x^2(x^4-1)(x^2+2) +1 = ...?

Bn có thể ghi rõ hơn không ạ.

b) 6x²-2x-6x²-6x-3+8x=-3, vậy bt không phụ thuộc vào biến x(đpcm)

câu a bn xem lại đề nhé

Chúc hk tốt!!!!

23568i

\(M=a^3+b^3+c\left(a^2+b^2\right)-abc\)

\(=a^3+b^3+a^2c+b^2c-abc\)

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

Do  \(a+b+c=0\)\(\Rightarrow\)\(\hept{\begin{cases}a+c=-b\\b+c=-a\end{cases}}\)

suy ra:  \(M=-a^2b-ab^2-abc\)

                   \(=-ab\left(a+b+c\right)=0\)   (do a+b+c = 0)

Ta có: \(N=\frac{a}{b+1}+\frac{b}{a+1}=\frac{a^2}{ab+a}+\frac{b^2}{ab+b}\)

             \(\ge\frac{\left(a+b\right)^2}{a+b+2ab}\ge\frac{1}{1+\frac{\left(a+b\right)^2}{2}}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\)

Dấu = xảy ra khi \(a=b=\frac{1}{2}\)

Lại có: \(\frac{a}{b+1}=\frac{a}{2-a}\)

Do \(a;b\ge0\);  a+b=1

\(\Rightarrow0\le a\le1\)

\(\Rightarrow2-a\ge1\)

\(\Rightarrow\frac{a}{2-a}\le a\left(a\ge0\right)\) 

Tương tự suy ra \(N\le a+b=1\)

Dấu = xảy ra khi \(\left(a;b\right)=\left(0;1\right);\left(1;0\right)\)

Vậy \(N_{Min}=\frac{2}{3}\Leftrightarrow a=b=\frac{1}{2}\)

    \(N_{Max}=1\Leftrightarrow\left(a;b\right)=\left(0;1\right);\left(1;0\right)\)

Ta có: \(a\sqrt{b+1}=\frac{a\sqrt{\left(b+1\right)2}}{\sqrt{2}}\le a\frac{b+1+2}{2\sqrt{2}}=\frac{ab+3a}{2\sqrt{2}}\)

Tương tự: \(b\sqrt{a+1}\le\frac{ab+3b}{2\sqrt{2}}\)

\(\Rightarrow M\le\frac{3\left(a+b\right)+2ab}{2\sqrt{2}}\le\frac{6+\frac{\left(a+b\right)^2}{2}}{2\sqrt{2}}=\frac{8}{2\sqrt{2}}=2\sqrt{2}\)

Dấu = khi a=b=1

Ta có: \(a+b=2\Rightarrow b=2-a\)

\(\Rightarrow a\sqrt{b+1}=a\sqrt{3-a}\)

Lại có: \(\hept{\begin{cases}a;b>0\\a+b=2\end{cases}}\Rightarrow0\le a;b\le2\)

Mặt khác: \(a\le2\Rightarrow3-a\ge1\)

\(\Rightarrow\sqrt{3-a}\ge1\)

\(\Rightarrow a\sqrt{3-a}\ge a\)     Do \(a\ge0\)

Tương tự suy ra \(M\ge a+b=2\)

Dấu = khi \(\left(a;b\right)=\left(0;2\right);\left(2;0\right)\)

Vậy \(M_{Max}=2\sqrt{2}\Leftrightarrow a=b=1\)

        \(M_{Min}=2\Leftrightarrow\left(a;b\right)=\left(0;2\right);\left(2;0\right)\)

a,<=>   x²-4x+2²+y²-8y+4²-14

<=> (x²-2x2+2²)+(y²-2x4+4²)-14

<=> (x-2)²+(y-4)²-14 

Vì (x-2)²+(y-4)²>= 0

=> F >= -14 => MIn F = -14 <=> x=2, y=4

b, <=> (x²+5²+(2y)²-4xy+10x-20y) +(y²-2y+1)+2

<=> (x+5-2y )²+(y-1)²+2 

Vì (x+5-2y) ²+(y-1)² >= 0

=> G >= 2 => Min =2 <=> y=1, x= -3

\(F=x^2-4x+y^2-8y+6\)

\(F=\left(x^2-2.2x+2^2\right)+\left(y^2-2.4.y+4^2\right)-14\)

\(F=\left(x-2\right)^2+\left(y-4\right)^2-14\)

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\left(y-4\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x\)

\(F=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy \(F_{min}=-14\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Ta có :

\(M=a^3+b^3+c\left(a^2+b^2\right)-abc\)

\(M=a^3+b^3+a^2c+b^2c-abc\)

\(=\left(a^3+a^2c\right)+\left(b^3+b^2c\right)-abc\)

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

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

\(=-ab\left(a+b+c\right)=0\)

Ta có: \(a+b+c=0\)

\(\Rightarrow a+b=-c;b+c=-a;a+c=-b\)

\(M=a^3+b^3+c.\left(a^2+b^2\right)-abc\)

\(M=a^3+b^3+ca^2+cb^2-abc\)

\(M=a^2.\left(a+c\right)+b^2.\left(b+c\right)-abc\)

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

\(M=-a^2b-b^2a\)

\(M=-ab.\left(a+b\right)\)

\(M=-ab.\left(-c\right)\)

\(M=abc\)

Tham khảo nhé~