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bn giải giúp mình đi

1)   P = \(3+15x-5x^2\)\(=-5x^2+15x+3=-5\left(x^2-3x-\frac{3}{5}\right)\)  \(=-5\left(x^2-2.\frac{3}{2}.x+\frac{9}{4}-\frac{9}{4}-\frac{3}{5}\right)\)= \(-5\left[\left(x-\frac{3}{2}\right)^2-\frac{57}{20}\right]=-5.\left(x-\frac{3}{2}\right)^2+\frac{57}{4}\)

vì \(\left(x-\frac{3}{2}\right)^2>=0\) => \(-5.\left(x-\frac{3}{2}\right)^2+\frac{57}{4}>=0\)  =>\(-5.\left(x-\frac{3}{2}\right)^2+\frac{57}{4}>=\frac{57}{4}\)

 => GTLN  của P là \(\frac{57}{4}\)tại x =\(\frac{3}{2}\)

2) GTNN của B là -36

Ta có : D = (x - 1).(x + 3).(x + 2).(x + 6)

=> D = [(x - 1)(x + 6)].[(x + 3).(x + 2)]

=> D = (x² + 5x - 6) . (x² + 5x + 6)

=> D = (x² + 5x)² - 36

=> D = [x(x + 5)]² - 36

Mà : [x(x + 5)]​²  \(\ge0\forall x\)

Suy ra : D = [x(x + 5)]​² - 36 \(\ge-36\forall x\)

Vậy D_min = -36 , dấu "=" xẩy ra khi và chỉ khi x = 0 hoặc -5

\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\\ =\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\\ =\left(x^2+5x-6\right)\left(x^2+5x+6\right)\\ =\left(x^2+5x\right)^2-36\ge-36\)

Dấu "=" xảy ra khi x^2+5x=0

\(\Rightarrow x\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Min A = -36 khi x=0 hoặc x=-5

Ta có:

(2x + \(\frac{1}{3}\))⁴ \(\ge\) 0 \(\forall\) x \(\in\) Z

=> (2x + \(\frac{1}{3}\))⁴ - 1 \(\ge\) -1 \(\forall\) x \(\in\) Z

=> A \(\ge\) -1 \(\forall\) x \(\in\) Z

Dấu "=" xảy ra khi (2x + \(\frac{1}{3}\))⁴ = 0

=> 2x + \(\frac{1}{3}\) = 0

=> 2x = 0 - \(\frac{1}{3}\)

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

=> x = \(\frac{-1}{6}\)

Vậy GTNN của A = -1 khi x = \(\frac{-1}{6}\).

b) Lại có:

- (\(\frac{4}{9}\)x - \(\frac{2}{15}\))⁶ \(\le\) 0 \(\forall\) x \(\in\) Z

=> - (\(\frac{4}{9}\)x - \(\frac{2}{15}\))⁶ + 3 \(\le\) 3 \(\forall\) x \(\in\) Z

=> B \(\le\) 3 \(\forall\) x \(\in\) Z

Dấu "=" xảy ra khi:

(\(\frac{4}{9}\)x - \(\frac{2}{15}\))⁶ = 0

=> \(\frac{4}{9}\)x - \(\frac{2}{15}\) = 0

=> \(\frac{4}{9}\)x = \(\frac{2}{15}\)

=> x = \(\frac{2}{15}\) : \(\frac{4}{9}\)

=> x = \(\frac{3}{10}\)

Vậy GTLN của B = 3 khi x = \(\frac{3}{10}\)

a)Ta thấy: \(\left(2x+\frac{1}{3}\right)^4\ge0\)

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

\(\Rightarrow A\ge-1\)

Dấu "=" xảy ra khi \(\left(2x+\frac{1}{3}\right)^4=0\Leftrightarrow x=-\frac{1}{6}\)

Vậy \(Min_A=-1\) khi \(x=-\frac{1}{6}\)

b)Ta thấy:\(\left(\frac{4}{9}x-\frac{2}{15}\right)^6\ge0\)

\(\Rightarrow-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le0\)

\(\Rightarrow-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\le3\)

\(\Rightarrow B\le3\)

Dấu "=" xảy ra khi \(-\left(\frac{4}{9}x-\frac{2}{15}\right)^6=0\Rightarrow x=\frac{3}{10}\)

Vậy \(Max_B=3\) khi \(x=\frac{3}{10}\)

a) GTNN nhé mk viết lộn

Đặt \(A=3.\left(6-\left|y-1\right|\right)-\left(x-2\right)^2\)

Vì \(\left|y-1\right|\ge0;\left(x-2\right)^2\ge0\forall x,y\)

\(\Rightarrow6-\left|y-1\right|\le6\forall y\Rightarrow3\left(6-\left|y-1\right|\right)\le18\forall x\)

\(\Rightarrow3\left(6-\left|y-1\right|\right)-\left(x-2\right)^2\le18\forall x,y\)

Dấu ''='' xảy ra \(\Leftrightarrow\left|y-1\right|=0;\left(x-2\right)^2=0\Rightarrow y=1;x=2\)

Vậy GTLN của A là 18 tại x = 2 , y = 1

`a)M=(x^4+2)/(x^6+1)+(x^2-1)/(x^4-x^2+1)-(x^2+3)/(x^4+4x^2+3)`

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

`=(x^4+2)/(x^6+1)+((x^2-1)(x^2+1))/(x^6+1)-1/(x^2+1)`

`=(x^4+2+x^4-1-x^4+x^2-1)/(x^2+1)`

`=(x^4+x^2)/(x^2+1)`

`=(x^2(x^2+1))/(x^2+1)`

`=x^2`

`b)` tìm gtnn chứ?

`M=x^2>=0`

Dấu '=" `<=>x=0`