\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2\)\(=4\left(2x^2+2-2013\right)\left(x^2-5x-2012\right)=0\)

=>\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)-\)\(4\left(2x^2+2-2013\right)\left(x^2-5x-2012\right)=0\)

=>\(\text{[}\left(2x^2+x-2013\right)^2-4\left(2x^2+2-2013\right)\left(x^2-5x-2012\right)\)\(+4\left(x-5x-2012\right)\text{] =0}\)

=>\(\text{[}\left(2x^2+x-2013\right)^2-2.2\left(2x^2+2-2013\right)\left(x^2-5x-2012\right)\)\(+2\left(x-5x-2012\right)^2=0\)

=>\(\text{[}\left(2x^2+x-2013\right)-\left(2x-10x-2012\right)^2\text{]}=0\)

=>\(\left(2x^2+x-2013-2x^2+10x+4024\right)^2=0\)

=>\(\left(11x+2011\right)^2=0\)

=> 11x+2011=0

=>11x=-2011

=>x=-2011/11

Vậy x=-2011/11

=> là <=> nhé bạn

(x-2)(x+2)(x²-10)=72

(x²-4)(x²-10)=72

 x⁴-10x²-4x²+40=72

x⁴-14x²+40-72=0

x⁴-14x²+49-81=0

(x²-7)²-9²=0

(x²-7+9)(x²-7-9)=0

(x²+2)(x²-16)=0

(x²+2)(x-4)(x+4)=0

=> x²+2=0 hoặc x-4=0 hoặc x+4=0

=>x²=-2( vô lí ) , x=4 hoặc x=-4

Vậy x =+-4

a) Ta có : (x−10)2−x(x+80)(x−10)2−x(x+80)

=x2−20x+100−x2−80x=x2−20x+100−x2−80x

=100−100x=100−100x

Thay x=0,98x=0,98 vào ta có :

100−100⋅0,98=100−98=2100−100⋅0,98=100−98=2

b) Ta có : (2x+9)2−x(4x+31)(2x+9)2−x(4x+31)

=4x2+36x+81−4x2−31x=4x2+36x+81−4x2−31x

=5x+81=5x+81

Thay x=−16,2x=−16,2 vào ta có :

5⋅(−16,2)+81=−81+81=05⋅(−16,2)+81=−81+81=0

N=(x+10)^2+x(80-x) khi x=2

Ta có: N=(x+10)^2+x(80-x)

=> N= (x^2+2.10.x+10^2)+80x-x^2

=> N= x^2+20x+100+80x-x^2

=> N= (x^2-x^2)+(20x+80x)+100

=> N= 100x+100

=> N=100.(x+1)

Thay x=2 vào biểu thức N=100.(x+1)

Ta có: N=100.(x+1)

        => N=100.(2+1)

       =>N=100.3=300

Vậy gt của bt N=300

3x (2x - 5 ) -6x² = -13x - 12

<=> 6x² - 15x - 6x² = -13x - 12

<=> 15x = -13x - 12 

<=> 15x + 13x = -12

<=> 18x = -12 

<=> x = -12 : 18

<=> x = \(-\frac{2}{3}\)

Vậy x = \(-\frac{2}{3}\)

Học tốt

#Gấu

Lộn đề :'((

Tìm x:

3x.(2x - 5) - 6x² = -13x + 12

ĐKXĐ : \(x\ne1;x\ne4\)

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

Ta có \(A=\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}\)

A \(\inℤ\Leftrightarrow\frac{2}{x-1}\inℤ\Leftrightarrow x-1\inƯ\left(2\right)\left(\text{Vì }x\inℤ\right)\)

<=> \(x-1\in\left\{1;-1;2;-2\right\}\)

<=> \(x\in\left\{2;0;3;-1\right\}\)

Vậy \(x\in\left\{2;0;3;-1\right\}\)thì A nguyên