1) \(2\left(3x-1\right)-3x=10\)
<=> \(6x-2-3x=10\)
<=>\(3x-2=10\)
<=> \(3x=12\)
<=> \(x=4\)
Vậy tập nghiệm của pt S={4}
2) \(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\)
ĐKXĐ: x khác 0; x khác 1,-1
<=> \(\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}+\dfrac{x\left(x+1\right)}{x\left(x+1\right)}\)= \(\dfrac{3x^2-x}{x\left(x+1\right)}+\dfrac{1}{x\left(x+1\right)}\)
=> \(\left(x+1\right)^2+x\left(x+1\right)\)= \(3x^2-x+1\)
<=> \(x^2+2x+1+x^2+x=3x^2-x+1\)
<=> \(x^2+x^2+2x+x-3x^2+x\)= \(1-1\)
<=> \(-x^2+4x=0\)
<=>\(4x=x^2\)
<=> \(4=x\) ( TMĐKXĐ)
Vậy tập nghiệm của pt S={4}
c) \(\dfrac{2x+1}{3}-\dfrac{3x-2}{2}>\dfrac{1}{6}\)
<=> \(\dfrac{4x+2}{6}-\dfrac{9x-6}{6}>\dfrac{1}{6}\)
<=> \(\dfrac{4x+2-9x+6}{6}-\dfrac{1}{6}>0\)
<=> \(\dfrac{-5x+7}{6}>0\)
Mà 6>0 . Nên \(-5x+7>0\)
Ta có \(-5x+7>0\)
<=> \(-5x>-7\)
<=> \(x< \dfrac{7}{5}\)
Vậy tập nghiệm của bất phương trình S={x thuộc R| \(x< \dfrac{7}{5}\)}
1)2.(3x-1)-3x=10
6x-2-3x =10
6x-3x =10+2
3x =12
x =4
Vậy S=4
2) \(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\)
Đkxđ: \(x\ne0\) và \(x\ne-1\)
MTC;x(x+1)
\(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\)
\(\Leftrightarrow\)\(\dfrac{\left(x+1\right)\left(x+1\right)+x\left(x+1\right)}{x\left(x+1\right)}=\dfrac{x\left(3x-1\right)+1}{x\left(x+1\right)}\)
\(\Leftrightarrow\)(x+1) (x+1)+x(x+1) = x (3x-1)+1
\(\Leftrightarrow\)x2+x+x+1+x2+x =3x2-x+1
\(\Leftrightarrow\)x2+x+x+1+x2+x-3x2+x-1=0
\(\Leftrightarrow\)-x24x=0
\(\Leftrightarrow\)4x-x2=0
\(\Leftrightarrow\)x(4-x)=0
\(\Leftrightarrow\)x=0 hoặc 4-x=0
\(\Leftrightarrow\)x=0 hoặc x =4
3)\(\dfrac{2x+1}{3}-\dfrac{3x-2}{2}>\dfrac{1}{6}\)
\(\Leftrightarrow\)\(\dfrac{2x+1}{3}6-\dfrac{3x-2}{2}6>\dfrac{1}{6}\)6
\(\Leftrightarrow\)2(2x+1)-3(3x-2)>1
\(\Leftrightarrow\)4x+2-9x+6>1
\(\Leftrightarrow\)4x-9x>1-2-6
\(\Leftrightarrow\)-5x>-7
\(\Leftrightarrow\)-5x.\(\dfrac{1}{-5}>-7.\dfrac{1}{-5}\)
\(\Leftrightarrow x>\dfrac{7}{5}\)
