b.\(\left(\dfrac{1}{2}-\dfrac{3}{2x-5^2-5}\right)=11\)

\(\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{3}{2x-30}\right)=11\);\(x\ne15\)

\(\Leftrightarrow-\dfrac{3}{2x-30}=\dfrac{21}{2}\)

\(\Leftrightarrow-\dfrac{3}{2\left(x-15\right)}=\dfrac{21}{2}\)

\(\Leftrightarrow-\dfrac{3}{2\left(x-15\right)}=\dfrac{21\left(x-15\right)}{2\left(x-15\right)}\)

\(\Leftrightarrow-3=21\left(x-15\right)\)

\(\Leftrightarrow-3=21x-315\)

\(\Leftrightarrow21x=312\)

\(\Leftrightarrow x=\dfrac{104}{7}\)

c.\(\dfrac{2}{3}-\left|4x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow-\left|4x+\dfrac{1}{2}\right|=-\dfrac{1}{6}\)

\(\Leftrightarrow\left|4x+\dfrac{1}{2}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\dfrac{1}{2}=\dfrac{1}{6}\\4x+\dfrac{1}{2}=-\dfrac{1}{6}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{12}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

a.\(\left(\dfrac{1}{2}-\dfrac{3}{2x-5^2-5}\right)=11\)

b.\(\dfrac{2}{3}-\left|4x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)

Đề là vậy đk bạn?

\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)

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

Dấu "=" \(\Leftrightarrow x=-1\)

b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)

c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)

Dấu "=" \(\Leftrightarrow x=2\)

\(B=\dfrac{2x^2-12x+25}{x^2-6x+12}=\dfrac{2\left(x^2-6x+12\right)+1}{x^2-6x+12}=2+\dfrac{1}{x^2-6x+9+4}=2+\dfrac{1}{\left(x-3\right)^2+4}\le2+\dfrac{1}{4}=\dfrac{9}{4}\)

Không có min nha bạn . Chỉ có max thôi 

Dấu = xảy ra khi x=3

\(M=x^2+2x+2=\left(x^2+x+x+1\right)+1\)

\(M=x\left(x+1\right)+1\left(x+1\right)+1=\left(x+1\right)\left(x+1\right)+1\)

\(M=\left(x+1\right)^2+1\)

Vì \(\left(x+1\right)^2\ge0\) với mọi x

=>\(\left(x+1\right)^2+1\ge1\) với mọi x

=>GTNN của M là 1

Dấu "=" xảy ra <=> x+1=0<=>x=-1

M_min=1 khi x=-1

a haha đoán bừa mà cũng đúng

Do l2x-22I \(\ge0\)

l12-xl\(\ge0\)

2lx-13l\(\ge0\)

Nên D=l2x-22l+l12-xl+2lx-13l\(\ge0\)

Min D = 0\(\Leftrightarrow\hept{\begin{cases}2x-22=0\\12-x=0\\x-13=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=11\\x=12\\x=13\end{cases}}}\)

             Vậy ko có gtri x thỏa mãn khi Min D =0

thanks

\(P=x^2+y^2-2x+6y+12\)

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

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

Vậy GTNN là 2 đạt được khi \(\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

GTNN của P=12

có cách giải k ạ

a) Ta có: \(\left(x+1\right)^2\ge0\forall x\)

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

Dấu " = " xảy ra khi 

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

\(x+1=0\)

\(x=-1\)

Vậy \(x=-1\)khi \(GTNN=-3\)

B:C: tương tự

d) Ta có: \(\left(2x-1\right)^{18}\ge0\forall x\)

              \(\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow D=\left(2x-1\right)^{18}+\left(y+2\right)^2+7\ge7\)

Dấu " = " xảy ra khi \(\hept{\begin{cases}\left(2x-1\right)^{18}=0\\\left(y+2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x-1=0\\y+2=0\end{cases}\Rightarrow}\hept{\begin{cases}2x=1\\y=-2\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=-2\end{cases}}}\)

Vậy \(x=\frac{1}{2};y=-2\)khi \(GTNN=7\)

e) \(\left|-2x+6\right|\ge0\)

\(\Rightarrow E=\left|-2x+6\right|+12\ge12\)

Dấu " = " xảy ra khi \(\left|-2x+6\right|=0\Rightarrow-2x=-6\Rightarrow x=3\)

Vậy x = 3 khi đạt GTNN = 12

F ; G tương tự

hok tốt!!

+) A=(x+1)² - 3  

Vì  (x+1)² \(\ge\)0 nên (x+1)² - 3 \(\ge\) - 3 .Dấu "=" xảy ra \(\Leftrightarrow\)(x+1)² = 0   \(\Leftrightarrow\)x = - 1

Vậy min A = - 3 khi x = -1

+) B=(2x-5)²⁰ + 9  

Vì (2x-5)²⁰ \(\ge\)0 nên (2x-5)²⁰+9\(\ge\)9.Dấu "=" xảy ra \(\Leftrightarrow\)(2x - 5)²⁰=0    \(\Leftrightarrow\)x=\(\frac{5}{2}\)

Vậy min B=9 khi x=\(\frac{5}{2}\)

Những phần khác cũng làm tương tự :

+) minC= - 5 khi x=\(\frac{4}{3}\)

+) minD= 7 khi x=\(\frac{1}{2}\)và y= - 2

+) minE=12 khi x=3

+) min F = -17 khi x=5

+) min G = -12 khi x= - 4

|2x-5|\(\ge\)0 mọi x. Dấu bằng xảy ra <=> x=5/2

=> |2x-5|-12\(\ge\)-12

Do đó B đạt GTNN bằng -12 <=> x=5/2

B = | 2x - 5 | - 12

 Ta có: | 2x - 5 | \(\ge\)0

=> | 2x - 5 | - 12 \(\ge\)-12

Hay               B \(\ge\)-12

Dấu '' = '' xảy ra \(\Leftrightarrow\)| 2x - 5 | = 0

                                => 2x - 5 = 0

                                =>      2x = 5

                                =>        x = \(\frac{5}{2}\)

Vậy GTNN của B bằng -12 khi x = \(\frac{5}{2}\)