\(a,x^2\left(5x^3-2x+3\right)=5x^5-2x^3+3x^2\\ b,\left(3x+2\right)\left(5x-3\right)=5x\left(3x+2\right)-3\left(3x+2\right)=15x^2+10x-9x-6=15x^2+x-6\\ c,\left(5x+y\right)^2=25x^2+10xy+y^2\)

Giúp mình đi mọi người.

Giúp mình nhanh nha mọi người

a)5-2(1/2)^2

=5-2*1

=5-2

=3

\(\left(\dfrac{1}{x}+x-2\right):\left(\dfrac{1}{x^2-x}+1-\dfrac{3}{x-1}\right)\)

\(=\dfrac{x^2-2x+1}{x}:\dfrac{1+x^2-x-3x}{x\left(x-1\right)}\)

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

a: =>x>=0 và x^2+x=x^2

=>x=0

b: =>x>=2 và x^2-4x-3=x^2-4x+4

=>-3=4(loại)

\(a)ĐK:x\ge0\)

\(pt\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x=0\left(tm\right)\end{matrix}\right.\)

Vậy, pt có nghiệm duy nhất là x=0

\(b)ĐK:x\ge2+\sqrt{7}\)

\(pt\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x^2-4x-3=(x-2)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x-3=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\-3=4\end{matrix}\right.\)(vô lý)

Vậy pt vô nghiệm

a) 4x^2 ( 5x^3 - 2x + 3 )

= 20x^5 - 8x^3 + 12x^2

b) ( 2x + 5 ) ( 3x - 2 )

= 6x^2 - 4x + 15x - 10

= 6x^2 + 11x - 10

c) ( 5x + y )^2

= ( 5x )^2 + 2 . 5x . y + y^2

= 24x^2 + 10xy + y^2

\(a,=\left(x^3+3x^2-x^2-3x+x+3\right):\left(x+3\right)\\ =\left(x+3\right)\left(x^2-x+1\right):\left(x+3\right)\\ =x^2-x+1\\ b,=\left(x^3+2x^2-x^2-2x+3x+6\right):\left(x+2\right)\\ =\left(x+2\right)\left(x^2-x+3\right):\left(x+2\right)\\ =x^2-x+3\)

a) \({x^2}.{x^4} = {x^{2 + 4}} = {x^6}\).

b) \(3{x^2}.{x^3} = 3.1.{x^{2 + 3}} = 3{x^5}\).

c) \(a{x^m}.b{x^n} = a.b.{x^{m + n}}\) (a ≠ 0; b ≠ 0; m, n \(\in\) N).

 \(A=3\div\left(\frac{-3}{2}\right)^2+\frac{1}{9}\times\sqrt{36}\)

\(\Rightarrow A=3\div\frac{9}{4}+\frac{1}{9}\times6\)

\(\Rightarrow A=\frac{4}{3}+\frac{1}{3}\)

\(\Rightarrow A=\frac{5}{3}\)

P/s : Nếu mình tính nhầm thì bỏ qua nhé bạn :P

                    GIẢI

A = 3 : ( -3/2)^2 + 1/9 +\(\sqrt{ }\)36

A = 3 : ( -9/4 ) +1/9 × 6 

A = 3 × 4/9 + 1/9 × 6

A = 4/3 + 1/9 × 6 

A = 4/3 + 6/9

A = 4/3 + 2/3 

A = 6/3

A = 2

HỌC TỐT ^_^

cảm ơn các bạn nhiều lắm ạ   ^_^

a) \(2xy\left(3x^2-5xy+4y^2\right)=6x^3y-10x^2y^2+8xy^3\)

b) \(\left(x-3\right)^2+\left(x+5\right)\left(5-x\right)=x^2-6x+9+25-x^2=34-6x\)

a: \(2xy\left(3x^2-5xy+4y^2\right)=6x^3y-10x^2y^2+8xy^3\)

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

\(=x^2-6x+9+25-x^2\)

=-6x+34