\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

 

a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

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

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

\(ĐTXR\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

\(=\left(2x+3\right)^2-8\ge-8\forall x\)

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

b: Ta có: \(4x^2-3x+10\)

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

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

\(A=\left|2x+1\right|+13\ge13\forall x\)

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

\(B=-\left(3x+5\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{3}\)

a, Vì |2x+1|≥0 với mọi 

⇒A≥13

Dấu = xảy ra ⇔2x+1=0⇔x=\(\dfrac{-1}{2}\)

b, Vì (3x+5)²≥0 với mọi x

⇒B≤9

Dấu = xảy ra ⇔3x+5=1⇔x=\(\dfrac{-5}{3}\)

 

1) Chỉ tìm được Max thôi nhé

a) \(C=\frac{4}{5}+\frac{20}{\left|3x+5\right|+\left|4y+5\right|+8}\le\frac{4}{5}+\frac{20}{8}=\frac{33}{10}\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|3x+5\right|=0\\\left|4y+5\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{5}{3}\\y=-\frac{5}{4}\end{cases}}\)

b) \(E=\frac{2}{3}+\frac{21}{\left(x+3y\right)^2+5\left|x+5\right|+14}\le\frac{2}{3}+\frac{21}{14}=\frac{13}{6}\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+3y\right)^2=0\\5\left|x+5\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=\frac{5}{3}\end{cases}}\)

2) Thì chỉ tìm được GTNN thôi nhé

a) \(A=5+\frac{-8}{4\left|5x+7\right|+24}\ge5-\frac{8}{24}=\frac{14}{3}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(4\left|5x+7\right|=0\Rightarrow x=-\frac{7}{5}\)

Vậy Min(A) = 14/3 khi x = -7/5

b) \(B=\frac{6}{5}-\frac{14}{5\left|6y-8\right|+35}\ge\frac{6}{5}-\frac{14}{35}=\frac{4}{5}\left(\forall y\right)\)

Dấu "=" xảy ra khi: \(5\left|6y-8\right|=0\Rightarrow x=\frac{4}{3}\)

Vậy Min(B)  = 4/5 khi x = 4/3

A = 5 + \(\frac{15}{4}\)|3x+7| + 3

Vì |3x+7| lớn hơn hoặc bằng 0                  Với mọi x

=>|3x+7| + 3  lớn hơn hoặc bằng 0 + 3          Với mọi x

=> \(\frac{15}{4}\)|3x+7| + 3 lớn hơn hoặc bằng 3     Với mọi x

=>5 + \(\frac{15}{4}\)|3x+7| + 3 lớn hơn hoặc bằng 5 + 3       Với mọi x

hay C lớn hơn hoặc bằng 8

Dấu = xảy ra <=> |3x+7| = 0

                    <=> 3x + 7 = 0

                    <=> 3x       = 0 + 7

                    <=> 3x       = 7

                    <=>  x        =  7 : 3

                    <=>  x        = \(\frac{7}{3}\)

Vậy biểu thức A đạt GTLN bằng 8 tại x =\(\frac{7}{3}\)

xong rùi đó 

thank zì^^

B=4/5+20/|3x+5|+|4y+5|+8 Ta có :|4y+5|lớn hơn hoặc bằng 0 suy ra |4y+5|+8 lớn hơn hoặc bằng 0+8=8 Suy ra 20/|3x+5|+|4y+5|+8 lớn hơn hoặc bằng 20+|24x+40| Vì |24x+40| lớn hơn hoặc bằng 0 suy ra 20+|24x+40| lớn hơn hoặc bằng 0+20=20 suy ra 4/5+20/|3x+5|+|4y+5|+8 lớn hơn hoặc bằng 20+4/5 =20,8 Còn x tự làm

a, bậc 6 

b, bậc 6 

c, bậc 12 

d, bậc 9 

e, bậc 8 

A=\(\frac{\left(2^2\right)^5\cdot\left(3^3\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\frac{2^{10}\cdot3^{12}-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\frac{2^{10}\cdot3^8\left(3^4-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\frac{78}{6}=13\)

cho mk hoi chan chay thuoc loai tu gi vay cac ban

a) Vì \(\sqrt{x-5}\) ≥0

⇒ \(\sqrt{x-5}+7\) ≥ 7

Min A=7⇔x-5=0

             ⇔x=5

b) Vì \(\sqrt{3x-5}\) ≥0

⇒ 8-\(\sqrt{3x-5}\) ≤8

Max=8⇔3x-5\(=\)0

           ⇔\(x=\dfrac{5}{3}\)