a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)

=>3(3x+2)-4(3x+1)=10

=>9x+6-12x-4=10

=>-3x+2=10

=>-3x=8

=>x=-8/3

b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)

=>(x-1)(x-2)-x(x+2)=-9x+10

=>x^2-3x+2-x^2-2x=-9x+10

=>-5x+2=-9x+10

=>x=2(loại)

\(1,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\\ \Leftrightarrow\dfrac{4\left(3x+2\right)}{24}-\dfrac{6\left(3x-2\right)}{24}-\dfrac{45}{24}=0\\ \Leftrightarrow12x+24-18x+12-45=0\\ \Leftrightarrow-6x-9=0\\ \Leftrightarrow x=-\dfrac{3}{2}\)

2, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{x\left(3+x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{8x-6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6-3x-x^2-8x+6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow-2x^2-10x+12=0\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow\left(x-1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

\(a,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\)

\(\Leftrightarrow4\left(3x+2\right)-6\left(3x-2\right)=45\)

\(\Leftrightarrow12x+8-18x+12=45\)

\(\Leftrightarrow12x-18x=45-12-8\)

\(\Leftrightarrow-6x=25\)

\(\Leftrightarrow x=\dfrac{-25}{6}\)

Vậy \(S=\left\{\dfrac{-25}{6}\right\}\)

\(b,\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)

\(\Leftrightarrow\left(x+2\right)\left(3-x\right)-x\left(3+x\right)=8x-6\)

\(\Leftrightarrow3x-x^2+6-2x-3x-x^2=8x-6\)

\(\Leftrightarrow-x^2-x^2+3x-2x-3x-8x=-6+6\)

\(\Leftrightarrow-2x^2-10x=0\)

\(\Leftrightarrow-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;5\right\}\)

= 3x^2-2x+2

k mình nha

\(\left(x+2\right)-2=0\)

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

chiu

a: =>2x-3x^2-x<15-3x^2-6x

=>x<-6x+15

=>7x<15

=>x<15/7

b: =>4x^2-24x+36-4x^2+4x-1>=12x

=>-20x+35>=12x

=>-32x>=-35

=>x<=35/32

\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\\ \Leftrightarrow2x-3x^2-x< 15-3x^2-6x\\ \Leftrightarrow3x^2-3x^2+2x+6x-x< 15\\ \Leftrightarrow7x< 15\\ \Leftrightarrow x< \dfrac{15}{7}\)

Vậy S={-∞; 15/7}

\(b,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12x\\ \Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\ge0\\ \Leftrightarrow4x^2-4x^2-24x+4x-12x\ge-36+1\\ \Leftrightarrow-32x\ge-35\\ \Leftrightarrow x\le\dfrac{35}{32}\)

Vậy S={-∞; 35/32]

Căn thức cuối cùng là \(\sqrt{1+x^2}\) hay \(\sqrt{1-x^2}\) vậy nhỉ?

\(\sqrt{1+x^2}\) thì bài này ko giải được

a: =>4/3x=7/9-4/9=1/3

=>x=1/4

b: =>5/2-x=9/14:(-4/7)=-9/8

=>x=5/2+9/8=29/8

c: =>3x+3/4=8/3

=>3x=23/12

hay x=23/36

d: =>-5/6-x=7/12-4/12=3/12=1/4

=>x=-5/6-1/4=-10/12-3/12=-13/12

em moi lop 4 mà

\(d,-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-4}{12}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{13}{12}\)

\(\Leftrightarrow x=\dfrac{13}{12}\)

\(c,3x+\dfrac{3}{4}=2\dfrac{2}{3}\)

\(\Leftrightarrow3x+\dfrac{3}{4}=\dfrac{8}{3}\)

\(\Leftrightarrow3x=\dfrac{8}{3}-\dfrac{3}{4}\)

\(\Leftrightarrow3x=\dfrac{23}{12}\)

\(\Leftrightarrow x=\dfrac{23}{12}:3\)

\(\Leftrightarrow x=\dfrac{23}{36}\)