Bài 2:

Vì x<0

nên x<1

=>|x-1|+|x|+x=1-x-x+x=-x+1

\(M=\dfrac{-x+1}{3x^2-4x+1}=\dfrac{-\left(x-1\right)}{\left(3x-1\right)\left(x-1\right)}=\dfrac{-1}{3x-1}\)

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

2x^2-3x+2=2(x^2-3/2x+1)

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

MinC = 3/4 (khi x = -3/2)

\(C=\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{x^4+3x^2-3x^3-9x+3x^2+9}=\dfrac{\left(x^2+3\right)\left(x^4+6x^2+9-9x^2\right)}{\left(x^2+3\right)\left(x^2-3x+3\right)}\\ C=\dfrac{\left(x^2+3\right)^2-9x^2}{x^2-3x+3}=\dfrac{\left(x^2-3x+3\right)\left(x^2+3x+3\right)}{x^2-3x+3}\\ C=x^2+3x+3=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

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

1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)

Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)

2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)

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

\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

Suy ra: \(9x^2+6x+24x+16=9x^2\)

\(\Leftrightarrow30x+16=0\)

\(\Leftrightarrow30x=-16\)

hay \(x=-\dfrac{8}{15}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)

giúp mình với ạ câu nào cũng được

Đặt A = \(9x+\dfrac{4x+3}{2x+1}\) (x\(\ne\dfrac{-1}{2}\))

= \(\dfrac{18x^2+13x+3}{2x+1}\)

<=> 2A = \(\dfrac{36x^2+26x+6}{2x+1}\)

= \(\dfrac{9\left(2x+1\right)^2-5\left(2x+1\right)+2}{2x+1}\)

= \(9\left(2x+1\right)+\dfrac{2}{2x+1}-5\)

Áp dụng bdt Co-si, ta có: 

\(9\left(2x+1\right)+\dfrac{2}{2x+1}\ge2\sqrt{9\left(2x+1\right).\dfrac{2}{2x+1}}=6\sqrt{2}\)

=> 2A \(\ge6\sqrt{2}-5\)

<=> A \(\ge3\sqrt{2}-\dfrac{5}{2}\)

Dấu "=" xảy ra <=> x = \(\dfrac{-3+\sqrt{2}}{6}\)