\(Q=2x^2-6x\)

\(Q=2.(x^2 - 2.\dfrac{3}{2}.x+\dfrac{9}{4}\text{)}-\dfrac{9}{2} \)

\(Q=2.(x-\dfrac{3}{2})^2-\dfrac{9}{2}\ge\dfrac{-9}{2}\)

\(\Rightarrow Min_A=\dfrac{-9}{2}\) khi \(x=\dfrac{3}{2}\) .

\(M=x^2+y^2-x+6y+10\)

\(M=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\)

\(M=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)

\(M=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow Min_M=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2},y=-3.\)

Đặt \(A=x^2+y^2-x+6y+10\)

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

\(\left(x-\frac{1}{2}\right)^2\ge0;\)\(\left(y-3\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}\)

\(\Rightarrow A\ge\frac{3}{4}\)

Dấu"=" xảy ra khi \(\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x=\frac{1}{2};\left(y-3\right)^2=0\Leftrightarrow y=3\)

Vậy \(MinA=\frac{3}{4}\Leftrightarrow x=\frac{1}{2};y=3\)

x+y-x+6y+10= x²-x+\(\frac{1}{4}\)+y²+6y+9+\(\frac{3}{4}\)=(x-\(\frac{1}{2}\))²+(y+3)²⁺\(\frac{3}{4}\) ≥\(\frac{3}{4}\)

Daauus bằng xảy ra khi và chỉ khi x=\(\frac{1}{2}\) và y= -3

Suy ra Min= \(\frac{3}{4}\)

xét x²  + y² - x + 6y + 10

= ( x² - 2 . x .1/2 + 1/4) + ( y² + 2 .y .3 + 9) + 3/4

= (x + 1/2)² + (y + 3)² + 3/4

Vì (x + 1/2) ² > 0 vói mọi x

( y + 3)² > vưới mọi x

3/4 > 0

=> (x + 1/2)² + (y+3)² + 3/4

=> M có GTNN là 3/4 <=> (x+1/2)² = 0 -> x + 1/2=0 -> x = -1/2

và (y + 3)^{2 = 0} -> y +3 = 0 -> y =-3

Vậy M có GTNN là 3/4 khi x = -1/2 và y =-3

đấy là 1 cahs tách cậu có thể tìm và tham khảo các cách khác : '> đừng thụ động quá nhé

x^2+y^2-x+6y+10

=(x^2-x+1/4)+(y^2+6y+9)+3/4

=(x-1/2)^2+(y+3)^2+3/4

Mmin=3/4 khi x=1/2; y=-3

Bài 1 :

\(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^2=0^2\)

\(a^2+b^2+c^2+2ab+2ac+2bc=0\)

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)

\(\Rightarrow\left[a^2+b^2+c^2\right]^2=\left[-2\left(ab+bc+ac\right)\right]^2\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=4\left(a^2b^2+b^2c^2+a^2c^2+2ab.bc+2bc.ac+2ab.ac\right)\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=4a^2b^2+4b^2c^2+4a^2c^2+8abc\left(a+b+c\right)\)

Mà \(a+b+c=0\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=4a^2b^2+4b^2c^2+4a^2c^2\)

Bớt cả 2 vế đi \(2a^2b^2+2a^2c^2+2b^2c^2;\)có :

\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2\)

Cộng cả 2 vế với \(a^4+b^4+c^4;\)có :

\(2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)( Hằng đẳng thức bình phương tổng 3 hạng tử )

Vậy ...

Bình phương cả 2 vế của a + b + c = 0,ta có :

a² + b² + c² + 2(ab + bc + ca) => a² + b² + c² = -2(ab + bc + ca).Bình phương cả 2 vế của đẳng thức bên,ta có :

a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + a²c²) = 4[a²b² + b²c² + a²c² + 2abc(a + b + c)] = 4(a²b² + b²c² + a²c²)

=> a⁴ + b⁴ + c⁴ = 2(a²b² + b²c² + a²c²) 

=> (a² + b² + c²)² = a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + a²c²) = a⁴ + b⁴ + c⁴ + a⁴ + b⁴ + c⁴ = 2(a⁴ + b⁴ + c⁴) 

Bạn ko hiểu chỗ nào thì hỏi mình nhé!

Bài 2 :

Ta có :

\(F=x^2+5y^2+4xy-6x+6y-10\)

Bài này khó ở chỗ \(4xy\)dùng để đánh lừa mình :)

\(=x^2-6x+4xy+9+4y^2-6y+y^2-1\)

\(=\left[x^2-2x\left(4-2y\right)+\left(3^2+4y^2-6y\right)\right]+y^2-1\)

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

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

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

Có \(\left(x-3+2y\right)^2\ge0\)

\(y^2\ge0\)

\(\Rightarrow F\ge0+0-1=-1\)

Dấu bằng xảy ra khi :

\(\hept{\begin{cases}x-3+2y=0\\y=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=3\\y=0\end{cases}}\)

\(A=x^2-5x+12\\ A=x^2-5x+\dfrac{25}{4}+\dfrac{23}{4}\\ A=\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{23}{4}\\ A=\left[x^2-2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{23}{4}\\ A=\left(x-\dfrac{5}{2}\right)^2+\dfrac{23}{4}\\ Do\text{ }\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{5}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-\dfrac{5}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{5}{2}=0\\ \Leftrightarrow x=\dfrac{5}{2}\\ \text{Vậy }A_{\left(Min\right)}=\dfrac{23}{4}\text{ }khi\text{ }x=\dfrac{5}{2}\)

\(B=2x^2-14x+5\\ \\ A=2x^2-14x+\dfrac{49}{2}-\dfrac{39}{2}\\ A=\left(2x^2-14x+\dfrac{49}{2}\right)-\dfrac{39}{2}\\ A=2\left(x^2-7x+\dfrac{49}{4}\right)-\dfrac{39}{2}\\ A=\left[x^2-2\cdot x\cdot\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2\right]-\dfrac{39}{2}\\ A=\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\\ Do\text{ }\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\ge-\dfrac{39}{2}\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-\dfrac{7}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{7}{2}=0\\ \Leftrightarrow x=\dfrac{7}{2}\\ \text{Vậy }B_{\left(Min\right)}=-\dfrac{39}{2}\text{ }khi\text{ }x=\dfrac{7}{2}\)

\(B=2x^2-14x+5\\ B=2x^2-14x+\dfrac{49}{2}-\dfrac{39}{2}\\ B=\left(2x^2-14x+\dfrac{49}{2}\right)-\dfrac{39}{2}\\ B=2\left(x^2-7x+\dfrac{49}{4}\right)-\dfrac{39}{2}\\ B=2\left[x^2-2\cdot x\cdot\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2\right]-\dfrac{39}{2}\\ B=2\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\\ \)

Do \(\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\)

\(\Rightarrow B=2\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\ge-\dfrac{39}{2}\forall x\)

Dấu \("="\) xảy ra khi :

\(\left(x-\dfrac{7}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{7}{2}=0\\ \Leftrightarrow x=\dfrac{7}{2}\)

Vậy \(B_{\left(Min\right)}=-\dfrac{39}{2}\) khi \(x=\dfrac{7}{2}\)

Do máy bị lỗi nên câu B bị trục trặc.

Mk xin lỗi.