Giups mik vs

b/ \(3-100x+8x^2=8x^2+x-300\)

\(\Leftrightarrow-101x=-303\)

\(\Rightarrow x=3\)

c/ \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow25x+10-80x+10=24x+12-150\)

\(\Leftrightarrow-79x=-158\)

\(\Rightarrow x=2\)

d/ \(3\left(3x+2\right)-\left(3x+1\right)=12x+10\)

\(\Leftrightarrow9x+6-3x-1=12x+10\)

\(\Leftrightarrow-6x=5\)

\(\Rightarrow x=-\frac{5}{6}\)

e/ \(30x-6\left(2x-5\right)+5\left(x+8\right)=210+10\left(x-1\right)\)

\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)

\(\Leftrightarrow13x=130\)

\(\Rightarrow x=10\)

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

\(\Rightarrow A_{min}=-3\) khi \(x=2\)

\(B=4x^2+4x+11=\left(2x+1\right)^2+10\ge10\)

\(\Rightarrow B_{min}=10\) khi \(x=-\frac{1}{2}\)

\(C=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

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

\(\Rightarrow C_{min}=-36\) khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(D=-x^2-8x-16+21=21-\left(x+4\right)^2\le21\)

\(\Rightarrow C_{max}=21\) khi \(x=-4\)

\(E=-x^2+4x-4+5=5-\left(x-2\right)^2\le5\)

\(\Rightarrow E_{max}=5\) khi \(x=2\)

\(A=x^2-3x+5\)

\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)

\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)

\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)

Dấu "=" xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)

Vậy Min A = \(\frac{11}{4}\Leftrightarrow x=\frac{3}{2}\)

a) \(A=x^2-3x+5\)

\("="\Leftrightarrow x=\frac{11}{4}\Rightarrow x=\frac{3}{2};\frac{11}{4}\)

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

\("="\Leftrightarrow x=5\Rightarrow x=0;5\)

c) \(C=4x-x^2+3\)

\("="\Leftrightarrow x=7\Rightarrow x=2;7\)

d) \(D=x^4+x^2+2\)

\("="\Leftrightarrow x=2\Rightarrow x=0;2\)

a, A <=> \(x^2-2x\frac{3}{2}+\left(\frac{3}{2}\right)^2+2,75\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+2,75\)

ta có \(\left(x-\frac{3}{2}\right)^2\ge0\)

\(\Rightarrow A\ge2,75\) 

=> Min A =2,75 \(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x=\frac{3}{2}\)

b, \(B\Leftrightarrow4x^2-4x+1+x^2+4x+4\)

\(B\Leftrightarrow5x^2+5\) 

Ta có \(5x^2\ge0\Rightarrow B\ge5\)

=> Min B = 5 <=> x=0

c,\(C\Leftrightarrow-\left(x^2-4x+4-7\right)\)

\(C\Leftrightarrow7-\left(x-2\right)^2\)

Ta có \(\left(x-2\right)^2\ge0\Rightarrow C\le7\)

=> Max C=7 <=> ( x - 2 )² = 0 <=> x=2

d, \(C=x^4+x^2+2\)

Lại có \(x^4+x^2\ge0\)

\(\Rightarrow C\ge2\). Để Min C= 2 <=> \(x^4+x^2=0\Leftrightarrow x^2\left(x^2+1\right)=0\)\(\Leftrightarrow x=0\)

f,F \(F\Leftrightarrow-\left(\left(3x\right)^2-2.3x.2+2^2-19\right)\)

\(F\Leftrightarrow19-\left(3x-2\right)^2\)

ta có \(\left(3x-2\right)^2\ge0\)

=> \(F\le19\)

Để Max F =19 <=> x=\(\frac{2}{3}\)

mình 2k4 ko bt làm

 a)    \(B=\frac{3x^2+6x+10}{x^2+2x+5}\)

\(\Leftrightarrow B=3-\frac{5}{x^2+2x+5}\)

\(\Leftrightarrow B=3-\frac{5}{5\left(\frac{x^2}{5}+\frac{2x}{5}+\frac{5}{5}\right)}\Leftrightarrow B=3-\frac{1}{\frac{\left(x^2+2x+1\right)}{5}+\frac{4}{5}}\)( cho \(\left(x+1\right)^2=0\))

\(\Leftrightarrow maxB=3-\frac{1}{\frac{4}{5}}=\frac{7}{4}\)   KHI X= -1

c)  \(D=x^2-2x+y^2+4y+7\)

\(\Leftrightarrow D=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+2\)

\(\Leftrightarrow D=\left(x-1\right)^2+\left(y+2\right)^2+2\)

\(\Leftrightarrow minD=2\)KHI X= 1 và Y= -2

e) Câu này đề có vẻ sai bạn kiểm tra lại giúp mk ! mk làm theo đề đúng nka !

         \(E=\frac{x^2-4x+1}{x^2}\)

\(\Leftrightarrow E=\frac{x^2\left(1-\frac{4}{x}+\frac{1}{x^2}\right)}{x^2}=1-\frac{4}{x}+\frac{1}{x^2}\)

ĐẶT    \(y=\frac{1}{x}\)\(\Leftrightarrow minE=-3\)KHI X = 1/2

Hai câu còn lại tối mk giải tiếp mk bận đi học rùi bạn thông cảm 

566889954+657565=

\(A=3x^2-5x+3=3(x^2-\frac{5}{3}x)+3\)

\(=3(x^2-\frac{5}{3}x+\frac{5^2}{6^2})+\frac{11}{12}=3(x-\frac{5}{6})^2+\frac{11}{12}\)

Vì \((x-\frac{5}{6})^2\geq 0, \forall x\Rightarrow A\geq 3.0+\frac{11}{12}=\frac{11}{12}\)

Vậy A(min)$=\frac{11}{12}$ khi $x=\frac{5}{6}$

\(B=2x^2+2x+1=2(x^2+x+\frac{1}{4})+\frac{1}{2}\)

\(=2(x+\frac{1}{2})^2+\frac{1}{2}\geq 2.0+\frac{1}{2}=\frac{1}{2}\)

Vậy \(B_{\min}=\frac{1}{2}\) tại \((x+\frac{1}{2})^2=0\Leftrightarrow x=\frac{-1}{2}\)

C)

\(C=2x^2+y^2+10x-2xy+27\)

\(=(x^2+10x+25)+(x^2+y^2-2xy)+2\)

\(=(x+5)^2+(x-y)^2+2\)

Vì \((x+5)^2\ge 0, (x-y)^2\geq 0\Rightarrow C\geq 0+0+2=2\)

Vậy \(C_{\min}=2\) tại \(\left\{\begin{matrix} (x+5)^2=0\\ (x-y)^2=0\end{matrix}\right.\Leftrightarrow x=y=-5\)

E)

\(E=(x-3)^2+(x-5)^2\)

\(=x^2-6x+9+x^2-10x+25\)

\(=2x^2-16x+34=2(x^2-8x+16)+2\)

\(=2(x-4)^2+2\geq 2.0+2=2\)

Vậy $E_{\min}=2$ tại $(x-4)^2=0$ hay $x=4$

H)

\(H=(x-7)(x-5)(x-4)(x-2)-70\)

\(=[(x-7)(x-2)][(x-5)(x-4)]-70\)

\(=(x^2-9x+14)(x^2-9x+20)-70\)

\(=a(a+6)-70\) (đặt \(a=x^2-9x+14)\)

\(=(a+3)^2-79\geq 0-79=-79\)

Vậy $H_{\min}=-79$ tại \(a=x^2-9x+14=-3\Leftrightarrow x=\frac{9\pm \sqrt{13}}{2}\)

a) \(\sqrt{x-2}+\dfrac{1}{x-5}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)

b) \(\sqrt{\left(2x-6\right)\left(7-x\right)}=\sqrt{2\left(x-3\right)\left(7-x\right)}\) có nghĩa khi:

\(\left(x-3\right)\left(7-x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\7-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\7-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\ge7\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow3\le x\le7\)

c) \(\sqrt{4x^2-25}=\sqrt{\left(2x-5\right)\left(2x+5\right)}\) có nghĩa khi:

\(\left(2x-5\right)\left(2x+5\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\2x+5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ge-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\)

d) \(\dfrac{2}{x^2-9}-\sqrt{5-2x}=\dfrac{2}{\left(x+3\right)\left(x-3\right)}-\sqrt{5-2x}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\5-2x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\le\dfrac{5}{2}\end{matrix}\right.\)

e) \(\dfrac{x}{x^2-4}+\sqrt{x-2}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\sqrt{x-2}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x-2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow x>2\)

\(A=x^2+3x+7\)

\(=x^2+2.1,5x+2,25+4,75\)

\(=\left(x+1,5\right)^2+4,75\ge4,75\)

Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)

\(B=2x^2-8x\)

\(=2\left(x^2-4x\right)\)

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

\(=2\left[\left(x-2\right)^2-4\right]\)

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

Vậy \(B_{min}=-8\Leftrightarrow x=2\)