$\textbf{1)}$

$(4x-1)^2-2(4x-1)(3x-7)+(7-3x)^2$

$=(4x-1)^2-2(4x-1)(3x-7)+(3x-7)^2$

$=\left[(4x-1)-(3x-7)\right]^2$

$=(x+6)^2.$

Tại $x=44$: $(44+6)^2=50^2=2500.$

$\textbf{2)}$

$(2x-5)^2-2(2x-5)(3x-4)+(4-3x)^2$

$=(2x-5)^2-2(2x-5)(3x-4)+(3x-4)^2$

$=\left[(2x-5)-(3x-4)\right]^2$

$=(x+1)^2.$

Tại $x=24$: $(24+1)^2=625.$

1,(2x + 3 ) \(^{^{ }2}\)=\(\left(2x\right)^2+2.2x.3+3^2\)

=\(4x^2+12x+9\)

2, ( 3x + 2y )\(^2=\left(3x\right)^2+2.3x.2y+\left(2y\right)^2\)

=\(9x^2+12xy+4y^2\)

3,(3a -1 )\(^2=\left(3a\right)^2-2.3a.1+1^2\)

\(=9a^2-6a+1\)

4, (a - 2 )\(^2=a^2-2.a.2+2^2\)

=\(a^2-4a+4\)

5, ( 1 - 5a )\(^2=1^2-2.1.5a+\left(5a\right)^2\)

=\(1-10a+25a\)

6, ( x - 4 )\(^3=x^3-3x^24+3x4^2-4^3\)

=\(x^3-12x^2+48x-64\)

a)x³-7x+6

=x³+0x²-7x+6

=x³-x²+x²-x-6x+6

=(x³-x²)+(x²-x)-(6x-6)

=x²(x-1)+x(x-1)-6(x-1)

=(x-1)(x²+x-6)

=(x-1)(x²-2x+3x-6)

=(x-1)[x(x-2)+3(x-2)]

=(x-1)(x+3)(x-2)

1) \(\left(5x-4\right)\left(4x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)

2) \(\left(4x-10\right)\left(24+5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)

3) \(\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)

$(5x-4)(4x+6)=0$

$5x-4=0$ hoặc $4x+6=0$

$x=\dfrac45$ hoặc $x=-\dfrac32$

Vậy $x\in\left{\dfrac45,-\dfrac32\right}$.

$(4x-10)(24+5x)=0$

$4x-10=0$ hoặc $24+5x=0$

$x=\dfrac52$ hoặc $x=-\dfrac{24}{5}$

Vậy $x\in\left{\dfrac52,-\dfrac{24}{5}\right}$.

a) Ta có: \(-3x^2\left(2x^2-\frac{1}{3}x+2\right)\)

\(=-6x^4+x^3-6x^2\)

b) Ta có: \(2xy^2\left(x-3y+xy\right)\)

\(=2x^2y^2-6xy^3+2x^2y^3\)

c) Ta có: \(\left(5x^2-4x\right)\left(x-2\right)\)

\(=5x^3-10x^2-4x^2+8x\)

\(=5x^3-14x^2+8x\)

d) Ta có: \(-\left(2-x\right)\left(2x+3\right)\)

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

\(=2x^2+3x-4x-6\)

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

e) Ta có: \(\left(3x^3-2x^2+x\right):\left(-2x\right)\)

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

f) Ta có: \(\left(15x^2y^2-21x^3y+2x^2y\right):\left(3x^2y\right)\)

\(=5y-7x+\frac{2}{3}\)

g)

1,4x².(5x³+2x-1)

=4x².5x³+4x².2x-4x².1

20x⁵+8x³-4x²

2,4x³y²:x²

=4xy²

3,(15x²y³-10x³y³+6xy):5xy

15x²y³:5xy-10x³y³:5xy+6xy:5xy

3xy²-2x²y²+\(\dfrac{6}{5}\)

1: \(=20x^5+8x^3-4x^2\)

2: \(=4xy^2\)

3: \(=3xy^2-2x^2y^2+\dfrac{6}{5}\)

4: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

5: \(=\dfrac{7}{2x}+\dfrac{11}{3y^2}=\dfrac{21y^2+22x}{6xy^2}\)

6: \(=\dfrac{4x^2-7x+3}{\left(4x-7\right)\left(x+2\right)}\)

7: \(=\dfrac{3x+3y-2x^3+2x^2y}{\left(x-y\right)\left(x+y\right)}\)

8: \(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\dfrac{1}{2}x^2y^4\)

9: \(=\left(x-\dfrac{1}{4}\right)\left(4x-1\right)=4\left(x-\dfrac{1}{4}\right)^2=4\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)

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

10: \(=\dfrac{3x^2+6-x}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

11: \(=\dfrac{x+1}{2}-\dfrac{3}{x-1}\)

\(=\dfrac{x^2-7}{2\left(x-1\right)}\)

12: \(=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)

15:=x^3-y^3+2