câu 1 

ta có .....

lười viết Min - cốp xki nha

DKXD của A, ta có \(x^{2\le5\Rightarrow-\sqrt{5}\le x\le\sqrt{5}}\)

mà \(3x\ge-3\sqrt{5}\)

mặt kkhác \(\sqrt{5-x^2}\ge0\Rightarrow A=3x+x\sqrt{5-x^2}\ge-3\sqrt{5}\)

min A= \(-3\sqrt{5}\)\(\Leftrightarrow x=-\sqrt{5}\)

ta có \(A^2\le25\)và ta cx có \(-5\le A\le5\)

nhưng dễ thấy \(A=-5\)không xảy ra, vô lí nên ...........bạn xem đoạn sau nhé ( tiếp phần kia )

Lời giải:
Đặt $t=\frac{2x}{x^2+1}$

$t+1=\frac{(x+1)^2}{x^2+1}\geq 0\Rightarrow t\geq -1$

$1-t=\frac{(x-1)^2}{x^2+1}\geq 0\Rightarrow t\leq 1$

Vậy $-1\leq t\leq 1$

$y=t^2-4t+25=(t+1)(t-5)+30$

Vì $-1\leq t\leq 1$ nên $t+1\geq 0; t-5\leq 0\Rightarrow (t+1)(t-5)\leq 0$

$\Rightarrow y\leq 30$

Vậy $y_{\max}=30$

B=(x^2-6x+9)-8

B=(x-3)^2-8

Vì (x-3)^2\(\ge0\forall x\)

-> (x-3)-8\(\ge-8\forall x\)

Dấu = xảy ra<=> x-3=0<=>x=3

C=2x^2-10x+1

C=2(x^2-5x+6,25)-11,5

C= 2(x-2,5)^2-11,5

Vì 2(x-2,5)^2\(\ge0\forall x\)

->2(x-2,5)^2-11,5\(\ge-11,5\forall x\)

Dấu = xẩy ra<=> x-2,5=0<=>x=2,5

Vậy Min C là -11,5 <=> x=2,5

D= x^2+10-25

D=(x^2+10+25)-50

D=(x+5)^2-50

Vì (x-5)^2 \(\ge0\forall x\)

-> (x-5)^2-50\(\ge-50\forall x\)

Dấu = xẩy ra <=> x-5=0<=>x=5

Vậy Min D là -50 <=>x=5

Tìm Max

B= 5x-x^2

B=-(x^2-5x+25/4)-25/4

B= -(x-5/2)^2-25/4

Vì -(x-5/2)^2\(\le0\forall x\)

-> -(x-5/2)^2-25/4\(\le\)-25/4

Dấu = xẩy ra <=> x-5/2=0<=>x=5/2

Vậy Max B là -25/4 <=> x=5/2

C=-x^2-6x+10

C=-(x^2+6x+9)+19

C= -(x+3)^2+19

Vì -(x+3)^2\(\le\)0

=> -(x+3)^2+19\(\le\)19

Dấu = xảy ra <=> x+3=0<=>x=-3

D= -2x^x+8x+12

D=-2(x^2-4x+4)+20

D=-2(x-2)^2 +20

 Vì -2(x-2)^2\(\le\)0

=> -2(x-2)^2+20\(\le\)20

Dấu= xẩy ra<=> x-2=0<=>x=2

Vậy Max D là 20<=>x-2

A = \(11-10x-x^2\)

\(A=-\left(x^2+10x-11\right)\)

\(A=-\left(x^2+2x5+25-11-25\right)\)

\(A=-\left(x+5\right)^2+36\)

Dấu "=" xảy ra khi \(x+5=0\)\(\Leftrightarrow x=-5\)

Vậy Max A= 36 khi x = -5

B bn làm tương tự nha, k cho mình nha bn <3 

C = \(-3x\left(x+3\right)-7\)

\(C=-\left(3x^2+9x+7\right)\)

\(C=-\left(căn3x+2.căn3x.\frac{3căn3}{2}+\frac{27}{4}+7-\frac{27}{4}\right)\)

\(C=-\left(căn3x+\frac{3căn3}{2}\right)^2-\frac{1}{4}\)

Dấu "=" xảy ra khi .... =0   \(\Leftrightarrow\)x= ... (bn tự bấm máy tính nha).

Vậy Max C = -1/4 khi x =.....

^.^ 

Tìm min:

$F=3x^2+x-2=3(x^2+\frac{x}{3})-2$

$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$

$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$

Vậy $F_{\min}=\frac{-25}{12}$. Giá trị này đạt tại $x+\frac{1}{6}=0$
$\Leftrightarrow x=\frac{-1}{6}$

Tìm min

$G=4x^2+2x-1=(2x)^2+2.2x.\frac{1}{2}+(\frac{1}{2})^2-\frac{5}{4}$

$=(2x+\frac{1}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}$ (do $(2x+\frac{1}{2})^2\geq 0$ với mọi $x$)

Vậy $G_{\min}=\frac{-5}{4}$. Giá trị này đạt tại $2x+\frac{1}{2}=0$

$\Leftrightarrow x=\frac{-1}{4}$

Tìm min

$H=5x^2-x+1=5(x^2-\frac{x}{5})+1$

$=5[x^2-\frac{x}{5}+(\frac{1}{10})^2]+\frac{19}{20}$

$=5(x-\frac{1}{10})^2+\frac{19}{20}\geq \frac{19}{20}$
Vậy $H_{\min}=\frac{19}{20}$. Giá trị này đạt tại $x-\frac{1}{10}=0$

$\Leftrightarrow x=\frac{1}{10}$

Ta có : \(P=2x^2-8x+1=2\left(x^2-4x\right)+1=2\left(x^2-4x+4-4\right)+1=2\left(x-2\right)^2-7\)

Vì \(2\left(x-2\right)^2\ge0\forall x\) 

Nên : \(P=2\left(x-2\right)^2-7\ge-7\forall x\in R\)

Vậy \(P_{min}=-7\) khi x = 2

a) ĐKXĐ : \(3\le x\le7\)

Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)

\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)

Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)

\(1,\\ a,A\le\sqrt{\left(x-3+7-x\right)\left(1+1\right)}=\sqrt{8}=2\sqrt{2}\\ A^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Leftrightarrow A\ge2\\ \Leftrightarrow2\le A\le2\sqrt{2}\\ \left\{{}\begin{matrix}A_{min}\Leftrightarrow\left(x-3\right)\left(7-x\right)=0\Leftrightarrow...\\A_{max}\Leftrightarrow x-3=7-x\Leftrightarrow x=5\end{matrix}\right.\)

\(B=\dfrac{\dfrac{5}{2}\left(4x^4+4x^2+1\right)+2\left(x^4-x^2+\dfrac{1}{4}\right)}{\left(2x^2+1\right)^2}\\ B=\dfrac{\dfrac{5}{2}\left(2x^2+1\right)^2+2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}=\dfrac{5}{2}+\dfrac{2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}\ge\dfrac{5}{2}\)

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

Vì \(-\dfrac{4x^2}{\left(1+2x^2\right)^2}\le0\Leftrightarrow B\le3\)

\(\Leftrightarrow\left\{{}\begin{matrix}B_{min}\Leftrightarrow x^2=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{2}}\\B_{max}\Leftrightarrow x=0\end{matrix}\right.\)

\(2,\)

Ta có \(\left(y-2x\right)^2=\left(-2x+y\right)^2=\left[\dfrac{1}{3}\left(-6x\right)+\dfrac{1}{4}\left(4y\right)\right]^2\)

\(\Leftrightarrow\left(y-2x\right)^2\le\left[\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2\right]\left[\left(-6x\right)^2+\left(4y\right)^2\right]=\dfrac{5^2}{3^2\cdot4^2}\left(36x^2+16y^2\right)=\dfrac{5^2}{4^2}\\ \Leftrightarrow\left|y-2x\right|\le\dfrac{5}{4}\\ \Leftrightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\\ \Leftrightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(Max\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{9}{20}\end{matrix}\right.\\ Min\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{9}{20}\end{matrix}\right.\)

nhân chéo dùng delta đi bạn