\(x\left(x-1\right)+y\left(y-1\right)+z\left(z-1\right)< =\frac{4}{3}\Rightarrow x^2-x+y^2-y+z^2-z< =\frac{4}{3}\)

\(\Rightarrow3x^2-3x+3y^2-3y+3z^2-3z< =4\Rightarrow3\left(x^2+y^2+z^2\right)-3\left(x+y+z\right)< =4\)

\(3\left(x^2+y^2+z^2\right)=\left(1+1+1\right)\left(x^2+y^2+z^2\right)>=\left(x+y+z\right)^2\)

\(\Rightarrow\left(x+y+z\right)^2-3\left(x+y+z\right)< =3\left(x^2+y^2+z^2\right)-3\left(x+y+z\right)< =4\)

\(\Rightarrow\left(x+y+z\right)^2-3\left(x+y+z\right)< =4\Rightarrow\left(x+y+z\right)\left(x+y+z-3\right)< =4\)

\(x+y+z>4\Rightarrow x+y+z-3>1\Rightarrow\left(x+y+z\right)\left(x+y+z-3\right)>4\cdot1=4\)(loại)

\(x+y+z=4\Rightarrow x+y+z-3=1\Rightarrow\left(x+y+z\right)\left(x+y+z-3\right)=4\cdot1=4\left(tm\right)\)

\(x+y+z< 4\Rightarrow x+y+z-3< 1\Rightarrow\left(x+y+z\right)\left(x+y+z-3\right)< 4\cdot1=4\left(tm\right)\)

\(\Rightarrow x+y+z< =4\)thì \(\left(x+y+z\right)\left(x+y+z-3\right)< =4\)

dấu = xảy ra khi \(x=y=z=\frac{4}{3}\)

vậy \(x\left(x-1\right)+y\left(y-1\right)+z\left(z-1\right)< =\frac{4}{3}\Rightarrow x+y+z< =4\)dấu = xảy ra khi \(x=y=z=\frac{4}{3}\)

\(x,y,z>0\Rightarrow\left(x+y\right)+z>=2\sqrt{\left(x+y\right)z}\Rightarrow1>=2\sqrt{\left(x+y\right)z}\Rightarrow1>=4\left(x+y\right)z\)(bđt cosi)

\(M=\frac{x+y}{xyz}=\frac{1\left(x+y\right)}{xyz}>=\frac{4\left(x+y\right)z\left(x+y\right)}{xyz}=\frac{4\left(x+y\right)^2z}{xyz}>=4\cdot\frac{\left(2\sqrt{xy}\right)^2z}{xyz}=\frac{4\cdot4xyz}{xyz}=4\cdot4=16\)

dấu = xảy ra khi \(\hept{\begin{cases}x=y=\frac{1}{4}\\z=\frac{1}{2}\end{cases}}\)

vậy min M là 16 khi \(x=y=\frac{1}{4}:z=\frac{1}{2}\)

c)x²-2xy+y²+3x-3y-10

=(x-y)²+3(x-y)-10

=(x-y)²+2(x-y).3/2+9/4-49/4

=(x-y+3/2)²-(7/2)²

=(x-y+3/2+7/2)(x-y+3/2-7/2)

=(x-y+5)(x-y-2)

a Đặt \(x^2\)=t[t\(\ge\)0}

6t^2-11t+3=6t^2-3t-9t+3=2t[3t-1] -3[3t-1]=[3t-1][2t-3]=[3x^2-1][2x^2-3]

b Đặt x^2+x=t[t\(\ge\)0]

t^2+3t+2=[t+1][t+2]

Đến đó Dương làm tương tự như câu a nhé

             \(\text{ab = -18 và a - b = 9}\)

\(=>\text{ab = -3 . 6 = 3 . -6}\)

\(\text{Vậy a + b với a = -3 ; b = 6 thì : }\)

                            \(\text{a + b = -3 + 6 = 3}\)

\(\text{Vậy a + b với a = 3 ; b = -6 thì : }\)

                            \(\text{a + b = 3 + -6 = -3}\)

\(\left(a+b\right)^2=\left(a-b\right)^2+4ab=9^2+4.\left(-18\right)=9\)
\(\Rightarrow a+b=3\)

1/

\(B=\frac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)

\(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)

\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)

\(=\frac{\left(3^8-1\right)\left(3^8+1\right)-3^{16}}{4}\)

\(=\frac{3^{16}-1-3^{16}}{4}=\frac{-1}{4}\)

2/

a, (x-5)²-(x+3)²=1

<=>(x-5+x+3)(x-5-x-3)=1

<=>-16.(x-1)=1

<=>x-1=-1/16

<=>x=15/16

b, (2x-1)²-(2x-3)²=4

<=>(2x-1+2x-3)(2x-1-2x+3)=4

<=>-8(x-1)=4

<=>x-1=-1/2

<=>x=1/2

\(x^2+4xy+5y^2=\text{[}x^2+4xy+\left(2y\right)^2\text{]}+y^2\)

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

Ta có: \(\left(x+2y\right)^2\ge\forall x;y\)

          \(y^2\ge0\forall y\)

\(\Rightarrow\left(x+2y\right)^2+y^2\ge0\forall x;y\)

\(\Rightarrow x^2+4xy+5y^2\) không có giá trị lớn nhất

\(x^2+4xy+5y^2=0\Leftrightarrow\orbr{\begin{cases}\left(x+2y\right)^2=0\\y^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+2y=0\\y=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\y=0\end{cases}}}\)

KL:.........................................

Không có giá trị nhỏ nhất vì:

- Không có số x ; y nhỏ nhất

Không có giá trị lớn nhất vì:

- Không có số x ; y lớn nhất

( Đây là phép cộng )