Question

tim x

-2x(6x-2)+3x(4x-7)=8

(7x-2)(2x+3)-(3x-5)(4x+6)=2x²-5

+(-2x+3)-6x(3x-4)=-22x2+7x

Answer 1

⇔(2 x ) ²+2.2 x .1+1 ²=0 ... c ,(3 x −4) ²−14(3 x −4)(6+3 x )+49(3 x +6)=16 ... ⇔9 x ²−24 x +16−126 x ²−252 x +168 x +336+147 x +294=16.

Answer 2

https://olm.vn/hoi-dap/detail/192758180810.html

Answer 3

\(-2x\left(6x-2\right)+3x\left(4x-7\right)=8\)

<=> \(-12x^2+4x+12x^2-21x=8\)

<=> \(-17x=8\)

<=> \(x=-\frac{8}{17}\)

\(\left(7x-2\right)\left(2x+3\right)-\left(3x-5\right)\left(4x+6\right)=2x^2-5\)

<=> \(14x^2+17x-6-12x^2+2x+30=2x^2-5\)

<=> \(14x^2+17x-6-12x^2+2x+30-2x^2+5=0\)

<=> \(19x+29=0\)

<=> \(19x=-29\)

<=> \(x=-\frac{29}{19}\)

Ý cuối mình k biết -22x2 là -22.2 hay -22x² nữa :)

Answer 4

a, \(-2x\left(6x-2\right)+3x\left(4x-7\right)=8\)

\(\Leftrightarrow-12x^2+4x+12x^2-21x=8\)

\(\Leftrightarrow-17x=8\Leftrightarrow x=-\frac{8}{27}\)

b, \(\left(7x-2\right)\left(2x+3\right)-\left(3x-5\right)\left(4x+6\right)=2x^2-5\)

\(\Leftrightarrow2x^2+19x+24=2x^2-5\Leftrightarrow19x+29=0\Leftrightarrow x=-\frac{29}{19}\)