Bài 4: 

Ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)

\(\Leftrightarrow-62x=-92\)

hay \(x=\dfrac{46}{31}\)

2) \(P=\left(2x+1\right)\left(4x^2-2x+1\right)=8x^3+1=8.\left(\dfrac{1}{2}\right)^3+1=8.\dfrac{1}{8}+1=2\)

\(Q=\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3=1^3+27.\left(\dfrac{1}{3}\right)^3=1+27.\dfrac{1}{27}=2\)

3) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow-24x^2+2x+2+24x^2-64x+10=-50\)

\(\Leftrightarrow-62x=-62\Leftrightarrow x=1\)

sos

A=(2x+1-2x+1)^2+xy

=xy+4

=2023+4

=2027

\(A=(2x+1-2x+1)^2+xy\)

\(=xy+4\)

\(=2023+4\)

\(=2027\)

a: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\dfrac{1}{2};-2\right\}\)

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

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

\(a)\)

\(\left(2x+3\right)^2+\left(2x-3\right)^2-\left(2x+3\right)\left(4x-6\right)+xy\)

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

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

\(=6^2+2\left(-1\right)\)

\(=36-2\)

\(=34\)

\(b)\)

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

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

\(=x^2-5x+5\)

Thay \(x=-2\)vào ta có:

\(\left(-2\right)^2-5\left(-2\right)+5\)

\(=4+10+5\)

\(=19\)

b) Thay x=-1 vào biểu thức \(B=\dfrac{2x^2+5x+4}{x^2-4x+3}\), ta được:

\(B=\dfrac{2\cdot\left(-1\right)^2+5\cdot\left(-1\right)+4}{\left(-1\right)^2-4\cdot\left(-1\right)+3}=\dfrac{2\cdot1-5+4}{1+4+3}=\dfrac{1}{8}\)

Vậy: Khi x=-1 thì \(B=\dfrac{1}{8}\)

Ta có:

|x| = \(\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{1}{3};x=-\dfrac{1}{3}\)

a: A=(x-1)^2=(-1-1)^2=4

b: B=(2x+1)^2=2^2=4

chịu