\(a,x^2+x+1=x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+1\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\left(\forall x\right)=>pt\) vô nghiệm

\(b,A=26x+3y+2015z=17x+9x+3y+1008z+1007z\)

\(=8x+9x+3y+1008z+9x+1007z\)

\(=29+9+9x+1008z-z\)

\(=38+9-z=47-z\)\(\le47\)

dấu'=' xảy ra\(< =>z=0\)

\(=>Max\left(A\right)=47< =>z=0\left(x,y,z\ge0\right)\)

Ta có: x+3z+x+2y=8+9

⇒2x+2y+3z=17

⇒2x+2y+2z+z=17

⇒2(x+y+z)=17−z

Mà x+y+z có GTLN

⇒17−z cũng có GTLN

Mà z≥0⇒−z≤0

⇒17−z≤17

⇒17−z đạt GTLN là 17 tại z=0

+) x+3z=8

Thay z=0

⇒x+0=8

⇒x=8

+) x+2y=9

Thay x=8

⇒8+2y=9

⇒2y=1

⇒y=12

Vậy x=8;y=12;z=0

okie mơn bn nha~

mk sẽ cho~

a) \(x^2+x+1\)

\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)

Do đó đa thức vô nghiệm

 x=1, y=7

các bạn sửa A=26x+3y+2015z nha

\(A\le\sqrt{3\left(x+y+y+z+z+x\right)}=\sqrt{6\left(x+y+z\right)}\le\sqrt{6.\sqrt{3\left(x^2+y^2+z^2\right)}}=\sqrt{6\sqrt{3}}\)

\(A_{max}=\sqrt{6\sqrt{3}}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)

Do \(x^2+y^2+z^2=1\Rightarrow0\le x;y;z\le1\)

\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\) \(\Rightarrow x+y+z\ge x^2+y^2+z^2=1\)

\(A^2=2\left(x+y+z\right)+2\sqrt{\left(x+y\right)\left(x+z\right)}+2\sqrt{\left(x+y\right)\left(y+z\right)}+2\sqrt{\left(y+z\right)\left(z+x\right)}\)

\(A^2=2\left(x+y+z\right)+2\sqrt{x^2+xy+yz+zx}+2\sqrt{y^2+xy+yz+zx}+2\sqrt{z^2+xy+yz+zx}\)

\(A^2\ge2\left(x+y+z\right)+2\sqrt{x^2}+2\sqrt{y^2}+2\sqrt{z^2}=4\left(x+y+z\right)\ge4\)

\(\Rightarrow A\ge2\)

\(A_{min}=2\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị

Áp dụng bất đẳng thức Cosi ta có:

1 32 32 x 29 x + 3 y  ≤  1 4 2 32 x + 29 x + 3 y 2 = 1 8 2 61 x + 3 y

Tương tự

1 32 32 y 29 y + 3 x  ≤  1 8 2 61 y + 3 x

=> P ≤  4 2 x + y  ≤  4 2 x 2 + 1 2 + y 2 + 1 2 = 8 2

Vậy P min =  8 2 <=> x = y = 1