bạn tách ra từng ít câu 1 thôi ạ

a: \(=5x\left(xy^2+3x+6y^2\right)\)

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

c: \(=\left(x-3\right)\left(x-4\right)\)

d: \(=x\left(x^2-2xy+y^2-9\right)\)

=x(x-y-3)(x-y+3)

e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)

f: \(=\left(x-4\right)\left(x+3\right)\)

b: \(\begin{cases}81x^3y^2-81x^2y^2+33xy^2-29y^2=4\left(1\right)\\ 25y^3+9x^2y^3-6xy^3-4y^2=24\left(2\right)\end{cases}\)

Khi y=0 thì (2) sẽ trở thành: \(25\cdot0^3+9x^2\cdot0^3-6\cdot x\cdot0^3-4\cdot0^2=24\)

=>0=24(loại)

=>y<>0

Ta có: \(81x^3y^2-81x^2y^2+33xy^2-29y^2=4\)

=>\(y^2(81x^3-81x^2+33x-29)=4\)

=>\(81x^3-81x^2+33x-29=\frac{4}{y^2}\)

Ta có: \(25y^3+9x^2y^3-6xy^3-4y^2=24\)

=>\(y^3(9x^2 - 6x + 25) - 4y^2 = 24\)

=>\(y(9x^2 - 6x + 25) - 4 = \frac{24}{y^2}\)

Đặt u=3x-1

=>3x=u+1

=>\(x=\frac{u+1}{3}\)

\(9x^2-6x+25=9x^2-6x+1+24=\left(3x-1\right)^2+24\)

\(=u^2+24\)

(1) sẽ tương đương: \(81\left(\frac{u+1}{3}\right)^3 - 81\left(\frac{u+1}{3}\right)^2 + 33\left(\frac{u+1}{3}\right) - 29 = \frac{4}{y^2}\)

=>\(3(u+1)^3 - 9(u+1)^2 + 11(u+1) - 29 = \frac{4}{y^2}\)

=>\(3u^3+2u-24=\frac{4}{y^2}\quad(3)\)

THay (3) vào phương trình (2), ta được:

\(y(u^2+24)-4=\frac{24}{y^2}\quad(4)\)

Ta có: \(3u^3+2u-24=\frac{4}{y^2}\quad(3)\)

=>\(18u^3+12u-144=\frac{24}{y^2}\)

=>\(y(u^2 + 24) - 4 = 18u^3 + 12u - 144\)

=>\(y(u^2+24)=18u^3+12u-140\)

=>\(y=\frac{18u^3 + 12u - 140}{u^2 + 24}\)

Thay lại vào (3), ta được: \(\frac{4}{\left(\frac{18u^3 + 12u - 140}{u^2 + 24}\right)^2} = 3u^3 + 2u - 24\)

=>\((9u^3 + 6u - 70)^2(3u^3 + 2u - 24) = (u^2 + 24)^2\)

=>u=2

\(y=\frac{18\cdot2^3+12\cdot2-140}{2^2+24}=\frac{144 + 24 - 140}{28}=\frac{28}{28}=1\)

3x-1=2

=>3x=3

=>x=1

\(\left\{{}\begin{matrix}5x^2+5y^2-6xy=2\\2x^2+3x-2y^2-y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x^2+5y^2-6xy=2\\4x^2+6x-4y^2-2y=6\end{matrix}\right.\)

\(\Rightarrow9x^2+y^2-6xy+6x-2y+1=9\)

\(\Leftrightarrow\left(3x-y+1\right)^2=9\)

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

Đến đây chia 2 trường hợp và thế vào 1 trong 2 pt để giải

\(\left\{{}\begin{matrix}\left(x-y\right)\left(2x+3y\right)=12\\\left(x-y\right)\left(xy+6\right)=12\end{matrix}\right.\)

Trừ trên cho dưới:

\(\left(x-y\right)\left(2x+3y-xy-6\right)=0\Leftrightarrow\left(x-y\right)\left(x-3\right)\left(2-y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\x=3\\y=2\end{matrix}\right.\)

TH1: \(x=y\) thay vào pt đầu ta được \(0=12\) (vô nghiệm)

TH2: \(x=3\Rightarrow-3y^2+3x+6=0\Rightarrow\left[{}\begin{matrix}y=-1\\y=2\end{matrix}\right.\)

TH3: \(y=2\Rightarrow2x^2+2x-24=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

Vậy pt có 3 cặp nghiệm \(\left(x;y\right)=\left(3;-1\right);\left(3;2\right);\left(-4;2\right)\)

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

2:

a: \(3xy^2-3x^3-6xy+3x\)

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

\(=3x\left\lbrack\left(y-1\right)^2-x^2\right\rbrack\)

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

b: \(3x^2+11x+6\)

\(=3x^2+9x+2x+6\)

=3x(x+3)+2(x+3)

=(x+3)(3x+2)

c: \(-x^3-4xy^2+4x^2y+16x\)

\(=x\left(16+4xy-4y^2-x^2\right)\)

\(=x\cdot\left\lbrack4^2-\left(x^2-4xy+4y^2\right)\right\rbrack=x\cdot\left\lbrack4^2-\left(x-2y\right)^2\right\rbrack\)

=x(4-x+2y)(4+x-2y)

d: \(xz-x^2-yz+2xy-y^2\)

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

=\(z\left(x-y\right)-\left(x-y\right)^2\)

=(x-y)(z-x+y)

e: \(4x^2-y^2-6x+3y\)

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

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

f: \(x^4-x^3-10x^2+2x+4\)

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

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

g: \(\left(x^3-x^2+x\right)\left(121-25y^2-10y\right)-\left(x^3-x^2+x\right)-\left(121-25y^2-10y\right)+1\)

\(=\left(x^3-x^2+x\right)\left(121-25y^2-10y-1\right)-\left(121-25y^2-10y-1\right)\)

\(=\left(x^3-x^2+x-1\right)\left\lbrack121-\left(25y^2+10y+1\right)\right\rbrack\)

\(=\left(x-1\right)\left(x^2+1\right)\left\lbrack121-\left(5y+1\right)^2\right\rbrack\)

=(x-1)(x^2+1)(11-5y-1)(11+5y+1)

=(x-1)(x^2+1)(10-5y)(12+5y)

=5(2-y)(x-1)(x^2+1)(5y+12)