Question

[Lương Cẩm Tú]

Bạn nào làm giúp mình với,sẵn ôn thi luôn

Answer 1

[Võ Đông Anh Tuấn]

Bài 1 : \(2\left(3x-1\right)-3x=10\)

\(\Leftrightarrow6x-2-3x=10\)

\(\Leftrightarrow3x=12\)

\(\Leftrightarrow x=4\)

Vậy...................

b ) \(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\left(1\right)\)

ĐKXĐ : \(x\ne0;x\ne-1\)

\(\left(1\right)\Rightarrow\left(x+1\right)^2+x\left(x+1\right)=x\left(3x-1\right)+1\)

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

\(\Leftrightarrow-x^2+4x=0\)

\(\Leftrightarrow x\left(-x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTMĐKXĐ\right)\\x=4\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy .......................

c ) \(\dfrac{2x+1}{3}-\dfrac{3x-2}{2}>\dfrac{1}{6}\)

\(\Leftrightarrow2\left(2x+1\right)-3\left(3x-2\right)>1\)

\(\Leftrightarrow4x+2-9x+6>1\)

\(\Leftrightarrow-5x>-7\)

\(\Leftrightarrow x< \dfrac{7}{5}.\)

Vậy .......

Answer 2

[Võ Đông Anh Tuấn]

a ) \(A=\left(\dfrac{x^2-3}{x^2-9}+\dfrac{1}{x-3}\right):\dfrac{x}{x+3}.ĐKXĐ:x\ne3;x\ne-3\)

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

\(A=\dfrac{x^2-3x+x^2+3x}{x\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x}\)

\(A=\dfrac{x+1}{x-3}\)

b ) \(\left|A\right|=3.\) thì x là ?

\(\left|\dfrac{x+1}{x-3}\right|=3\)

Kẻ bảng ra làm nha :D