trong R
vì lớp 8 chưa học đến số phức
\(x^4+x^3=-x^2-x-1\)
\(x^4+x^3+\frac{1}{4}x^2=\left(\frac{1}{4}x^2-x^2\right)-x-1\)
\(\left(x^2+\frac{1}{2}x\right)^2=-\frac{3}{4}x^2-x-1\)
\(4\left(x^2+\frac{1}{2}x\right)^2=-3x^2-4x-4\)
\(\Delta`=\left(-2\right)^2-\left(-4\right).\left(-3\right)=4-12< 0\)
denta < 0 x vô nghiệm
vậy đa thức trên ko thể phân tích và nó là 1 đa thức bất khả Q
c) ,
\(\left(6x^4-12x^3\right)+\left(17x^3-34x^2\right)-\left(4x^2-8x\right)-\left(3x-6\right)\)
\(6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)
\(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)\)
\(\left(x-2\right)\left\{\left(6x^3+18x^2\right)-\left(x^2+3x\right)-\left(x+3\right)\right\}\)
\(\left(x-2\right)\left\{6x^2\left(x+3\right)-x\left(x+3\right)-\left(x+3\right)\right\}\)
\(\left(x-2\right)\left(x+3\right)\left(6x^2-x-1\right)\)
\(\left(x-2\right)\left(x+3\right)\left\{\left(6x^2+\frac{6}{3}x\right)-\left(\frac{9}{3}x+\frac{9}{9}\right)\right\}\)
\(\left(x-2\right)\left(x+3\right)\left\{6x\left(x+\frac{1}{3}\right)-\frac{9}{3}\left(x+\frac{1}{3}\right)\right\}\)
\(\left(X-2\right)\left(X+3\right)\left(X+\frac{1}{3}\right)\left(6x-1\right)\)
d)
\(\left(x^4-x^3\right)+\left(6x^3-6x^2\right)-\left(6x^2-6x\right)-\left(x-1\right)\)
\(x^3\left(x-1\right)+6x^2\left(x-1\right)-6x\left(x-1\right)-\left(x-1\right)\)
\(\left(x-1\right)\left(x^3+6x^2-6x-1\right)\)
\(\left(x-1\right)\left\{\left(x^3-x^2\right)+\left(7x^2-7x\right)+\left(x-1\right)\right\}\)
\(\left(x-1\right)^2\left(x^2+7x+1\right)\)
\(\Delta=49-4=45\)
\(x1,2=\frac{-7+\sqrt{45}}{2},\frac{-7-\sqrt{45}}{2}\)
\(\left(x-1\right)^2\left(x-\frac{7+\sqrt{45}}{2}\right)\left(x-\frac{7-\sqrt{45}}{2}\right)\)
3x3 - 5x2 + 2 - 3x(2x2 - 1) + 6(x2 - 2x +3) = x2 - 5x
<=> 3x3 - 5x2 + 2 - 3x.2x2 + 3x + 6x2 - 6.2x + 6.3 = x2 - 5x
<=> 3x3 - 5x2 + 2 - 6x3 + 3x +6x2 - 12x + 18 - x2 + 5x = 0
<=> (3x3 - 6x3) + (6x2 - 5x2 - x2) - (12x - 3x - 5x) + (18 + 2 ) = 0
<=> ...
rút gọn biểu thức
(6x +1) ^2 +(6x-1)^2-2(1+6x)(6x-1)
x(2x^2-3)-x^2(5x+1)+x^2
3x(x-2)-5x(1-x)-8(x^2-3)
\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
\(=4\)
\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=\left(2x^3-5x^3\right)+\left(x^2-x^2\right)-3x\)
\(=-3x^3-3x\)
\(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2+5x^2-8x^2\right)-\left(6x+5x\right)+24\)
\(=-11x+24\)
rút gọn biểu thức
(6x +1) ^2 +(6x-1)^2-2(1+6x)(6x-1)
x(2x^2-3)-x^2(5x+1)+x^2
3x(x-2)-5x(1-x)-8(x^2-3)
rút gọn biểu thức
(6x +1) ^2 +(6x-1)^2-2(1+6x)(6x-1)
x(2x^2-3)-x^2(5x+1)+x^2
3x(x-2)-5x(1-x)-8(x^2-3)
rút gọn biểu thức
(6x +1) ^2 +(6x-1)^2-2(1+6x)(6x-1)
x(2x^2-3)-x^2(5x+1)+x^2
3x(x-2)-5x(1-x)-8(x^2-3)
a) (6x+1)2 + (6x-1)2 - 2(1+6x)(6x-1)
= (6x+1)2 - 2(6x+1)(6x-1) + (6x-1)2
= [ (6x+1) - (6x-1) ]2 = 22 = 4
b) x(2x2 - 3) - x2(5x+1) + x2
= 2x3 - 3x -5x3 -x2 + x2
= -3x3 - 3x = -3x(x2 -1) = -3x(x-1)(x+1)
c) 3x(x-2) - 5x(1-x) - 8(x2 -3)
= 3x2 - 6x - 5x + 5x2 -8x2 + 24
= -11x +24
phan tich da thuc thanh nhan tu : a) 3x^2 - 22xy + 4x + 8y + 7x^2 + 1 ; b) 12x^2 + 5x - 12y^2 + 12y - 10xy - 3 ; c)x^4 + 6x^3 + 11x^2 + 6x + 1
Giai phường trình sau:
a, \(3x^2+2x-1=0\) e, \(4x^2-12x+5=0\) i,\(2x^2+5x-3=0\)
b,\(x^2-5x+6=0\) f, \(2x^2+5x+3=0\) j,\(x^2+6x-16=0\)
c,\(x^2-3x+2=0\) g,\(x^2+x-2=0\)
d,\(2x^2-6x+1=0\) h, \(x^2-4x+3=0\)
a) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)
b) Ta có: \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: S={2;3}
c) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: S={1;2}
d) Ta có: \(2x^2-6x+1=0\)
\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)
mà \(2\ne0\)
nên \(x^2-3x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)
e) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)