a,(25x^2-2x+4/25)+171/25

(5x-2/5)^2+171/25

vi (5x-2/5)^2>=0

suy ra H>=171/25

dau bang say ra khi ma chi khi 5x-2/5=0 suy ra x=2/25

vay gia tr nho nhat cua bieu thuc H=2/25 khi x=2/25

b,2(x^2-6x+9)+13

2(x-3)^2+13

vi2(x-3)^2>=0

suy ra K>=13

dau bang say ra khy va chi khy x-3=0 suy ra x=3

vay gia chi nho nhat bieu thuc K=13 khi x=3

Giúp mình với ạ,cảm ơn mọi người

b: Ta có: \(B=x^2+4x+9y^2-6y-1\)

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

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

Dấu '=' xảy ra khi x=-2 và \(y=\dfrac{1}{3}\)

`y=sin^2 5x+cos^2 5x+3sin 2x-2`       `TXĐ: R`

`y=1+3sin 2x-2`

`y=3sin 2x-1`

Ta có: `-1 <= sin 2x <= 1`

`<=>-3 <= 3sin 2x <= 3`

`<=>-4 <= y <= 2`

    `=> y_[mi n]=4<=>sin 2x =-1<=>x=-\pi/4 + k\pi`  `(k in ZZ)`

         `y_[max] = 2<=>sin 2x=1<=>x=\pi/4+k\pi`  `(k in ZZ)`

Ta có: \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3-3x^2+16x^2-8x-6x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[3x^2\left(2x-1\right)+8x\left(2x-1\right)-3\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(3x^2+8x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(3x^2+9x-x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left[3x\left(x+3\right)-\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(x+3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\2x-1=0\\x+3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\2x=1\\x=-3\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\\x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{1}{2};-3;\dfrac{1}{3}\right\}\)

\(=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)= \(\left|2-5x\right|+\left|5x\right|\ge2+5x-5x=2\)

min A=2 \(\Leftrightarrow\hept{\begin{cases}2-5x\ge0\\5x\ge0\end{cases}\Leftrightarrow0\le x\le\frac{2}{5}}\)

Chuẩn đấy

Bạn dùng phương pháp phân tích đa thức thành nhân tử sẽ đc :

(2x-1).(x+3).(x+2).(3x-1) = 0

<=> x=1/2 hoặc x=-3 hoặc x=-2 hoặc x=1/3

Vậy .............

Tk mk nha

giải ra

             \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow\)\(6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow\)\(\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\)\(\left(x+2\right)\left(6x^3+18x^2-5x^2-15x+x+3\right)=0\)

\(\Leftrightarrow\)\(\left(x+2\right)\left(x+3\right)\left(6x^2-5x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x+2\right)\left(x+3\right)\left(6x^2-2x-3x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x+2\right)\left(x+3\right)\left(2x-1\right)\left(3x-1\right)=0\)

P/S:lm tiếp nha   x = -2;  x = -3;  x = 1/2;  x = 1/3

\(A=xy\left(x-2\right)\left(y+6\right)+12x^2-24x+3y^2+18y+2047\)

   \(=xy\left(x-2\right)\left(y+6\right)+12\left(x^2-2x\right)+3y\left(y+6\right)+2047\)

   \(=y\left(y+6\right)\left(x^2-2x\right)+12\left(x^2-2x+3\right)+3y\left(y+6\right)+2011\)

   \(=y\left(y+6\right)\left(x^2-2x+3\right)+12\left(x^2-2x+3\right)+2011\)

   \(=\left(x^2-2x+3\right)\left(y^2+6y+12\right)+2011\)

   \(=\left[\left(x-1\right)^2+2\right].\left[\left(y+3\right)^2+3\right]+2011\ge2.3+2011=2017\)

Dấu "=" xảy ra khi: 

\(\hept{\begin{cases}x-1=0\\y+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}}\)

Vậy GTNN của A là 2017 khi \(x=1,y=-3\)

\(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow\)  \(6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow\)  \(6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\)  \(\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\)  \(\left(x+2\right)\left(6x^3+18x^2-5x^2-15x+x+3\right)=0\)

\(\Leftrightarrow\)  \(\left(x+2\right)\left[6x^2\left(x+3\right)-5x\left(x+3\right)+x+3\right]=0\)

\(\Leftrightarrow\)  \(\left(x+2\right)\left(x+3\right)\left(6x^2-5x+1\right)=0\)

\(\Leftrightarrow\)  \(\left(x+2\right)\left(x+3\right)\left(2x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\)   \(x+2=0\)  hoặc  \(x+3=0\)  hoặc  \(2x-1=0\)  hoặc  \(3x-1=0\)

\(\Leftrightarrow\)   \(x=-2\)  hoặc \(x=-3\)  hoặc  \(x=\frac{1}{2}\)  hoặc  \(x=\frac{1}{3}\)

Vậy, tập nghiệm của pt là  \(S=\left\{-2;-3;\frac{1}{2};\frac{1}{3}\right\}\)