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1, B=\(\frac{3}{4x^2-4x+5}\)
=\(\frac{3}{\left(4x^2-2.2x+4\right)+5-4}\)
=\(\frac{3}{\left(2x-2\right)^2+1}\le\frac{3}{1}=3\)
Để B=3 thì : (2x-2)2=0
\(\Leftrightarrow2x-2=0\)
\(\Leftrightarrow x=1\)
Vậy Max B =3 \(\Leftrightarrow x=1\)
\(A=-2x^2+5x-8\)
\(A=-2\left(x^2-\frac{5}{2}\cdot x+4\right)\)
\(A=-2\left(x^2-2\cdot x\cdot\frac{5}{4}+\frac{25}{16}+\frac{39}{16}\right)\)
\(A=-2\left[\left(x-\frac{5}{4}\right)^2+\frac{39}{16}\right]\)
\(A=-2\left(x-\frac{5}{4}\right)^2-\frac{39}{6}\le\frac{-39}{6}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{5}{4}\)
\(B=-x^2-y^2+xy+2x+2y\)
\(2B=-2x^2-2y^2+2xy-4x-4y\)
\(2B=-\left(2x^2+2y^2-2xy+4x+4y\right)\)
\(2B=-\left(x^2-2xy+y^2+x^2+4x+4+y^2+4y+4-8\right)\)
\(2B=-\left[\left(x-y\right)^2+\left(x+2\right)^2+\left(y+2\right)^2-8\right]\)
\(B=-\frac{\left(x-y\right)^2+\left(x+2\right)^2+\left(y+2\right)^2}{2}+4\le4\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=-2\)
\(C=\frac{3}{4x^2-4x+5}=\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)
\(D=\frac{x^2-6x+14}{x^2-6x+12}=\frac{x^2-6x+12+2}{x^2-6x+12}\)
\(=1+\frac{2}{\left(x-3\right)^2+3}\le1+\frac{2}{3}=\frac{5}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\)
a.\(\frac{3x-1}{3x+1}+\frac{x-3}{x+3}=2\)
\(\frac{\left(3x-1\right)\left(x+3\right)+\left(3x+1\right)\left(x-3\right)}{\left(3x+1\right)\left(x+3\right)}=\frac{3x^2+8x-3+3x^2-8x-3}{\left(3x+1\right)\left(x+3\right)}=\frac{6x^2-6}{\left(3x+1\right)\left(x+3\right)}=2\)
\(6x^2-6=2\left(3x^2+10x+3\right)\)
\(6x^2-6=6x^2+20x+6\)
-20x-12=0
x=\(\frac{-3}{5}\)
a) ĐKXĐ: \(x\ne\left\{-3;-\frac{1}{3}\right\}\)
Ta có: \(\frac{3x-1}{3x+1}+\frac{x-3}{x+3}=\)\(\frac{\left(3x-1\right)\left(x+3\right)+\left(x-3\right)\left(3x+1\right)}{\left(3x+1\right)\left(x+3\right)}\)=\(\frac{3x^2+9x-x-3+3x^2+x-9x-3}{3x^2+9x+x+3}\)
= \(\frac{6x^2-6}{3x^2+10x+3}\)
=> \(\frac{6x^2-6}{3x^2+10x+3}=2\)
<=> \(6x^2-6=6x^2+20x+6\)
<=> 20x=12
<=>x=\(\frac{12}{20}=\frac{3}{5}\)
Vậy x=3/5
\(a.=\frac{4x\left(x^2-2x+1\right)}{x^2-1x-5x+5}\)
\(=\frac{4x\left(x-1\right)^2}{x\left(x-1\right)-5\left(x-1\right)}\)
\(=\frac{4x\left(x-1\right)^2}{\left(x-5\right)\left(x-1\right)}\)
\(=\frac{4x\left(x-1\right)}{x-5}\)
b) \(\frac{4x^3-64x}{x^2-7x+12}\)
\(=\frac{4x\left(x^2-16\right)}{x^2-3x-4x+12}\)
\(=\frac{4x\left(x+4\right)\left(x-4\right)}{x\left(x-3\right)-4\left(x-3\right)}\)
\(=\frac{4x\left(x+4\right)\left(x-4\right)}{\left(x-4\right)\left(x-3\right)}\)
\(=\frac{4x\left(x+4\right)}{x-3}=\frac{4x^2+16x}{x-3}\)
c) \(\frac{x^2-6x+8}{x^3-8}\)
\(=\frac{x^2-2x-4x+8}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\frac{x\left(x-2\right)-4\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\frac{\left(x-4\right)\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\frac{x-4}{x^2+2x+4}\)
\(A=\frac{4x^2-12x+15}{x^2-3x+3}=4+\frac{3}{x^2-3x+3}=4+\frac{3}{\left(x-\frac{3}{2}\right)^2+\frac{3}{4}}\le8\)
dau '=' xay ra khi \(x=\frac{3}{2}\)
\(B=\frac{4x^2-8x+12}{x^2-2x+5}=4-\frac{8}{x^2-2x+5}=4-\frac{8}{\left(x-1\right)^2+4}\le2\)
dau '=' xay ra khi \(x=1\)
\(A=\frac{x^2}{2}-\frac{x}{6}+3\)
\(2A=x^2-\frac{x}{3}+6=x^2-2.x\frac{1}{6}+\frac{1}{36}+\frac{35}{36}\)
\(2A=\left(x+\frac{1}{6}\right)^2+\frac{35}{36}\ge\frac{35}{36}\)
\(\Rightarrow A\ge\frac{35}{72}\)Dấu "=" xảy ra khi \(x=\frac{-1}{6}\)
b)\(B=x^4-4x^3+6x^2-4x+5\)
\(B=\left(x^4-4x^3+4x^2\right)+\left(2x^2-4x+2\right)+3\)
\(B=\left(x^2-2x\right)^2+2\left(x+1\right)^2+3\ge3\)
Dấu "=" xảy ra khi:\(x=0;-1;2\)
\(B=\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\Rightarrow B_{max}=\frac{3}{4}\) khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\)
2/ Xem lại đề bài, đề bài này thì ko có max, 12 ở mẫu là dấu + thì may ra làm được
ở 12 là dấu cộng bạn ạ