\(B=\frac{x^2-6x+14}{x^2-6x+12}\)

\(B=\frac{x^2-6x+12+2}{x^2-6x+12}\)

\(B=1+\frac{2}{\left(x-3\right)^2+3}\le1+\frac{2}{3}\)

\(B=1+\frac{2}{\left(x-3\right)^2+3}\le\frac{5}{3}\)

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

B=\(\frac{x^2-6x+14}{x^2-6x+12}\)

=\(\frac{x^2-6x+9+3+2}{x^2-6x+9+3}\)

=\(\frac{\left(x^2-6x+9\right)+3+2}{\left(x^2-6x+9\right)+3}\)

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

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

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

*Ta có:(x-3)² \(\ge\) 0;với mọi x;cộng 3 vào 2 vế

\(\Rightarrow\)(x-3)²+3 \(\ge\) 0+3;với mọi x

\(\Rightarrow\)(x-3)²+3 \(\ge\) 3;với mọi x

\(\Rightarrow\)\(\frac{2}{\left(x-3\right)^2+3}\) \(\ge\) 0;với mọi x;lấy hai vế cộng cho1​​

\(\Rightarrow\)​\(1+\frac{2}{\left(x-3\right)^2+3}\)​ \(\ge\)1+0;với mọi x

Vậy .................................

\(B=\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\Rightarrow B_{max}=\frac{3}{4}\) 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

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 \(\Leftrightarrow x=1\)

Rút gọn nha các cậu

\(A=\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\times\frac{x^2-36}{12x^2+12}\)

\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\times\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)

\(A=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}\times\frac{1}{12\left(x^2+1\right)}\)

\(A=\frac{12\left(x^2+1\right)}{x}\times\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)

\(B=\frac{x^2-6x+14}{x^2-6x+12}=\frac{x^2-6x+12+2}{x^2-6x+12}=1+\frac{2}{x^2-6x+12}\)

ta có: \(x^2-6x+12=x^2-2.3.x+3^2+4=\left(x-3\right)^2+4\ge4\)

để Bmax => \(\left(\frac{2}{x^2-6x+12}\right)max\Rightarrow x^2-6x+12min\)và lớn hơn 0 vì 2>0

mà \(\left(x-3\right)^2+4\) \(\ge\)4

dấu = xảy ra khi x-3=0

=> x=3

Vậy \(MaxB=\frac{3}{2}\)khi x=3

Ta có: \(x^2-6x-2+\frac{14}{x^2-6x+7}=0\)

\(\Leftrightarrow\frac{\left(x^2-6x-2\right)\left(x^2-6x+7\right)+14}{x^2-6x+7}=0\)

\(\Leftrightarrow x^4-12x^3+41x^2-30x-14+14=0\)

\(\Leftrightarrow x^4-12x^3+41x^2-30x=0\)

ĐKXĐ : \(x^2-6x+7\ne0\)

=> \(x^2-6x+9-2\ne0\)

=> \(\left(x-3\right)^2\ne2\)

=> \(\left[{}\begin{matrix}x-3\ne-\sqrt{2}\\x-3\ne\sqrt{2}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x\ne3-\sqrt{2}\\x\ne3+\sqrt{2}\end{matrix}\right.\)

- Ta có : \(x^2-6x-2+\frac{14}{x^2-6x+7}=0\)

Đặt : \(a=x^2-6x+7\)

=> \(a-9=x^2-6x-2\)

- Thay \(a-9=x^2-6x-2\), \(a=x^2-6x+7\) vào phương trình ta được : \(a-9+\frac{14}{a}=0\)

=> \(\frac{a^2}{a}-\frac{9a}{a}+\frac{14}{a}=0\)

=> \(a^2-9a+14=0\)

=> \(a^2-7a-2a+14=0\)

=> \(a\left(a-2\right)-7\left(a-2\right)=0\)

=> \(\left[{}\begin{matrix}a-7=0\\a-2=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}a=7\\a=2\end{matrix}\right.\)

- Thay \(a=x^2-6x+7\) vào phương trình trên ta được :

\(\left[{}\begin{matrix}x^2-6x+7=7\\x^2-6x+7=2\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x^2-6x=0\\x^2-6x=5\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x\left(x-6\right)=0\\x^2-5x-x-5=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x\left(x-6\right)=0\\x\left(x-1\right)-5\left(x-1\right)=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x\left(x-6\right)=0\\\left(x-1\right)\left(x-5\right)=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\x-6=0\\x-5=0\\x-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\x=6\\x=5\\x=1\end{matrix}\right.\) ( TM )

Vậy phương trình có nghiệm là x = 0, x = 6, x = 5, 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) ĐKXĐ: $x\neq 1$

PT \(\Leftrightarrow \frac{x^2+x+1+2(x-1)}{(x-1)(x^2+x+1)}=\frac{3x^2}{x^3-1}\)

\(\Leftrightarrow \frac{x^2+3x-1}{x^3-1}=\frac{3x^2}{x^3-1}\)

\(\Rightarrow x^2+3x-1=3x^2\Leftrightarrow 2x^2-3x+1=0\)

\(\Leftrightarrow (x-1)(2x-1)=0\)

Mà $x\neq 1$ nên $2x-1=0\Rightarrow x=\frac{1}{2}$ là nghiệm

b) ĐK: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{3-x}{2-x}=\frac{1}{x+2}-\frac{6-x}{3x^2-12}\)

\(\Leftrightarrow \frac{1}{x+2}-\frac{3-x}{2-x}=\frac{6-x}{3(x^2-4)}\)

\(\Leftrightarrow \frac{1}{x+2}+\frac{3-x}{x-2}=\frac{6-x}{3(x-2)(x+2)}\)

\(\Leftrightarrow \frac{-x^2+2x+4}{(x-2)(x+2)}=\frac{6-x}{3(x-2)(x+2)}\)

\(\Rightarrow 3(-x^2+2x+4)=6-x\)

\(\Leftrightarrow -3x^2+7x+6=0\)

\(\Leftrightarrow (x-3)(3x+2)=0\Rightarrow x=3\) hoặc $x=-\frac{2}{3}$

Vậy........

c) ĐK: $x\neq \pm 2$

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

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

\(\Leftrightarrow \frac{x^2-7x-2}{(x-2)(x+2)}=\frac{2x-22}{(x-2)(x+2)}\)

\(\Rightarrow x^2-7x-2=2x-22\)

\(\Leftrightarrow x^2-9x+20=0\Leftrightarrow (x-4)(x-5)=0\Rightarrow x=4\) hoặc $x=5$

(đều thỏa mãn)

d) ĐK: \(x^2-6x+7\neq 0\)

PT \(\Leftrightarrow (x^2-6x+7)+\frac{14}{x^2-6x+7}-9=0\)

\(\Rightarrow (x^2-6x+7)^2-9(x^2-6x+7)+14=0\)

\(\Leftrightarrow (x^2-6x+7-2)(x^2-6x+7-7)=0\)

\(\Leftrightarrow (x^2-6x+5)(x^2-6x)=0\)

\(\Leftrightarrow (x-1)(x-5)x(x-6)=0\)

\(\Rightarrow x\in \left\{1;5;0;6\right\}\) (đều thỏa mãn)

Vậy.........

\(=\frac{x-12}{6\left(x-6\right)}-\frac{6}{x\left(x-6\right)}\)

\(=\frac{x^2-12x-36}{6x\left(x-6\right)}\)

\(=\frac{\left(x-6-6\sqrt{2}\right)\left(x-6+6\sqrt{2}\right)}{6x\left(x-6\right)}\)

\(\frac{x-12}{6x-36}-\frac{6}{x^2-6x}=\frac{\left(x-12\right)\left(x^2-6x\right)}{\left(6x-36\right)\left(x^2-6x\right)}-\frac{6\left(6x-36\right)}{\left(x^2-6x\right)\left(6x-36\right)}\)

\(=\frac{x^3-6x^2-12x^2+72x}{\left(6x-36\right)\left(x^2-6x\right)}-\frac{36x-216}{\left(x^2-6x\right)\left(6x-36\right)}\)

\(=\frac{x^3-18x^2+72x-36x+216}{\left(6x-36\right)\left(x^2-6x\right)}=\frac{x^3-18x^2+36x+216}{\left(6x-36\right)\left(x^2-6x\right)}\)