a, SBC=\(\sqrt{xy}\)(3\(\sqrt{x}\)-4\(\sqrt{y}\)+5\(\sqrt{xy}\))

câu b bn lmf tương tự nhé,mấy bài này liên quan đến phân tích đa thức bằng nhân tử đó bn:))

\(a,=5^3:5^2=5\\ b,=\left(\dfrac{3}{4}\right)^{5-3}=\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\\ c,=1728-512=1216\\ d,=x^{10}:x^8=x^2\\ e,=\left(-x\right)^{5-3}=\left(-x\right)^2=x^2\\ f,=\left(-y\right)^{5-4}=-y\)

`a, 20x^3y^5 : 5x^2y^2`

`= (20:5)x^(3-2) . y^(5-2)`

`= 4xy^3`

`b, 18x^3y^5 : (3(-x^3)y^2)`

`= -(18:3)y^(5-3)`

`= -6y^2`

Thực hiện phép chia:

a) (-y^2):y^4=\(\dfrac{-1}{y^2}\)

b) (-x)^5:(-x)^3=(-x)^2

a) (-y²) : y⁴=

a) \(\left(-y^2\right):y^4=\dfrac{-1}{y^2}\)

b) \(\left(-x\right)^5:\left(-x\right)^3=\left(-x\right)^2=x^2\)

`a, (4x^3y^2 - 8x^2y + 10xy) : 2xy`

`= 2x^2y - 4x + 5`.

`b, 7x^4y^2 - 2x^2y^2 - 5x^3y^4 : 3x^2y`

`= 7/3 x^2y - 3/2y - 5/3xy^3`

b: \(=\dfrac{x^3+6x^2-25}{x\left(x+5\right)\left(x-2\right)}+\dfrac{x+5}{x\left(x-2\right)}\)

\(=\dfrac{x^3+6x^2-25+x^2+10x+25}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x^3+7x^2+10x}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x+2}{x-2}\)

`a, (5ab - 2a^2):a`

`= 5b - 2a`

`b, (6x^2y^2-xy^2+3x^2y) : (-3xy)`

`= -2xy + y/3 - x`.

\(\left\{{}\begin{matrix}\left(x-15\right)\left(y+2\right)=xy\\\left(x+15\right)\left(y-1\right)=xy\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}xy+2x-15y-30-xy=0\\xy-x+15y-15-xy=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}2x-15y=30\\-x+15y=15\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}2x-15=30\\3x=45\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=45\\y=4\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y) = (45;4)

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=5\\\dfrac{2}{x}+\dfrac{5}{y}=7\end{matrix}\right.\) (ĐK: x,y >0)

⇔\(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{5}{y}=25\\\dfrac{2}{x}+\dfrac{5}{y}=7\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{5}{y}=25\\\dfrac{3}{x}=18\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=\dfrac{6}{29}\end{matrix}\right.\) (TM)

Vậy HPT có nghiệm (x;y) = (\(\dfrac{1}{6};\dfrac{6}{29}\))

Bài 8: Cho a+b= 1 nha ( mk thiếu đề)

Bài 1:

Theo bài ra ta có:

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

\(=\left(5-y\right)^2-2\times2+\left(5-x\right)^2\)

\(=5^2-2\times5y+y^2-4+5^2-2\times5x+x^2\)

\(=25-10y+y^2+25-10x+x^2-4\)

\(=\left(25+25\right)-\left(10x+10y\right)+x^2+y^2-4\)

\(=50-10\left(x+y\right)+x^2+2xy+y^2-2xy-4\)

\(=50-10\times5+\left(x+y\right)^2-2\times2-4\)

\(=50-50+5^2-4-4\)

\(=25-8=17\)

Vậy giá trị của \(\left(x-y\right)^2\)là 17

ta có : M=2.(a^3  +b^3) -3.(a^2 + b^2)

       <=>M=2.(a+b)(a^2  -ab  +b^2)  - 3(a^2  +3b^2)

      <=>M=2(a^2  -ab  +b^2)  -3(a^2 +b^2)               vì a+b=1(gt)

      <=>M=-(a^2 +b^2 +2ab)

      <=>M=-(a+b)^2

      <=>M=-1  (vì a+b=1)