Gọi 5 số tự nhiên liên tiếp là : k;k+1;k+2;k+3

Có k(k+1)(k+2)(k+3)+1

=k(k+3)(k+1)(k+2)+1

=(k²+3k)(k²+3k+2)+1

Đặt k²+3k=A

=A(A+2)+1

=A²+2A+1

=(A+1)²

ĐPCM

\(\left(3^{n+1}-2.2^n\right)\left(3.3^n+2^{n+1}\right).3^{2n+2}+\left(8.2^{n-2}.3^{n+1}\right)^2\)

\(=\left(3^{n+1}-2^{n+1}\right)\left(3^{n+1}+2^{n+1}\right).3^{2n+2}+\left(2^{n+1}.3^{n+1}\right)^2\)

\(=\left(3^{2n+2}-2^{2n+2}\right).3^{2n+2}+2^{2n+2}.3^{2n+2}\)

\(=3^{2\left(2n+2\right)}-2^{2n+2}.3^{2n+2}+2^{2n+2}.3^{2n+2}\)

\(=3^{2\left(2n+2\right)}=\left(3^{2n+2}\right)^2\).

Ta thấy \(\left(3^{2n+2}\right)^2\)luôn là 1 số chính phương với mọi n\(\in\)N

Nên ta có ĐPCM.

\(a,\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2-9-x^2-5x+2x+10=6\)

\(\Leftrightarrow-3x+1=6\Leftrightarrow x=\frac{-5}{3}\)

Vậy x =\(\frac{-5}{3}\)

\(b,\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)

\(\Leftrightarrow18x+16=7\Leftrightarrow x=\frac{-1}{2}\)

Vậy x =\(\frac{-1}{2}\)

a/ \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=6\)

<=> \(x^2-9-\left(x^2+3x-10\right)=6\)

<=> \(x^2-9-x^2-3x+10=6\)

<=> \(-3x+1=6\)

<=> \(-3x=5\)

<=> \(x=-\frac{5}{3}\)

b/ \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

<=> \(6x^2+31x+18-\left(6x^2+13x+2\right)=x+1-x+6\)

<=> \(6x^2+31x+18-6x^2-13x-2=7\)

<=> \(18x+16=7\)

<=> \(18x=-9\)

<=> \(x=-\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)(Đúng)

"=" khi a=b=c

Ta có BĐT \(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow}\)BĐT luôn đúng

\(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Rightarrow2\left(a^2+b^2\right)-\left(a+b\right)^2\ge0\)

\(2\left(a^2+b^2\right)- \left(a^2+b^2+2ab\right)=2\left(a^2+b^2\right)-a^2-b^2-2ab\)

\(2\left(a^2+b^2\right)-\left(a^2+b^2\right)-2ab=a^2+b^2-2ab=\left(a-b\right)^2\)

\(\Rightarrow\left(a-b\right)^2\ge0\)

2.

Áp dụng bất đẳng thức Cauchy - schwarz ( hay còn gọi là bất đẳng thức Cosi ):

\(\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}=\frac{\left(x+y+z\right)^2}{x+y+z+3}=\frac{9}{3+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi x = y = z = 1

1: 

Áp dụng bất đẳng thức Cô si:

\(x\left(y+\frac{x}{1+y}\right)+y\left(z+\frac{y}{1+z}\right)+z\left(x+\frac{z}{1+x}\right)\)

\(=\left(x+y+z\right)\left[\left(y+\frac{x}{1+y}\right)+\left(z+\frac{y}{1+z}\right)+\left(x+\frac{z}{1+x}\right)\right]\)

\(=1\left[\left(x+y+z\right)+\left(\frac{x}{1+y}+\frac{y}{1+z}+\frac{z}{1+x}\right)\right]\)

\(=1\left[1+\left(\frac{x+y+z}{1+y+1+z+1+x}\right)\right]\)

\(=1\left[1+\left(\frac{1}{3+\left(x+y+z\right)}\right)\right]\)

\(=1\left[1+\frac{1}{4}\right]\)

\(=1+\frac{5}{4}=\frac{9}{4}\)

Dấu "=" xảy ra khi x = y = z = \(\frac{1}{3}\)

2. áp dạng bất đẳng thức cauchy - schwarz dạng engel

\(\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3}=\frac{3^2}{3+3}=\frac{9}{6}=\frac{3}{2}\)

dấu bằng xay ra khi x=y=z=1

\(M=31^2+2.31.19+19^2\)

\(\Rightarrow M=\left(31+19\right)^2\)

\(\Rightarrow M=50^2\)

\(\Rightarrow M=2500\)

\(N=45^2-90.35+25^2\)

\(\Rightarrow N=45^2-2.45.35+25^2\)

\(\Rightarrow N=\left(45-25\right)^2\)

\(\Rightarrow N=20^2=400\)

\(P=51^2-50^2+49^2-48^2+...+3^2-2^2+1^2\)

\(\Rightarrow P=\left(51-50\right)\left(51+50\right)+\left(49-48\right)\left(49+48\right)+...+\left(3-2\right)\left(3+2\right)+1\)

\(\Rightarrow P=101+97+...+5+1\)

\(\Rightarrow P=\frac{\left(101+1\right)\left[\left(101-1\right):2+1\right]}{2}\)

\(\Rightarrow P=102.51:2=51.51=51^2\)

\(K=x^2+2y^2-2xy+2x-6y+8\)

\(K=x^2+2x\left(y-1\right)-2y^2-6y+8\)

\(K=x^2+2x\left(y-1\right)-y^2-2y+1+y^2-4y+4+4\)

\(K=x^2+2x\left(y-1\right)-\left(y-1\right)^2+\left(y-2\right)^2+4\)

\(K=\left(x+y-1\right)^2+\left(y-2\right)^2+4\ge4\forall x;y\)

Dấu "=" xảy ra khi x = -3; y = 4

A = ( 2x + 1 ).( x - 2) - x.( 2x - 3 ) + 7

A = 2x.x + 2x.(-2) + 1x + 1.(-2) - x.2x + x.(-3) + 7

A = \(2x^2\) - 4x + x - 2 - \(2x^2\) - 3x + 7

A = 6x + 5

ko biết đúg k ạk

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

B = \(x^2\)- 2x.1 + \(1^2\) -  \(2.x^2\) - 2.1 + \(x^2\)+ 2x.1 + \(1^2\)

B =  \(x^2\)- 2x + 1 - \(2x^2\) - 2 + \(x^2\)+ 2x  + 1

B = \(2x^2\) - 2x + 2 - \(2x^2\) + 1

B = - 2x

>< k bt đúg k -.-