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 \(x=-\dfrac{5}{4}\)

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

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

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

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

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

Ta có : \(x^4-3x^3+4x^2-3x+10.\)

\(=\left(x^4-2x^3+x^2\right)-\left(x^3-3x^2+3x-1\right)+9\)

\(=x^2\left(x-1\right)^2-\left(x-1\right)^3+9\)

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

Mà \(\left(x-1\right)^2\ge0\)

\(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

\(\Rightarrow\left(x-1\right)^2\left(x^2-x+1\right)\ge0\)

\(\Rightarrow\left(x-1\right)^2\left(x^2-x+1\right)+9\ge9\)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vậy GTNN cảu \(x^4-3x^3+4x^2-3x+10.\)là 9 <=> \(x=1\)

A = (x^2-6x+9)+1

   = (x-3)^2+1 >= 1

Dấu "=" xảy ra <=> x-3=0 <=> x=3

Vậy GTNN của A = 1 <=> x=3

Tk mk nha

\(A=x^2-6x+10\)

\(=\left(x^2-3x\right)-\left(3x-9\right)+1\)

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

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

\(=\left(x-3\right)^2+1\)   \(\ge1\)    (vi    (x-3)²  \(\ge0\))

Dấu  "="  xảy ra  \(\Leftrightarrow\)\(x-3=0\)

                          \(\Leftrightarrow\) \(x=3\)

Vậy   \(Min\)  \(A=1\) \(\Leftrightarrow\)\(x=3\)

Ta có: \(A=\left|x-2\right|+\left|x-10\right|=\left|x-2\right|+\left|10-x\right|\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(A\ge\left|x-2+10-x\right|=\left|-8\right|=8\)

Dấu " = " khi \(\left\{{}\begin{matrix}x-2\ge0\\10-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\le10\end{matrix}\right.\Rightarrow2\le x\le10\)

Vậy \(MIN_A=8\) khi \(2\le x\le10\)

GTNN là gì vậy bạn?

phải nói thì mình mới giúp đc!

tìm GTNN đâu phải trong chương trình lớp 7

\(C=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)

Đặt \(x^2-7x=t\),khi đó:

\(C=\left(t+10\right).\left(t-10\right)=t^2-10^2=t^2-100\)

Vì \(t^2\ge0=>t^2-100\ge-100\) (với mọi t)

Dấu "=" xảy ra\(< =>t=0< =>x^2-7x=0< =>x\left(x-7\right)=0< =>\orbr{\begin{cases}x=0\\x=7\end{cases}}\)

Vậy minC=-100 khi x=0 hoặc x=7

Áp dụng 1/x + 1/y >= 4/(x+y) với x,y>0(đề có thiếu ko nhỉ)

Dấu "=" xảy ra khi x=y 

ta co 10=x+y>=2can xy

100=(x+y)²>=4xy

100/xy>=4

10/xy>=4/10

1:x+1:y>=2/5