Cho a,b dương thỏa mãn \(a^3+b^3=a^5+b^5.\) CMR : \(a^2+b^2\le1+ab\)

Vì thời gian tim làm việc ( pha nhĩ co+ pha thất co =0,4s ) = thời gian tim nghỉ ngơi hoàn toàn ( pha giãn chung = 0,4s ) nên tim làm việc suốt đời mà ko mỏi .

Vì trong hình thang đường trung bình có độ dài = 1 nửa tổng 2 đáy nên :

Độ dài của đường trung bình EF là : (12+8):2 =10 cm

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Ta có : \(\left(x^3+ax+b\right)⋮\left(x^2+x-2\right)\)

Gọi ( x+k) là thương của đa thức trên .Ta có :

\(\left(x^3+ax+b\right)=\left(x+k\right)\left(x^2+x-2\right)\)

\(=>x^3+ax+b=x^3+kx^2+x^2+kx-2x-2k\)

\(=>x^3+ax+b=x^3+x^2\left(k+1\right)+x\left(k-2\right)-2k\)

Đồng nhất các hệ số ta có :

\(\left\{{}\begin{matrix}k+1=0\\k-2=a\\b=\left(-2k\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k=\left(-1\right)\\a=\left(-3\right)\\b=2\end{matrix}\right.\)

Vậy : a= (-3) : b= 2

a) x^5- 3x^4+ 3x^3- x^2

\(=x^2\left(x^3-3x^2+3x-1\right)\)

\(=x^2\left(x^3-3.x^2.1+3.x.1^2-1^3\right)\)

\(=x^2\left(x-1\right)^3\)

b)3y^2- 3z^2+ 3x^2+ 6xy

\(=3\left(y^2-z^2+x^2+2xy\right)\)

\(=3\left(\left(x^2+2.x.y+y^2\right)-z^2\right)\)

\(=3\left(\left(x+y\right)^2-z^2\right)\)

\(=3\left(x+y+z\right)\left(x+y-z\right)\)

c)x^2- 6x +5

\(=x^2-6x+9-4\)

\(=\left(x^2-2.x.3+3^2\right)-2^2\)

\(=\left(x-3\right)^2-2^2\)

=(x-3-2)(x-3+2)

=(x-5)(x-1)

d)x^2 + 5x -6

\(=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{49}{2}\)

\(=\left(x+\dfrac{5}{2}\right)^{2 }-\left(\dfrac{7}{2}\right)^2\)

\(=\left(x+\dfrac{5}{2}+\dfrac{7}{2}\right)\left(x+\dfrac{5}{2}-\dfrac{7}{2}\right)\)

=(x+6)(x-1)

e)x^2 - 5x + 4

=\(x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{9}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2-\left(\dfrac{3}{2}\right)^2\)

\(=\left(x-\dfrac{5}{2}-\dfrac{3}{2}\right)\left(x-\dfrac{5}{2}+\dfrac{3}{2}\right)\)

=(x-4)(x-1)

a, \(x^2+10x+26+y^2+2y\)

\(=\left(x^2+2.x.5+5^2\right)+\left(1^2+2.1.y+y^2\right)\)

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

b, \(x^2-2xy+2y^2+2y+1\)

\(=x^2-2xy+y^2+y^2+2y+1\)

\(=\left(x^2-2.x.y+y^2\right)+\left(y^2+2.y.1+1^2\right)\)

\(=\left(x-y\right)^2+\left(y+1\right)^2\)

c,\(z^2 -6z+5-t^2-4t\)

\(=-\left(t^2+4t-z^2+6z-5\right)\)

\(=-\left(t^2+2.t.2+2^2-z^2+2.z.3-3^2\right)\)

\(=-\left(\left(t^2+2.t.2+2^2\right)-\left(z^2-2.z.3+3^2\right)\right)\)

\(=-\left(\left(t+2\right)^2-\left(z-3\right)^2\right)\)

\(=\left(z-3\right)^2-\left(t+2\right)^2\)

d, Không biết làm hihi :)

Tên:Trần Minh Anh

Lớp:7a1

Lick của góc học tập:https://hoc24.vn/vip/kudo110

P=\(12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

2P=24\(\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

2P=\(\left(5^{16}-1\right)\left(5^{16}+1\right)\)

2P=\(3^{32}-1\)

Ta có :

a, M =\(\left(x+y+z\right)^2+\left(y+z\right)^2-2\left(y+z\right)\left(x+y+z\right)\)

\(=\left(x+y+z\right)^2-2\left(y+z\right)\left(x+y+z\right)+\left(y+z\right)^2\)

\(=\left(x+y+z-y-z\right)^2\)

\(=x^2\)

b,\(N=\left(x-1\right)^3+\left(x+1\right)^3\)

\(=\left(x-1+x+1\right)\left(\left(x-1\right)^2-\left(x-1\right)\left(x+1\right)-\left(x+1\right)^2\right)\)

\(=2x\left(x^2-2x+1-x^2+x-x+1+x^2+2x+1\right)\)

=\(2x\left(x^2+3\right)=2x^3+6x\)