(x-1/2)=19/4:5/3

(x-1/2)=19/4x3/5

x-1/2=57/20

x=57/20+1/2

x=67/20

tk cho mk nha

\(\frac{19}{4}:\left(x-\frac{1}{2}\right)=\frac{5}{3}\)

\(x-\frac{1}{2}=\frac{19}{4}:\frac{5}{3}=\frac{57}{20}\)

\(x=\frac{57}{20}+\frac{1}{2}\)

\(x=\frac{67}{20}\)

thank kiu các bn nhìu nha

\(-3^2-5.\left(x-1\right)=4x+7\)

\(\Rightarrow-9-5x+5=4x+7\)

\(\Rightarrow-9+5-7=5x+4x\)

\(\Rightarrow9x=-11\)

\(\Rightarrow x=\frac{-11}{9}\)

Vậy....

\(-3^2=-9\)hay \(\left(-3\right)^2=9\)mk làm bài này cho

A=(x+y)\(^3\)+z\(^3\)+3(x+y)z(x+y+z)-x\(^3\)-y\(^3\)-z\(^3\)                                                                                                    

   =x\(^3\)+y\(^3\)+3xy(x+y)+3(x+y)z(x+y+z)-x\(^3\)-y\(^3\)

   =3(x+y)(xy+zx+zy+z\(^2\))

    =3(x+y)\([\)x(y+z)+z(y+z)\(]\)

    =3(x+y)(y+z)(x+z)

  (bạn chỉ cần dùng hằng đẳng thức là ok ngay)

\(!x+\frac{1}{x}!\ge2\Rightarrow!a!\ge2\\ \)

Với IaI>=2

ta có: \(\left(x+\frac{1}{x}\right)^2=a^2\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

số nào nhân với 0 cũng bằng 0.nên =>(x-34).15=0 
=> 0 .15=0 
=>x-34=0 =>x=34

số nào nhân với 0 cũng bằng 0.nên =>(x-34).15=0 
=> 0 .15=0 
=>x-34=0 =>x=34

( x - 34 ) . 15 = 0

( x - 34 )   = 0 . 15

( x - 34 )   = 0

   x      = 0 + 34

   x      = 34

k nh

Đặt \(x^2+x+1=t\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

= \(\left(x^2+x+1\right)\left[\left(x^2+x+1\right)+1\right]-12\)

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

= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)

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

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

bn ghi lung tung quá..............

Ghi chú: Dấu "!" là giá trị tuyệt đối.

a) ta có : \(x^3-64=x^3-4^3=\left(x-4\right)\left(x^2+4x+16\right)\)

b) ta có : \(125x^3+y^6=\left(5x\right)^3+\left(y^2\right)^3=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)

c) ta có : \(125\left(x+1\right)^3-1=\left(5x+5\right)^3-1\)

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

\(=\left(5x+4\right)\left(25x^2+55x+31\right)\)

d) ta có : \(2x\left(x+1\right)+2\left(x+1\right)=\left(2x+2\right)\left(x+1\right)=2\left(x+1\right)\left(x+1\right)\)