P=n³+4n-5=n³-n+5n-5=n(n²-1)+5(n-1)

=n(n-1)(n+1)+5(n-1)=(n-1)[n(n+1)+5]

=(n-1)(n²+n+5)

Vì n \(\in\) N nên n²+n+5 > 1

Để P là số nguyên tố thì n-1=1=>n=2

Thử lại thấy n=2 thỏa mãn

Vậy n=2

1) a)  x  =  -7 / 44

    b)  x  =  -1 / 8

a) \(A=5-8x-x^2=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left[\left(x+4\right)^2-21\right]\)

\(=-\left(x+4\right)^2+21\le21\)

Vậy \(A_{max}=21\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

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

\(=-3\left(x^2-\frac{5}{3}x+\frac{35}{36}-\frac{25}{36}\right)\)

\(=-3\left[\left(x-\frac{5}{6}\right)^2-\frac{25}{36}\right]\)

\(=-3\left[\left(x-\frac{5}{6}\right)^2\right]+\frac{25}{12}\le\frac{25}{12}\)

Vậy \(B_{min}=\frac{25}{12}\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)

\(A=x^2-5x+1=x^2-2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{21}{4}=\left(x-\frac{5}{2}\right)^2-\frac{21}{4}\)

Vì \(\left(x-\frac{5}{2}\right)^2\ge0\)

nên \(\left(x-\frac{5}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)

Vậy \(Min_{x^2-5x+1}=-\frac{21}{4}\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\).

\(B=1-x^2+3x=-\left(x^2-3x-1\right)=-\left[x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{13}{4}\right]=-\left[\left(x-\frac{3}{2}\right)^2-\frac{13}{4}\right]=-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\)Vì \(\left(x-\frac{3}{2}\right)^2\ge0\)

nên \(-\left(x-\frac{3}{2}\right)^2\le0\)

do đó \(-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\le\frac{13}{4}\)

Vậy \(Max_{1-x^2+3x}=\frac{13}{4}\)khi \(x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

Tuyet Anh Nguyen

1.a)(3x-2)(4x+5)=0 
12x^2+7x-10=0>>x1=2/3,x2=-5/4 
b)4x^3+2x^2+4x+2=0>>x=-1 
c)0,23x^2-4,21x-13,8=0>>x1=21,14,x2=-2,8... 
d)10x^3-13x^2-178x-35=0>>x1=5,x2=-1/5 
b2/a)2x^3+5x^2-3x=0>>x1=1/2,x2=-3 
b)(3x-1)(x^2-7x+12)=0>>x1=1/3,x2=4,x3=... 
b3/ 
a)x^2+x-2=0>>x1=1,x2=-2 
b)x1=-1,x2=-6 
b4/a)0,5x^2-1,5x-1,5x^2+x+4,5x-3=0>>-x... 
b)3x/7-1=3x/7-x>>x=1 
c)2x^2-13x+15=0>>x1=5,x2=3/2

P/s: Tham khảo nha

a)0

b)-1

c)-2,8......

k cho tui

4.

= x\(^2\)-2.\(\dfrac{5}{2}\)x+\(\dfrac{25}{4}\)-\(\dfrac{13}{4}\)

= (x-\(\dfrac{5}{2}\))\(^2\)-\(\dfrac{13}{4}\)lớn hơn hoặc bằng -\(\dfrac{13}{4}\) với mọi x

=> min= -\(\dfrac{13}{4}\) <=> x = 5/2

5.

= 2( x\(^2\)-\(\dfrac{5}{2}\)x-\(\dfrac{1}{2}\))

=2( x\(^2\)-2.\(\dfrac{5}{4}\)+\(\dfrac{25}{4}\)-\(\dfrac{27}{4}\))

=2( x-\(\dfrac{5}{4}\))\(^2\)-\(\dfrac{27}{2}\) lớn hơn hoặc bằng -27/2 với mọi x

vậy min = -\(\dfrac{27}{2}\) <=> x= 5/4

1.

= x\(^2\)- 2.3x+9-3

= (x-3)\(^2\)-3 lớn hơn hoặc bằng -3 với mọi x

=> min = -3 <=> x=3