\(\dfrac{5}{x+2}-\dfrac{x-1}{x-2}=\dfrac{12}{x^2-4}+1\left(x\ne-2;x\ne2\right)\)

\(< =>\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

suy ra

`5x-10-(x^2 +2x-x-2)=12+x^2 -4`

`<=>5x-10-x^2 -2x+x+2-12-x^2 +4=0`

`<=>-x^2 -x^2 +5x-2x+x-10+2+4=0`

`<=>-x^2 +4x-4=0`

`<=>x^2 -4x+4=0`

`<=>(x-2)^2 =0`

`<=>x-2=0`

`<=>x=2(ktmđk)`

vậy phương trình vô nghiệm

ĐKXĐ: \(x\ne\pm2\)

\(\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow5\left(x-2\right)-\left(x-1\right)\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow5x-10-\left(x^2+x-2\right)=12+x^2-4\)

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

\(\Leftrightarrow2x^2-4x+16=0\)

\(\Leftrightarrow2\left(x-1\right)^2+14=0\)

Do \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\\14>0\end{matrix}\right.\) ;\(\forall x\)

\(\Rightarrow2\left(x-1\right)^2+14>0\)

Vậy phương trình đã cho vô nghiệm

Bạn cần gấp thì cũng nên viết đề chỉn chu, đầy đủ để mọi người hiểu đề mới giúp được bạn chứ?

3/5 - 1/3 x (2,48 + 0,52) x y : 60 : 5 = 1/5

       1/3 x (2,48+0,52) x y : 60 : 5 = 2/5

    1/3 x 3 x y : 60 : 5                     = 2/5

                      y: 60 : 5                    =2/5

                      y: 60                          =2/5 x 5

                      y : 60                         =2 

                      y                                =  2 x 60

                      y                                  =120

Ta có: \(\dfrac{3}{5}-\dfrac{1}{3}\left(2.48+0.52\right)\cdot y:60:5=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{5}-\dfrac{1}{3}\cdot3\cdot y\cdot\dfrac{1}{60}\cdot\dfrac{1}{5}=\dfrac{1}{5}\)

\(\Leftrightarrow y\cdot\dfrac{1}{300}=\dfrac{2}{5}\)

hay y=120

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)

Đặt \(x^2+7x=t\)

\(\left(t+10\right)\left(t+12\right)-8=t^2+22t+120-8\)

\(=t^2+22t+112=\left(t+8\right)\left(t+14\right)\)

Theo cách đặt \(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

CHAO BAN

1+1bang may

Đề bài yêu cầu gì?

=20,21*(4,8-3,7)-20,21

=20,21*1,1-20,21

=20,21*0,1

=2,021

Giúp mình vs ạ. Cần gấp

\(\dfrac{3}{4}\)+\(\dfrac{1}{4}\): x=\(\dfrac{-2}{5}\)

⇒\(\dfrac{1}{4}\):x=\(\dfrac{-2}{5}\)-\(\dfrac{3}{4}\)

⇒ x=\(\dfrac{1}{4}\): \(\dfrac{-23}{20}\)

⇒ x=\(\dfrac{-5}{23}\)

Lời giải:

a. $x\in \left\{-6; -5; -4; -3; -2; -1; 0; 1;2;3;4;5\right\}$

Tổng các số nguyên $x$ là:

$(-6)+(-5)+....+0+1+2+...+5=-6$

b. $x\in \left\{-6; -5; -4; -3; -2; -1;0; 1;2;3;4\right\}$

Tổng các số nguyên $x$ là:

$(-6)+(-5)+....+0+1+...+4=-11$

giúp mình đi mọi người T.T

a) x = 6/5 - 5/6

    x =  11/30

vậy x =...

b) x = 13/4 - 3/5 

    x = 53/20

vậy x =...

a. x=6/5-5/6=11/30

b.13/4-3/5=53/20