khó quá

xin loi to h cau 3 lan het luot roi\

de mai nha

Bn giải câu của mih trc đc k ?

Tìm x: 4x^2(x-5)-(5-x)^2 =0 Mình cảm ơn

ta có:

x+2x+3x+...+18x=1368

(1+2+3+...+18)x=1368

x=1368:(1+2+3...+18)

x=1368:171

x=8

=> x(1+2+3+...+18)=1368

=> 171x = 1368

=> x = 8

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1/3x+2/5(x+1)=0

1/3x+2/5x+2/5=0

11/15x            =-2/5

 x                   =-2/5:11/15

x                     =-6/11

chia ra 2 trường hợp bạn ơi

TH1: x+5<0

TH2: x+5>=0

chia rra 2 truong hop roi chuyen ve

khi x<-5 thì => -(x+5)= 3x-2  <=> x= -0,75 ( không thõa mãn vì > -5)

Khi x> hoặc = -5 => x+5= 3x-2  <=> x= 3,5 ( Thõa mãn vì lớn hơn -5)

vậy giải đc x= 3,5

3 + x - 5 = 6x -4

3 - 5 + 4 = 6x -x 

2             = 5x 

x = 2 / 5 

\(3+\left(x-5\right)=2\left(3x-2\right)\)

\(3+x-5=6x-4\)

\(x-6x=-4+5-3\)

\(-5x=-2\)

\(x=\frac{2}{5}\)

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

3 + ( x - 5 ) = 2 . 3x - 2. 2

3 + x + ( -5 ) = 6x - 4

3 + ( -5 ) + x = 6x - 4

-2 + x = 6x - 4

- 6x + x = 2 - 4

- 5 x = -2

x = -2 : ( -5 ) 

x = 0,4

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

\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}-\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(3x+1\right)-x\right]=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(6x+2-x\right)=0\)

Bỏ vế đằng trước \(x^2+x+1=0\) do vô nghiệm

\(\Leftrightarrow6x+2-x=0\)

\(\Leftrightarrow5x=-2\)

\(\Leftrightarrow x=-\dfrac{2}{5}\left(tmdk\right)\)

Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)

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

hay \(\left(x^2+x+1\right).\dfrac{3x+1}{x+2}=\left(x^2+x+1\right).\dfrac{x}{2\left(x+2\right)}\)

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

\(\Rightarrow3x^2-3^2-3x^2+15x=1\)

\(\Rightarrow3x^2-9-3x^2+15x=1\)

\(\Rightarrow-9+15x=1\)

\(\Rightarrow15x=-8\)

\(\Rightarrow x=\frac{-8}{15}\)