\(A=-x^2+3x-5\)\(=-\dfrac{11}{4}-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)=-\dfrac{11}{4}-\left(x-\dfrac{3}{2}\right)^2\le-\dfrac{11}{4}\) với mọi x

\(\Rightarrow A_{max}=-\dfrac{11}{4}\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

\(B=5x-4x^2-3=-\dfrac{23}{16}-\left(4x^2-2.\dfrac{5}{4}.2x+\dfrac{25}{16}\right)\)\(=-\dfrac{23}{16}-\left(2x-\dfrac{5}{4}\right)^2\)\(\le-\dfrac{23}{16}\forall x\)

\(\Rightarrow B_{max}=-\dfrac{23}{16}\Leftrightarrow2x-\dfrac{5}{4}=0\Leftrightarrow x=\dfrac{5}{8}\)

\(C=5-4x-25x^2=\dfrac{129}{25}-\left(25x^2+2.5x.\dfrac{2}{5}+\dfrac{4}{25}\right)\)\(=\dfrac{129}{25}-\left(5x+\dfrac{2}{5}\right)^2\le\dfrac{129}{25}\forall x\)

\(\Rightarrow C_{max}=\dfrac{129}{25}\Leftrightarrow5x+\dfrac{2}{5}=0\Leftrightarrow x=-\dfrac{2}{25}\)

\(D=3x-2x^2=-2\left(x^2-\dfrac{3}{2}x\right)=-2\left(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{9}{8}\)\(=\dfrac{9}{8}-2\left(x-\dfrac{3}{4}\right)^2\le\dfrac{9}{8}\) với mọi x

\(\Rightarrow D_{max}=\dfrac{9}{8}\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)

\(E=2+6x-\dfrac{1}{4}x^2=-\dfrac{1}{4}\left(x^2-24x\right)+2=-\dfrac{1}{4}\left(x^2-2.12x+144\right)+38\)\(=38-\dfrac{1}{4}\left(x-12\right)^2\le38\forall x\)

\(\Rightarrow E_{max}=38\Leftrightarrow x-12=0\Leftrightarrow x=12\)

\(F=-5x^2+4x=-5\left(x^2-\dfrac{4}{5}x\right)=-5\left(x^2-2.\dfrac{2}{5}x+\dfrac{4}{25}\right)+\dfrac{4}{5}\)\(=\dfrac{4}{5}-5\left(x-\dfrac{2}{5}\right)^2\le\dfrac{4}{5}\forall x\)

\(\Rightarrow F_{max}=\dfrac{4}{5}\Leftrightarrow x-\dfrac{2}{5}=0\Leftrightarrow x=\dfrac{2}{5}\)

\(A=-2x^2+5x-8=-2\left(x^2-\frac{5}{2}x+4\right)\)

\(=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}+\frac{39}{16}\right)=-2\left(x-\frac{5}{2}\right)^2-\frac{39}{8}\)

Vì: \(-2\left(x-\frac{5}{2}\right)^2-\frac{39}{8}\le\frac{39}{8}\forall x\)

GTLN  của bt là 39/8 tại \(-2\left(x-\frac{5}{2}\right)^2=0\Rightarrow x=\frac{5}{2}\)

cn lại lm tg tự  nha bn

giúp mình với ,mình cần gấp

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

1/

a, \(A=4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\ge4\)

Dấu "=" xảy ra khi x=1/2

Vậy Amin=4 khi x=1/2

b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)

Dấu "=" xảy ra khi x=-1

Vậy Bmin = -4 khi x=-1

2/

a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)

Dấu "=" xảy ra khi x=3

Vậy Amax = 19 khi x=3

b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmax = 31/8 khi x=5/4

1.

A =\(2x^2-8x+10=\left(x^2-2x+1\right)+\left(x^2-6x+9\right)\)

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

Có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(3-x\right)^2\ge0\end{matrix}\right.\forall x\)

<=> \(\left|x-1\right|+\left|x-3\right|\)

Áp dụng bđt |a| + |b| \(\ge\) |a + b| có:

\(\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\)

đẳng thức xảy ra khi \(1\le x\le3\)

Vậy ................

1.

a)

\(A=2x^2-8x+10=2\left(x^2-4x+4\right)+2\ge=2\left(x-2\right)^2+2\ge2\)

Đẳng thức xảy ra \(\Leftrightarrow x=2\)

b)

\(B=3x^2-x+20=3\left(x^2-\dfrac{1}{3}x+\dfrac{1}{36}\right)+\dfrac{239}{12}=3\left(x-\dfrac{1}{6}\right)^2+\dfrac{239}{12}\ge\dfrac{239}{12}\)

Đẳng thức xảy ra \(\Leftrightarrow x=\dfrac{1}{6}\)

c) ĐK: \(x\ne-1\)

\(C=\dfrac{x^2+x+1}{x^2+2x+1}=\dfrac{4x^2+4x+4}{4x^2+8x+4}\)

\(=\dfrac{3x^2+6x+3}{4x^2+8x+4}+\dfrac{x^2-2x+1}{4x^2+8x+4}\)

\(=\dfrac{3\left(x^2+2x+1\right)}{4\left(x^2+2x+1\right)}+\dfrac{\left(x-1\right)^2}{4x^2+8x+4}=\dfrac{3}{4}+\dfrac{\left(x-1\right)^2}{4x^2+8x+4}\ge\dfrac{3}{4}\)

Đẳng thức xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

2.

\(A=-2x^2+3x+1=-2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{17}{8}=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\)

Đẳng thức xảy ra \(\Leftrightarrow x=\dfrac{3}{4}\)

\(B=-5x^2+4x-19=-5\left(x^2-\dfrac{4}{5}x+\dfrac{4}{25}\right)-\dfrac{91}{5}=-5\left(x-\dfrac{2}{5}\right)^2-\dfrac{91}{5}\le\dfrac{-91}{5}\)

Đẳng thức xảy ra \(\Leftrightarrow x=\dfrac{2}{5}\)

\(C=\dfrac{3}{4x^2-4x+5}=\dfrac{3}{\left(4x^2-4x+1\right)+4}=\dfrac{3}{\left(2x-1\right)^2+4}\le\dfrac{3}{4}\)

Đẳng thức xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

\(a.A=-2x^2+5x-8=-2\left(x^2-2.\dfrac{5}{4}x+\dfrac{25}{16}\right)-\dfrac{39}{8}=-2\left(x-\dfrac{5}{4}\right)^2-\dfrac{39}{8}\text{≤}-\dfrac{39}{8}\) ⇒ \(A_{Max}=-\dfrac{39}{8}."="\) ⇔ \(x=\dfrac{5}{4}\)

\(b.B=3-x^2+4x=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\) ≤ 7

⇒ \(B_{Max}=7."="\) ⇔ \(x=2\)

\(c.C=-2x^2+3x+1=-2\left(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{17}{8}=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\text{≤}\dfrac{17}{8}\)

⇒ \(C_{Max}=\dfrac{17}{8}."="\)⇔ \(x=\dfrac{3}{4}\)

\(d.D=-5x^2-4x-\dfrac{19}{5}=-5\left(x^2+2.\dfrac{2}{5}x+\dfrac{4}{25}\right)-3=-5\left(x+\dfrac{2}{5}\right)^2-3\text{≤}-3\)⇒ \(D_{Max}=-3."="\) ⇔ \(x=-\dfrac{2}{5}\)