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

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

\(\Leftrightarrow-6x^2=0\)

hay x=0

\(\left(1+\dfrac{1}{2x}\right).lg3+lg2=lg\left(27-3^{\dfrac{1}{x}}\right)\)
\(\Leftrightarrow lg3^{1+\dfrac{1}{2x}}+lg2=lg\left(27-3^{\dfrac{1}{x}}\right)\)
\(\Leftrightarrow lg\left(2.3^{1+\dfrac{1}{2x}}\right)=lg\left(27-3^{\dfrac{1}{x}}\right)\)
\(\Leftrightarrow2.3^{1+\dfrac{1}{2x}}=27-3^{\dfrac{1}{x}}\)
\(\Leftrightarrow2.3.\left(3^{\dfrac{1}{x}}\right)^2=27-3^{\dfrac{1}{x}}\)
Đặt \(3^{\dfrac{1}{x}}=t\left(t>0\right)\) phương trình trở thành:
\(2.3t^2=27-t\)
\(\Leftrightarrow\left[{}\begin{matrix}t_1=\dfrac{-1-\sqrt{649}}{12}\left(l\right)\\t_2=\dfrac{1+\sqrt{649}}{12}\left(tm\right)\end{matrix}\right.\)
Với \(t=\dfrac{-1-\sqrt{649}}{12}\Leftrightarrow3^{\dfrac{1}{x}}=\dfrac{-1-\sqrt{649}}{12}\)
\(\Leftrightarrow\dfrac{1}{x}=log^{\dfrac{-1-\sqrt{649}}{12}}_3\)
\(\Leftrightarrow x=log^3_{\dfrac{-1-\sqrt{649}}{12}}\).

\(\Leftrightarrow16-3\left(x+1\right)< 24+2\left(x-1\right)\)

=>16-3x-3<24+2x-2

=>-3x+13<2x+22

=>-5x<9

hay x>-9/5

|x-9|=2x+5

Xét 3 TH

TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)

TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)

TH3: x=9 =>0=23(L)

Vậy  x= 4/3

Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)

\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)

\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)

Ta có:

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\)

\(\dfrac{2\left(x+3\right)+3\left(x-3\right)}{x^2-9}=\dfrac{3x+5}{x^2-9}\)

\(5x-4=3x+5\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\)

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

BXD: 

Theo BXD, ta được; -1<x<3/2 hoặc x>2

a)\(\dfrac{7x-1}{2}+2x=\dfrac{16-x}{3}\)

\(\dfrac{\left(7x-1\right).3}{2.3}+\dfrac{2x.6}{6}=\dfrac{\left(16-x\right)2}{3.2}\)

khử mẫu 

=> (7x-1).3+12x=(16-x).2

=>21x-3+12x=-2x+32

=>21x-3+12x+2x-32=0

=>35x-35=0

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

ĐKXĐ: x khác +-2

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

khử mẫu

(x+1).(x+2)+(x-1)(x-2)=2x²+4

=>x²+x+2+x+2+x²-2x-x+2=2x²+4

=>x²+x+2+x+2+x²-2x-x+2-2x²-4=0

=>(x²+x²-2x²)+(x+x-2x-x)+(2+2+2-4)=0

=>-x+2=0

=>-x=-2

=>x=2(loại)

vậy pt vô nghiệm

ĐKXĐ : \(0\le x\le1\)

Đặt \(\sqrt{x}=a;\sqrt{1-x}=b\left(a;b\ge0\right)\)

Khi đó ta được a² + b² = 1 (1)

Lại có phương trình ban đầu trở thành 

\(\dfrac{2a^3}{a+b}+ab=1\) (2) 

Từ (1) ; (2) ta được \(\dfrac{2a^3}{a+b}+ab=a^2+b^2\)

\(\Leftrightarrow2a^3=\left(a+b\right).\left(a^2-ab+b^2\right)\)

\(\Leftrightarrow a^3=b^3\Leftrightarrow a=b\)

Khi đó \(\sqrt{x}=\sqrt{1-x}\Leftrightarrow x=1-x\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

Vậy tập nghiệm \(S=\left\{\dfrac{1}{2}\right\}\)