c.

\(4y^2+1=4y\)

\(\Leftrightarrow4y^2-4y+1=0\)

\(\Leftrightarrow4y^2-2y-2y+1=0\)

\(\Leftrightarrow2y\left(2y-1\right)-\left(2y-1\right)=0\)

\(\Leftrightarrow\left(2y-1\right)^2=0\)

\(\Leftrightarrow y=0\)

d.

\(y^2-2y=80\)

\(\Leftrightarrow y^2-2y-80=0\)

\(\Leftrightarrow y^2-10y+8y-80=0\)

\(\Leftrightarrow y\left(y-10\right)+8\left(y-10\right)=0\)

\(\Leftrightarrow\left(y+8\right)\left(y-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y+8=0\\y-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-8\\y=10\end{matrix}\right.\)

\(\dfrac{y}{2\left(y-3\right)}=\dfrac{2y}{\left(y+1\right)\left(y-3\right)}-\dfrac{y}{2y+2}\)

\(DKXD:y\ne-1;y\ne3\)

<=>\(\dfrac{y\left(y+1\right)}{2\left(y-3\right)\left(y+1\right)}=\dfrac{4y}{2\left(y-3\right)\left(y+1\right)}-\dfrac{y\left(y-3\right)}{2\left(y+1\right)\left(y-3\right)}\)

=>y²+y=4y-y²+3y

<=>2y²-6y=0

<=>2y(y-3)=0

\(\left[{}\begin{matrix}y=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=0\left(TM\right)\\y=3\left(KTM\right)\end{matrix}\right.\)

vay..........

\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)

\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+1+1}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

=>x+1=1 và y-2=1/2

=>x=0 và y=5/2

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{1}{2}-\dfrac{1}{18}=\dfrac{9}{18}-\dfrac{1}{18}=\dfrac{8}{18}=\dfrac{4}{9}\\\dfrac{2}{2x-y}=\dfrac{1}{18}+\dfrac{1}{x-2y}\end{matrix}\right.\)

=>x-2y=9 và 2/2x-y=1/18+1/9=1/18+2/18=3/18=1/6

=>x-2y=9 và 2x-y=12

=>x=5; y=-2

c: \(\Leftrightarrow\left\{{}\begin{matrix}10\left|x-6\right|+15\left|y+1\right|=25\\10\left|x-6\right|-8\left|y+1\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23\left|y+1\right|=23\\\left|x-6\right|=1\end{matrix}\right.\)

=>|x-6|=1 và |y+1|=1

=>\(\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)

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

\(=-n^2+4\)

2)\(\left(y+3\right)\left(y-3\right)\left(y^2+9\right)-\left(y^2-4\right)\left(y^2+4\right)=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-16\right)\)

\(=y^4-81-y^4+16=-65\)

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

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

4)\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

5)\(\left(a+b-c\right)^2=a^2+b^2+c^2+2ab-2bc-2ac\)

6)\(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab+2bc-2ac\)

Học tốt nha bạn !

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

⇔ 2y² + 2y - 3y - 3 + y² - 2y = 3(y² + 4y + 4)

⇔ 3y² - 3y - 3 = 3y² + 12y + 12

⇔ 3y² -3y - 3y² -12y = 12 + 3

⇔ -15y = 15

⇔ y = -1

Vậy S = {-1}

Đề yc làm gì em?

\(\left(2y-3\right).\left(y+1\right)+y.\left(y+2\right)=3\left(y+2\right)^2\\ \Leftrightarrow2y^2-y-3+y^2+2y=3y^2+12y+12\\ \Leftrightarrow3y^2-2y^2-y^2+12y-2y+y=-3-12\\ \Leftrightarrow11y=-15\\ \Leftrightarrow y=-\dfrac{15}{11}\\ Vậy:S=\left\{-\dfrac{15}{11}\right\}\)