a: \(=49x^2-64-10\left(4x^2+12x+9\right)+5x\left(9x^2-12x+4\right)+4x\left(x^2-10x+25\right)\)

\(=49x^2-64-40x^2-120x-90+45x^3-60x^2+20x+4x^3-40x^2+100x\)

\(=49x^3-91x^2-154\)

b: \(=27x^3+189x^2+441x+343-125x^3+y^3+x^3+6x^2y+12xy^2+8y^3\)

\(=-97x^3+189x^2+441x+6x^2y+12xy^2+9y^3+343\)

Answer:

Câu đầu bạn xem lại.

\(\left(3x+4\right)^2+\left(4x-3\right)^2+\left(2+5x\right).\left(2-5x\right)\)

\(=\left(3x\right)^2+2.2x.4+4^2+\left(4x\right)^2-2.4x.3+3^2+2^2-\left(5x\right)^2\)

\(=9x^2+24x+16+16x^2-24x+9+4-25x^2\)

\(=\left(9x^2+16x^2-25x^2\right)+\left(24x-24x\right)+\left(16+9+4\right)\)

\(=29\)

\(\left(5x+y\right).\left(25x^2-5xy+y^2\right)-\left(5x-y\right).\left(25x^2+5xy+y^2\right)\)

\(=\left(5x+y\right).[\left(5x\right)^2-5x.y+y^2]-\left(5x-y\right).[\left(5x\right)^2+5x.y+y^2]\)

\(=\left(5x\right)^3+y^3-[\left(5x\right)^3-y^3]\)

\(=\left(5x\right)^3+y^3-\left(5x\right)^3+y^3\)

\(=2y^3\)

help me!!

a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=24-11x\)

b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)

\(=8x^2y-6y^2-9x^2y+12y^2\)

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

c) \(3y^2\left[\left(2x-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

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

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

\(=4y^3+y^2+6xy^2\)

b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)

\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)

\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)

\(=-x^2+18xy\)

c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)

\(=\left(2a-3b\right)^2-16c^2\)

\(=4a^2-12ab+9b^2-16c^2\)

a. \(\left(20x^4y-25x^2y^2-3x^2y\right):5x^2y\)

\(=4x^2-5y-\frac{3}{5}\)

b. \(\left(15xy^2+17xy^3+18y^2\right):6y^2\)

\(=\frac{5}{2}x+\frac{17}{6}xy+3\)

c. \(\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(y-x\right)^2\)

\(=\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(x-y\right)^2\)

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

d. \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)

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

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

\(=y-x\)

\(\hept{\begin{cases}\frac{25x^2-y^2}{20x-4y-3\left(5x+y\right)}=3\\\frac{25x^2-y^2}{2\left(5x-y\right)+10x+2y}=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{\left(5x-y\right)\left(5x+y\right)}{4\left(5x-y\right)-3\left(5x+y\right)}=3\\\frac{\left(5x-y\right)\left(5x+y\right)}{2\left(5x-y\right)+2\left(5x+y\right)}=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{4\left(5x-y\right)-3\left(5x+y\right)}{\left(5x-y\right)\left(5x+y\right)}=\frac{1}{3}\\\frac{2\left(5x-y\right)+2\left(5x+y\right)}{\left(5x-y\right)\left(5x+y\right)}=1\end{cases}}\)

 \(\Leftrightarrow\hept{\begin{cases}\frac{4}{5x+y}-\frac{3}{5x-y}=\frac{1}{3}\\\frac{2}{5x+y}+\frac{2}{5x-y}=1\end{cases}}\) 

Đặt: \(\hept{\begin{cases}\frac{1}{5x+y}=a\\\frac{1}{5x-y}=b\end{cases}}\)thì hệ thành

\(\hept{\begin{cases}4a-3b=\frac{1}{3}\\2a+2b=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=\frac{11}{42}\\b=\frac{5}{21}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{5x+y}=\frac{11}{42}\\\frac{1}{5x-y}=\frac{5}{21}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{441}{550}\\y=-\frac{21}{110}\end{cases}}\)

PS: Bí thì bỏ chứ đăng lên làm gì :3

Em không thích bỏ đó được không? :3