Đặt nhan tử chung
\(x^4+2016x^2+2015x+2016\)
Phân tích x^4+2016x^2+2015x+2016 thành nhân tử
Phân tích thành nhân tử: x4+2016x2+2015x+2016
bạn có: x^4 + 2016x^2 + 2015x + 2016
= x^4 + x^3 + x^2 - x^3 - x^2 - x + 2016x^2 + 2016x + 2016
= x^2(x^2 + x + 1) - x(x^2 + x + 1) + 2016(x^2 + x + 1)
= (x^2 + x + 1)(x^2 - x + 2016)
\(x^4+2016x^2+2015x+2016\)
=\(x^4+x^3+x^2+2015x^2+2015x+2015+1-x^3\)
=\(x^2\left(x^2+x+1\right)+2015\left(x^2+x+1\right)+\left(1-x\right)\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^2+2015+1-x\right)\)
=\(\left(x^2+x+1\right)\left(x^2-x+2016\right)\)
\(^{2015x^4+2016x^2+x+2016}\)
Phân tích đa thức thành nhân tử
2015x4 + 2016x2 + x + 2016
= (2015x4 + 2015x3 + 2015x2) + (- 2015x3 - 2015x2 - 2015x) + (2016x2 + 2016x + 2016)
= (x2 + x + 1)(2015x2 - 2015x + 2016)
Vào câu trả lời tương tự đi có đáp án đó
\(X^4+2016x^2+2015x+2016\)Phân tích đa thức thành nha tử
Ta có: x^4 + 2016x^2 + 2015x + 2016
= x^4 + x^3 + x^2 - x^3 - x^2 - x + 2016x^2 + 2016x + 2016
= x^2(x^2 + x + 1) - x(x^2 + x + 1) + 2016(x^2 + x + 1)
= (x^2 + x + 1)(x^2 - x + 2016)
\(x^4+2016x^2+2015x+2016\)
\(=x^4-x+2016x^2+2016x+2016\)
\(=x\left(x^3-1\right)+2016\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)
\(x^4+2016x^2+2015x+2016\)
\(=x^4+2015x^2+x^2+2015x+2015+1\)
\(=\left(x^4+2x^2+1-x^2\right)+\left(2015x^2+2015x+2015\right)\)
\(=\left(x^4+2x^2+1\right)-x^2+2015\left(x^2+x+1\right)\)
\(=\left(x^2+1\right)^2-x^2+2015\left(x^2+x+1\right)\)
\(=\left(x^2+1+x\right)\left(x^2+1-x\right)+2015\left(x^2+x+1\right)\)
\(=\left(x^2+1+x\right)\left(x^2-x+2016\right)\)
tìm x,y,z t/m:
a, x^4-2014x^2+2015x-2014=0
b, 9x^2+y^2+2z^2-18x+4z-6y+20=0
c, x^4+2016x^2+2015x+2016=0
1) Tìm giá trị lớn nhất nhỏ nhất của hàm số: \(f\left(x\right)=x+\frac{4}{x}\)với \(1\le x\le3\)
2) Rút gọn \(A=\sqrt{\frac{2015x+2016}{2016x-2015}}+\sqrt{\frac{2015x+2016}{2015-2016x}}+2017\)
Tính giá trị biểu thức :
a, N = \(x^6-2017x^5+2017x^4-2017x^3+2017x^2-2017x+2025\)
tại x = 2016
b, Q = \(2017x^{2016}+2016x^{2015}+2015x^{2014}+...+3x^2+2x+1\)
tại x = ( -1 )
a/ Với \(x=2016\Rightarrow2017=x+1\)
\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+2025\)
\(A=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2025\)
\(A=2025-x=9\)
b/ Với \(x=-1\Rightarrow\left\{{}\begin{matrix}x^{2k}=1\\x^{2k+1}=-1\end{matrix}\right.\) ta có:
\(Q=2017-2016+2015-2014+...+3-2+1\)
\(Q=1+1+1+...+1+1\) (có \(\frac{2016}{2}+1=1009\) số 1)
\(Q=1009\)
Tim x thuoc Z de bieu thuc sau day dat gia tri lon nhat :
B= 2015x+1/2016x-2016 (x thuoc Z; x khac 1)
Phân tích x^4+2014x^2+2015x+2016 thành nhân tử
Sửa đề: 2016x^2
x^4+2016x^2+2015x+2016
=x^4+x^3+x^2-x^3-x^2-x+2016x^2+2016x+2016
=(x^2+x+1)(x^2-x+2016)