Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(x^4+6x^3+7x^2-6x+1\)
\(=x^4-2x^2+1+6x^3+9x^2+6x\)
\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)
\(=\left(x^2-1+3x\right)^2\)
b, \(x^4-7x^3+14x^2-7x+1\)
\(=x^4+2x^2+1+7x^3+12x^2-7x\)
\(=\left(x^2+1\right)^2-7x\left(x^2+1\right)+12^2\)
\(=\left(x^2-1+3x\right)^2\)
c, \(12x^2-11x-36\)
\(=12x^2-27x+16x-36\)
\(=3x\left(4x-9\right)+4\left(4x-9\right)\)
\(=\left(4x-9\right)\left(3x+4\right)\)
\(1.x^4+6x^3+11x^2+6x+1\)
\(=x^4+6x^3+9x^2+2x^2+6x+1\)
\(=x^4+9x^2+1+6x^3+2x^2+6x\)
\(=\left(x^2\right)^2+\left(3x\right)^2+1^2+2.x^2.3x+2.x^2.1+2.3x.1\)
\(=\left(x^2+3x+1\right)^2\)
\(2,6x^4+5x^3-38x^2+5x+6\)
\(=6x^4+6x^3+2x^3-3x^3-36x^2+2x^2-3x^2-x^2-12x+18x-x+6\)
\(=\left(6x^4+2x^3\right)+\left(6x^3+2x^2\right)-\left(3x^3+x^2\right)-\left(36x^2+12x\right)+\left(18x+6\right)-\left(3x^2+x\right)\)
\(=2x^3\left(3x+1\right)+2x^2\left(3x+1\right)-x^2\left(3x+1\right)-12x\left(3x+1\right)+6\left(3x+1\right)-x\left(3x+1\right)\)
\(=\left(3x+1\right)\left(2x^3+2x^2-x^2-12x+6-x\right)\)
\(=\left(3x+1\right)\left[\left(2x^3-x^2\right)+\left(2x^2-x\right)-\left(12x-6\right)\right]\)
\(=\left(3x+1\right)\left[x^2\left(2x-1\right)+x\left(2x-1\right)-6\left(2x-1\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x^2+x-6\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x^2+3x-2x-6\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\left[\left(x^2+3x\right)-\left(2x+6\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x+3\right)\left(x-2\right)\)
1. \(x^4+6x^3+11x^2+6x+1\)
\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2+2x^2+6x+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
3. \(x^4-7x^3+14x^2-7x+1\)
\(=x^2\left(x^2-7x+14-\dfrac{7}{x}+\dfrac{1}{x^2}\right)\)
\(=x^2\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(7x+\dfrac{7}{x}\right)+14\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}\right)^2-7\left(x+\dfrac{1}{x}\right)+12\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right).\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{1}{4}\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}-\dfrac{7}{2}\right)^2-\dfrac{1}{4}\right]\)
\(=\left(x^2+1-\dfrac{7}{2}x\right)^2-\left(\dfrac{1}{2}x\right)^2\)
\(=\left(x^2-3x+1\right)\left(x^2-4x+1\right)\)
Có thể phân tích thành HĐT tiếp hoặc không.
1/ \(\frac{x-3}{3xy}\)+\(\frac{5x+3}{3xy}\)= \(\frac{6x}{3xy}\)=\(\frac{3}{y}\)
2/\(\frac{5x-7}{2x-3}\)+\(\frac{4-3x}{2x-3}\)=\(\frac{2x-3}{2x-3}\)=1
3/\(\frac{11x-7}{3-5x}\)-\(\frac{6x+4}{5x-3}\)=\(\frac{11x-7}{3-5x}\)+\(\frac{6x+4}{3-5x}\)=\(\frac{17x-3}{3-5x}\)
4/\(\frac{3}{2x+6}\)-\(\frac{x-6}{2x^2+6x}\)=\(\frac{3x}{x\left(2x+6\right)}\)-\(\frac{x-6}{x\left(2x+6\right)}\)=\(\frac{2x-6}{x\left(2x+6\right)}\)
5/\(\frac{1}{2x-10}\)+\(\frac{2x}{3x^2-15x}\)=\(\frac{1}{2\left(x-5\right)}\)+\(\frac{2x}{3x\left(x-5\right)}\)=\(\frac{3x}{6x \left(x-5\right)}\)+\(\frac{4x}{6x\left(x-5\right)}\)
=\(\frac{7x}{6x\left(x-5\right)}\)=\(\frac{7}{6\left(x-5\right)}\)
a) \(4x^4+4x^3+5x^2+2x+1\)
= \(x^2\left(4x^2+4x+5+\frac{4}{x}+\frac{1}{x^2}\right)\)
=\(x^2\left[\left(4x^2+\frac{1}{x^2}\right)+2\left(2x+\frac{1}{x}\right)+5\right]\)(1)
Đặt \(2x+\frac{1}{x}=a\)thì \(\left(2x+\frac{1}{x}\right)^2=a^2\)\(\Rightarrow4x^2+\frac{1}{x^2}=a^2-4\)
Thay vào (1), ta có:
\(x^2\left(a^2-4+2a+5\right)\)
=\(x^2\left(a^2+2a+1\right)\)
=\(x^2\left(a+1\right)^2\)
=\(\left[x\left(a+1\right)\right]^2\)
=\(\left[x\left(2x+\frac{1}{x}+1\right)\right]^2\)
=\(\left(2x^2+1+x\right)^2\)
\(=\left(2x^2+x+1\right)^2\)
a) Đặt f(x) = 4x4 + 4x3 + 5x2 + 2x + 1
Sau khi phân tích thì đa thức có dạng ( 2x2 + ax + 1 )( 2x2 + bx + 1 )
=> f(x) = ( 2x2 + ax + 1 )( 2x2 + bx + 1 )
<=> f(x) = 4x4 + 2bx3 + 2x2 + 2ax3 + abx2 + ax + 2x2 + bx + 1
<=> f(x) = 4x4 + ( a + b )2x3 + ( ab + 4 )x2 + ( a + b )x + 1
Đồng nhất hệ số ta có : \(\hept{\begin{cases}a+b=2\\ab=1\end{cases}\Leftrightarrow}a=b=1\)
Vậy f(x) = 4x4 + 4x3 + 5x2 + 2x + 1 = ( 2x2 + x + 1 )2
b) 3x4 + 11x3 - 7x2 - 2x + 1
= 3x4 - x3 + 12x3 - 4x2 - 3x2 + x - 3x + 1
= x3( 3x - 1 ) + 4x2( 3x - 1 ) - x( 3x - 1 ) - ( 3x - 1 )
= ( 3x - 1 )( x3 + 4x2 - x - 1 )
câu này là câu b và c nhé nếu là câu a thì cái bt = cái khác
Gỉa sử : ( bt = biểu thức :D )
\(bt=\left(x^2+ax+b\right)\left(x^2+cx+d\right)=x^4+\left(a+c\right)x^3+\left(d+ac+b\right)x^2+\left(bc+ad\right)x+bd\)
Ta có : \(\hept{\begin{cases}a+c=-6\\d+ac+b=14\\bc+ad=-7and:bd=1\end{cases}}\)(do không có ngoặc 4
Đến đây thì giải ra như hpt thôi
Dạng này được cái không cần sáng tạo già cả chỉ cần theo công thức nhưng khá khó trong việc giải hệ
a) Giả sử
\(4x^4+4x^3+5x^2+2x+1=4\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)
Khai triển vế trái = \(4x^4+4\left(a+c\right)x^3+4\left(b+d+ac\right)x^2+4\left(ad+bc\right)x+4bd\)
Rồi sử dụng đồng nhất thức, ta có hpt gồm các pt
\(4\left(a+c\right)=4\),\(4b+4d+4ac=5\),\(4ad+4bc=2\),\(4bd=1\)
Rồi ...
Các câu còn lại tương tự:))
b: \(=x^4+x^2+36-2x^3+12x^2-12x+x^2-6x+9\)
\(=x^4-2x^3+14x^2-18x+45\)
\(=x^4+9x^2-2x^3-18x+5x^2+45\)
\(=\left(x^2+9\right)\left(x^2-2x+5\right)\)
d: \(=2x^4+2x^3+6x^2-x^3-x^2-3x+x^2+x+3\)
\(=\left(x^2+x+3\right)\left(2x^2-x+1\right)\)
e: \(=3x^4-3x^3-3x^2-2x^3+2x^2+2x+2x^2-2x-2\)
\(=\left(x^2-x-1\right)\left(3x^2-2x+1\right)\)
1:\(=3x^3-9x^2-5x^2+15x-x+3\)
\(=\left(x-3\right)\left(3x^2-5x-1\right)\)
2: \(=x^4-3x^3+x^2-3x^3+9x^2-3x+x^2-3x+1\)
\(=\left(x^2-3x+1\right)^2\)
a)
\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}=\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}\)
\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\)
MTC: \(2\left(x-1\right)\left(x+1\right)\left(x-5\right)\)
\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}\\ =\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}=\dfrac{2\left(x+1\right)\left(3x-6\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)
\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x-5\right)\left(5x-5\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)
câu c:x^4-2x^3-x^2+x^3-2x^2-x+5x^2-10x-5=x^2(x^2-2x-1)+x(x^2-2x-1)+5(x^2-2x-1)=(x^2-2x-1)(x^2+x+5)