a: Ta có: \(P=\left(x-1\right)^2-4x\left(x+1\right)\left(x-1\right)+3\)

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

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

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

b: Thay x=-2 vào P, ta được:

\(P=-4\cdot\left(-8\right)+4-4+4=36\)

Answer:

a) \(\frac{5x}{2x+2}+1=\frac{6}{x+1}\)

\(\Rightarrow\frac{5x}{2\left(x+1\right)}+\frac{2\left(x+1\right)}{2\left(x+1\right)}=\frac{12}{2\left(x+1\right)}\)

\(\Rightarrow5x+2x+2-12=0\)

\(\Rightarrow7x-10=0\)

\(\Rightarrow x=\frac{10}{7}\)

b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\left(ĐK:x\ne0\right)\)

\(\Rightarrow x^2-6=x^2+\frac{3}{2}x\)

\(\Rightarrow\frac{3}{2}x=-6\)

\(\Rightarrow x=-4\)

c) \(\frac{3x-2}{4}\ge\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\ge0\)

\(\Rightarrow9x-6-6x-6\ge0\)

\(\Rightarrow3x-12\ge0\)

\(\Rightarrow x\ge4\)

d) \(\left(x+1\right)^2< \left(x-1\right)^2\)

\(\Rightarrow x^2+2x+1< x^2-2x+1\)

\(\Rightarrow4x< 0\)

\(\Rightarrow x< 0\)

e) \(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}\le\frac{x^2}{7}-\frac{2x-3}{5}\)

\(\Rightarrow\frac{2x-3+5\left(x^2-2x\right)}{35}\le\frac{5x^2-7\left(2x-3\right)}{35}\)

\(\Rightarrow2x-3+5x^2-10x\le5x^2-14x+21\)

\(\Rightarrow6x\le24\)

\(\Rightarrow x\le4\)

f) \(\frac{3x-2}{4}\le\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\le0\)

\(\Rightarrow9x-6-6x-6\le0\)

\(\Rightarrow3x\le12\)

\(\Rightarrow x\le4\)

Giải:

a) \(\dfrac{5x-2}{3}< x+1\)

\(\Leftrightarrow5x-2< 3\left(x+1\right)\)

\(\Leftrightarrow5x-2< 3x+3\)

\(\Leftrightarrow5x-3x< 3+2\)

\(\Leftrightarrow2x< 5\)

\(\Leftrightarrow x< \dfrac{5}{2}\)

Vậy ...

b) \(\dfrac{x-1}{4-1}>\dfrac{x+1}{3+8}\)

\(\Leftrightarrow\dfrac{x-1}{3}>\dfrac{x+1}{11}\)

\(\Leftrightarrow\dfrac{11\left(x-1\right)}{33}>\dfrac{3\left(x+1\right)}{33}\)

\(\Leftrightarrow11\left(x-1\right)>3\left(x+1\right)\)

\(\Leftrightarrow11x-11>3x+3\)

\(\Leftrightarrow11x-3x>3+11\)

\(\Leftrightarrow8x>14\)

\(\Leftrightarrow x>\dfrac{14}{8}=\dfrac{7}{4}\)

Vậy ...

a) \(\dfrac{5x-2}{3}< x+1\)

\(\Leftrightarrow\dfrac{5x-2-3x-3}{3}< 0\)

\(\Leftrightarrow2x-5< 0\) (vì 3>0)

\(\Leftrightarrow x< \dfrac{5}{2}\)

Vậy tập nghiệm của pt là \(x< \dfrac{5}{2}\)

b) \(\dfrac{x-1}{4}-1>\dfrac{x+1}{3}+8\)

\(\Leftrightarrow\dfrac{3\left(x-1\right)-12-4\left(x+1\right)-96}{12}>0\)

\(\Leftrightarrow3x-3-12-4x-4-96>0\left(do12>0\right)\)

\(\Leftrightarrow-x-115>0\)

\(\Leftrightarrow x< 115\)

Vậy tập nghiệm của pt là x<115

Bài 1:

a) \(A=5\left(x-3\right)\left(x+3\right)+\left(2x+3\right)^2\)

\(A=5\left(x^2-3^2\right)+\left(4x^2+12x+9\right)\)

\(A=5x^2-45+4x^2+12x+9\)

\(A=9x^2+12x-36\)

b) Thay x = 1/3 vào A  ta có :

\(A=9\cdot\frac{1}{9}+\frac{12}{3}-36\)

\(A=1+4-36\)

\(A=-31\)

a: \(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

mà n là số nguyên

nên \(n\in\left\{0;-1;1\right\}\)

b: \(\Leftrightarrow10n^2-10n+11n-11+1⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1\right\}\)

hay \(n\in\left\{2;0\right\}\)

c: \(\Leftrightarrow x^4-x^3+5x^2+x^2-x+5+n-5⋮x^2-x+5\)

=>n-5=0

hay n=5

x^3+ y^3+ 3xy

=(x+y)(x^2 -xy + y^2 ) + 3xy
=x^2  -xy + y^2 + 3xy

=x^2 + 2xy + y^2

=(x+y)^2 =1

=> x^3+ y^3+ 3xy=1

còn câu b ai giúp m vs