a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

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

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

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

c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)

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

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

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

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

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

`a,-x^3/8 + 3/(4x^2) - 3/(2x) +1`

`=-(x^3/8 - 3/(4x^2) + 3/(2x) - 1)`

`=-(x/2 - 1)^3`

`b,x^6 - 3/(2x^{4} y) + 3/(4x^{2}y^{2}) - 1/(8y^{3})`

`=(x^3 - 1/(2y))^{3}`

Thay x = 1 ; y = -1 ta đc

\(\dfrac{1}{2}.1\left(-1\right)-\dfrac{3}{4}.1\left(-1\right)+1\left(-1\right)=-\dfrac{1}{2}+\dfrac{3}{4}-1=\dfrac{1}{4}-1=-\dfrac{3}{4}\)

a. \(\left(3b+\dfrac{5a}{6}\right)^2\)

= \(9b^2+15ab+\dfrac{25a^2}{36}\)

b. (5x - y)²

= 25x² - 10xy + y²

c. (2a + b - 5)(2a - b + 5)

= 4a² - (b - 5)²

d. \(\left(x^2+\dfrac{2}{5y}\right)\left(x^2-\dfrac{2}{5y}\right)\)

= \(x^4-\dfrac{4}{25y^2}\)

a: \(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=11+2y\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\4\left(\dfrac{2}{3}y+\dfrac{11}{3}\right)-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\\dfrac{8}{3}y+\dfrac{44}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\-\dfrac{7}{3}y=3-\dfrac{44}{3}=-\dfrac{35}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5\\x=\dfrac{2}{3}\cdot5+\dfrac{11}{3}=\dfrac{10}{3}+\dfrac{11}{3}=\dfrac{21}{3}=7\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=3-10=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=3\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\3x+5\left(2x+8\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x+8\\3x+10x+40=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\13x=-39\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-3\\y=2\cdot\left(-3\right)+8=8-6=2\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)

Cái biểu thức là như thế này phải k bạn?

\(\dfrac{1}{2}x^5y-\dfrac{3}{4}x^5y+x^5y\)

đúng không?

Thay x=1 và y= -1 vào biểu thức:\(\dfrac{1}{2}x^5y-\dfrac{3}{4}x^5y+x^5y\)

Ta có:

= \(\dfrac{1}{2}.1^5.\left(-1\right)-\dfrac{3}{4}.1^5.\left(-1\right)+1^5.\left(-1\right)\)

= \(-\dfrac{1}{2}+\dfrac{3}{4}-1\)

= \(-\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{4}{4}\)

= \(\dfrac{1}{4}-\dfrac{4}{4}=-\dfrac{3}{4}\)

=> Giá trị của biểu thức là \(-\dfrac{3}{4}\)

a) \(x^2-6x+9=x^2-2\cdot x\cdot3+3^2=\left(x-3\right)^2\)

b) \(4x^2-12xy+9y^2=\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2=\left(2x-3y\right)^2\)

c) \(4x^2-2x+1=\left(2x-1\right)^2\)

d) \(x^2+8xy+16y^2=\left(x+4y\right)^2\)