Ta có : x >=0

=>\(\frac{1}{x}\)>=0

=>\(\frac{3}{x}\)>=0

=>\(\frac{3}{x}\)+3 >= 3

Vậy Min B=3 <=> x=0

3 trên x+3 chứ ko phải là 3/x  +3

         \(x^6-2x^3-5\)

\(=x^6-2x^3+1-6=\left(x^3-1\right)^2-6\ge-6\)\(\forall x\)

Dấu "=" xảy ra khi: \(x^3-1=0\)

                            \(\Rightarrow x^3=1\Rightarrow x=1\)

Vậy GTNN của \(x^6-2x^3-5\) là -6 khi \(x=1\)

Chúc bạn học tốt.

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)

      \(\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\)

      \(\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\)

\(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}=\frac{x}{z}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{y}{x}+\frac{z}{x}\)

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

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

\(=y.\frac{-1}{y}+x.\frac{-1}{x}+z.\frac{-1}{z}=-1-1-1=-3\)

Vậy nên A = -3

Câu 4 : 

       Ta có : a+b+c=0

​​=> a+b=-c

Lại có : a³+b³=(a+b)³-3ab(a+b)

=> a³+b³+c³=(a+b)³-3ab(a+b)+c³

                    =-c³-3ab. (-c)+c³

                    =3abc

Vậy a³+b³+c³=3abc với a+b+c=0

\(x^4+x^5+1\)

\(=\left(x^5-x^3+x^2\right)+\left(x^4-x^2+x\right)+\left(x^3-x+1\right)\)

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

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

x⁴+x⁵+1=2x⁹+1

Bài 1:

a)  \(\left(n+2\right)^2-\left(n-2\right)^2=n^2+4n+4-\left(n^2-4n+4\right)=8n\)    \(⋮\)\(8\)   (đpcm)

b)  \(\left(n+7\right)^2-\left(n-5\right)^2=n^2+14n+49-\left(n^2-10n+25\right)=24n-24\)\(⋮\)\(24\) (đpcm)

Bài 2:

mk biến đổi về pt tích sau đó bạn giải nốt nhé

a)   \(\left(x-4\right)^2-36=0\)

<=>  \(\left(x-4-6\right)\left(x-4+6\right)=0\)

<=>  \(\left(x-10\right)\left(x+2\right)=0\)

................

b)  \(4x^2-12x=-9\)

<=> \(4x^2-12x+9=0\)

<=>  \(\left(2x-3\right)^2=0\)

..............

c) \(\left(x+8\right)^2=121\)

<=>  \(\left(x+8\right)^2-121=0\)

<=>  \(\left(x+8+11\right)\left(x+8-11\right)=0\)

<=> \(\left(x+19\right)\left(x-3\right)=0\)

...................

Bài 3:

a)  \(31,8^2-2\times31,8\times21,8+21,8^2\)

    \(=\left(31,8-21,8\right)^2=10^2=100\)

b)  mạo phép chỉnh đề

\(58,2^2+2\times58,2\times41,8+41,8^2\)

\(=\left(58,2+41,8\right)^2=100^2=10000\)

\(x^3+y^3+z^3-3xyz\)

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

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

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

Bài 3) 

Ta có :

\(x^3+y^3+z^3-3xyz\)

\(\Rightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(\Rightarrow\left(x+y+z\right)\left[\left(x+y^2\right)-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

P/s tham khảo nha

hok tốt