\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{1}{x^2+2x-3}=1.\)

\(ĐK:\hept{\begin{cases}x-1\ne0\\x+3\ne\\x^2+2x-3\ne0\end{cases}0}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\Leftrightarrow-3\end{cases}}\)

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)+4-x^2-2x+3=0\)

\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5+4-x^2-2x+3=0\)

\(\Leftrightarrow3x+9=0\)

\(\Leftrightarrow3x=-9\Leftrightarrow x=-3\) (loại)

 Vậy pt vô N^o

Số t tính đc rất thú dị :) 

\(x^4-2x^3+3x^2-2x+1=0\)

Chia cả hai vé cho \(x^2\)

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

\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)+1=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)+1=0\)

Đặt x+1/x = a, ta có:

\(a^2-2a+1=0\)

\(\Leftrightarrow\left(a-1\right)^2=0\)

\(\Leftrightarrow a=1\)

\(\Leftrightarrow x+\dfrac{1}{x}=1\)

\(\Leftrightarrow x^2+1=x\)

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

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+3>0\)

Do đó phương trình vô nghiệm

       1 + sin ( 3X) = sin( 2X)+ sin ( 3X)

<=>1 - sin( 2X) = sin ( 3X) - sin(3X)

<=>1 - sin (2X) = 0

<=> - sin ( 2X) = -1

<=>   sin ( 2X) = 1

<=>  X             = 45  ( dùng máy tính bấm ra nha ) 

Ta có:

$\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x+4}-\dfrac{12x-1}{4-4x}$

Điều kiện: $x\ne-1,\ x\ne1$.

Ta có:

$\dfrac{6}{(x-1)(x+1)}+5=\dfrac{8x-1}{4(x+1)}+\dfrac{12x-1}{4(x-1)}$

Quy đồng:

$\dfrac{6+5(x^2-1)}{x^2-1}=\dfrac{(8x-1)(x-1)+(12x-1)(x+1)}{4(x^2-1)}$

$\dfrac{5x^2+1}{x^2-1}=\dfrac{20x^2+2x-2}{4(x^2-1)}$

$4(5x^2+1)=20x^2+2x-2$

$20x^2+4=20x^2+2x-2$

$2x=6$

$x=3$

Vậy $x=3$.

Ta có:

$\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}$

Điều kiện: $x\ne\pm\dfrac12$.

Quy đồng:

$\dfrac{(2x+1)^2-(2x-1)^2}{4x^2-1}=\dfrac{8}{4x^2-1}$

$\dfrac{8x}{4x^2-1}=\dfrac{8}{4x^2-1}$

$8x=8$

$x=1$

Vậy $x=1$.

a) \(3x-5\le x+1\)

\(\Leftrightarrow3x-x\le1+5\)

\(\Leftrightarrow2x\le6\)

\(\Leftrightarrow x\le3\)

Vậy.............

b) \(\frac{2x-2}{2}>\frac{2-x+2}{3}\)

\(\Leftrightarrow3\left(2x-2\right)>2\left(2-x+2\right)\)

\(\Leftrightarrow6x-6>4-2x+4\)

\(\Leftrightarrow6x+2x>4+4+6\)

\(\Leftrightarrow8x>14\)

\(\Leftrightarrow x>\frac{7}{4}\)

Vậy..................

a,\(3x-5\le x+1\Leftrightarrow2x\le6\Leftrightarrow x\le3\)

b,\(\frac{2x-2}{3}>2-\frac{x+2}{2}\)

\(\Leftrightarrow2\left(2x-2\right)>12-3\left(x+2\right)\)

\(\Leftrightarrow4x-4>12-3x-6\)

\(\Leftrightarrow7x>10\Leftrightarrow x>\frac{10}{7}\)

câu c thì easy khỏi làm nha đã ib ở tin nhắn riêng rồi mà còn :)

\(2x+1\ge0\Leftrightarrow x\ge-\frac{1}{2}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-1=2x+1\\x^2-1=1-2x\end{cases}}\)Rồi tự giải nốt đi bạn hiện gợi ý cho tới đó rồi đấy