a. (2x+3y)²= (2x)²+2.2x.3y+(3y)²

=4x²+12xy+9y²

b. 2(\(\dfrac{1}{2}\)x²+y)(x²-2y)

=(x²+2y)(x²-2y)

=x⁴-4y²

c, (x+y+z)²= [(x+y)+z]²

=(x+y)²+2(x+y)z+z²

=x²+2xy+y²+2xz+2yz+z²

=x²+y²+z²+2xy+2yz+2xz

a) \({\left( {3x + y} \right)^4} = {\left( {3x} \right)^4} + 4.{\left( {3x} \right)^3}y + 6.{\left( {3x} \right)^2}{y^2} + 4.\left( {3x} \right){y^3} + {y^4}\)

\( = 81{x^4} + 108{x^3}y + 54{x^2}{y^2} + 12x{y^3} + {y^4}\)

b) \(\begin{array}{l}{\left( {x - \sqrt 2 } \right)^5} = \left( {x + (-\sqrt 2) } \right)^5 ={x^5} + 5.{x^4}.\left( { - \sqrt 2 } \right) + 10.{x^3}.{\left( { - \sqrt 2 } \right)^2} + 10.{x^2}.{\left( { - \sqrt 2 } \right)^3} + 5.x.{\left( { - \sqrt 2 } \right)^4} + 1.{\left( { - \sqrt 2 } \right)^5}\\ = {x^5} - 5\sqrt 2 .{x^4} + 20{x^3} - 20\sqrt 2 .{x^2} + 20x - 4\sqrt 2 \end{array}\)

a) \({\left( {x + 1} \right)^5} = {x^5} + 5.{x^4}.1 + 10.{x^3}{.1^2} + 10.{x^2}{.1^3} + 5.{x^1}{.1^4} +{1^5} = {x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 5x + 1\)

b) \(\begin{array}{l}{\left( {x - 3y} \right)^5} = {\left[ {x + \left( { - 3y} \right)} \right]^5} = {x^5} + 5{x^4}{\left( { - 3y} \right)^1} + 10{x^3}{\left( { - 3y} \right)^2} + 10{x^2}{\left( { - 3y} \right)^3} + 5{x^1}{\left( { - 3y} \right)^4} + {\left( { - 3y} \right)^5}\\ = {x^5} - 15{x^4}y + 90{x^3}{y^2} - 270{x^2}{y^3} + 405x{y^4} - 243{y^5}\end{array}\)

a) \({\left( {2x + 1} \right)^4} = {\left( {2x} \right)^4} + 4.{\left( {2x} \right)^3}{.1^1} + 6.{\left( {2x} \right)^2}{.1^2} + 4.\left( {2x} \right){.1^3} + {1^4} = 16{x^4} + 32{x^3} + 24{x^2} + 8x + 1\)

b) \(\begin{array}{l}{\left( {3y - 4} \right)^4} = {\left[ {3y + \left( { - 4} \right)} \right]^4} = {\left( {3y} \right)^4} + 4.{\left( {3y} \right)^3}.\left( { - 4} \right) + 6.{\left( {3y} \right)^2}.{\left( { - 4} \right)^2} + 4.{\left( {3y} \right)^1}{\left( { - 4} \right)^3} + {\left( { - 4} \right)^4}\\ = 81{y^4} - 432{y^3} + 864{y^2} - 768y + 256\end{array}\)

c) \({\left( {x + \frac{1}{2}} \right)^4} = {x^4} + 4.{x^3}.{\left( {\frac{1}{2}} \right)^1} + 6.{x^2}.{\left( {\frac{1}{2}} \right)^2} + 4.x.{\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^4} = {x^4} + 2{x^3} + \frac{3}{2}{x^2} + \frac{1}{2}x + \frac{1}{{16}}\)

d) \(\begin{array}{l}{\left( {x - \frac{1}{3}} \right)^4} = {\left[ {x + \left( { - \frac{1}{3}} \right)} \right]^4} = {x^4} + 4.{x^3}.{\left( { - \frac{1}{3}} \right)^1} + 6.{x^2}.{\left( { - \frac{1}{3}} \right)^2} + 4.x.{\left( { - \frac{1}{3}} \right)^3} + {\left( { - \frac{1}{3}} \right)^4}\\ = {x^4} - \frac{4}{3}{x^3} + \frac{2}{3}{x^2} - \frac{4}{27}x + \frac{1}{{81}}\end{array}\)

a) Khai triển biểu thức \({\left( {5{\rm{x}} - y} \right)^2}\)

• Sử dụng lệnh Expand(<biểu thức cần khai triển>).

• Nhập biểu thức trên dòng lệnh của cửa sổ CAS sau đó nhấn Enter, kết quả sẽ được hiển thị ngay bên dưới.

Vậy \({\left( {5{\rm{x}} - y} \right)^2} = 25{{\rm{x}}^2} - 10{\rm{x}}y + {y^2}\)

b) Khai triển biểu thức: \({\left( {\frac{1}{3}x + 2y} \right)^3}\)

• Sử dụng lệnh Expand(<biểu thức cần khai triển>).

• Nhập biểu thức trên dòng lệnh của cửa sổ CAS sau đó nhấn Enter, kết quả sẽ được hiển thị ngay bên dưới.

Vậy \({\left( {\frac{1}{3}x + 2y} \right)^3} = \frac{1}{{27}}{x^3} + \frac{2}{3}{x^2}y + 4{\rm{x}}{y^2} + 8{y^3}\)

a) Áp dụng công thức nhị thức Newton, ta có:

\(\begin{array}{l}{\left( {a - \frac{b}{2}} \right)^4} = C_4^0.{a^4}{\left( { - \frac{b}{2}} \right)^0} + C_4^1.{a^3}\left( { - \frac{b}{2}} \right) + C_4^2.{a^2}{\left( { - \frac{b}{2}} \right)^2} + C_4^3.a{\left( { - \frac{b}{2}} \right)^3} + C_4^4.{a^0}{\left( { - \frac{b}{2}} \right)^4}\\ = {a^4} - 2{a^3}b + \frac{3}{2}{a^2}{b^2} - \frac{1}{2}a{b^3} + \frac{1}{16}{b^4}\end{array}\)

b) Áp dụng công thức nhị thức Newton, ta có:

\(\begin{array}{l}{\left( {2{x^2} + 1} \right)^5} = C_5^0.{\left( {2{x^2}} \right)^5}{.1^0}  + C_5^1.{\left( {2{x^2}} \right)^4}.1 + C_5^2.{\left( {2{x^2}} \right)^3}{.1^2} + C_5^3.{\left( {2{x^2}} \right)^2}{.1^3} + C_5^4.\left( {2{x^2}} \right){.1^4} +C_5^5.{\left( {2{x^2}} \right)^0} {.1^5}\\ = 32{x^{10}} + 80{x^8} + 80{x^6} + 40{x^4} + 10{x^2} + 1\end{array}\).

a: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

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

\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=16x^2\)

b: \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

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

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

a) \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

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

\(=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2\)

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

\(=\left(4x\right)^2\)

\(=16x^2\)

b) \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=\left(x^3+2x^2-x-2\right)-\left(x^3-8\right)\)

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

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

\(a,\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

\(=\left(2x+1\right)^2+2\cdot\left[\left(2x\right)^2-1^2\right]+\left(2x-1\right)^2\)

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

\(=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2\)

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

\(=\left(4x\right)^2\)

\(=16x^2\)

\(---\)

\(b,\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

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

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

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

Giups mik vs

\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)

\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))