(10x+9)x-(5x-1)(2x+3)=8

<=>10x²+9x-(10x²+15x-2x-3)=8

<=>10x²+9x-10x²-15x+2x+3=8

<=>-4x+3=8

<=>-4x=5

<=>x=-5/4

\(5x\left(x-4y\right)-4y\)

Thay vào ta được:

\(5\left(\frac{-1}{5}\right)[\left(\frac{-1}{5}\right)-4\left(\frac{-1}{2}\right)]-4\left(\frac{-1}{2}\right)\)

\(=-[\left(\frac{-1}{5}\right)-2]-2\)

\(=\left(\frac{1}{5}-2\right)-2\)

\(=\frac{11}{5}-2\)

\(=\frac{1}{5}\)

\(5x\left(x-4y\right)-4y=5x^2-20xy-4y\)

thay   x= -1/5; y= -1/2 vào ta có:

\(5\left(-\frac{1}{5}\right)^2-20\left(-\frac{1}{5}\right)\left(-\frac{1}{2}\right)-4\left(-\frac{1}{2}\right)^2=\frac{5}{25}-\frac{20}{10}-\frac{4}{4}=\frac{1}{5}-2-1=\frac{1}{5}-\frac{15}{5}=-\frac{14}{5}\)

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

\(=x^2-4x+4-\left(4x^2-4x+1\right)+3x^2-16x+5\)

\(=\left(x^2-4x^2+3x^2\right)-\left(4x-4x+16x\right)+4-1+5\)

\(=-16x+8\)

VT = (12x – 5)(4x – 1) + (3x – 7)(1 – 16x)

     = 12x.(4x – 1) + (–5).(4x – 1) + 3x.(1 – 16x) + (–7).(1 – 16x)

     = 12x.4x+ 12x.(–1) + (–5).4x + (–5).(–1) + 3x.1 + 3x.(–16x) + (–7).1 + (–7).(–16x)

     = 48x2 – 12x – 20x + 5 + 3x – 48x2 – 7 + 112x

     = (48x2 – 48x2) + (– 12x – 20x + 3x + 112x) + (5 – 7)

     = 83x – 2

Vậy ta có:

83x – 2 = 81

       83x = 81 + 2

       83x = 83

           x = 83 : 83

           x = 1.

\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\Rightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\)

\(\Rightarrow\left(48x^2-48x^2\right)+\left(-12x-20x+3x+112x\right)+\left(5-7-81\right)=0\)

\(\Rightarrow83x-83=0\)

\(\Rightarrow x=1\)

a) =(2x+3x ² )+(2y+3xy)

    =x(2+3x)+y(2+3x)

    =(x+y)(2+3x)

b) = (2x)²-2.2x-9y+1

  =(2x)²-2.2x+1-9y

  =(2x-1)²-9y

các bài sau tương tự nha 

toi chua hoc

\(\left(3y-\frac{4}{3}xy\right)\left(-2x+2\right)\)

\(=-2x\left(3y-\frac{4}{3}xy\right)+2\left(3y-\frac{4}{3}xy\right)=-6xy+\frac{8}{3}x^2y+6y-\frac{8}{3}xy\)

bc=db+dc

cho dù tổng khoảng cách từ d đến hai cạnh bên trên đáy bc cũng ko hay đổi vì tổng của db và dc luôn bằng bc, nó nằm trên bc

\(5x^2+5y^2+17z^2-4xy-4yz-8zx+1\)

\(=4x^2-4xy+y^2+4y^2-4yz+z^2+16z^2-8zx+x^2+1\)

\(=\left(2x-y\right)^2+\left(2y-z\right)^2+\left(4z-x\right)^2+1\ge1\)