13/17

102/109

42/53

03/109

bạn tick mình nha ai thấy đúng tick ủng hộ tinh thần mình nha mình xin cảm ơn (^-^)1

Ôi má ơi, hoa mắt rồi 

1313/1717=13/17

102102/109109=102/109

424242 / 535353  =424/535

103 / 109

Ta có: \(4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2\)

\(=4\left[x\left(x+y+z\right)\right]\left[\left(x+y\right)\left(x+z\right)\right]+y^2z^2\)

\(=4\left(x^2+xy+zx\right)\left(x^2+xy+yz+zx\right)+y^2z^2\) \(\left(1\right)\)

Đặt \(\hept{\begin{cases}x^2+xy+zx=a\\yz=b\end{cases}}\)

Khi đó: \(\left(1\right)=4a\left(a+b\right)+b^2\)

\(=4a^2+4ab+b^2\)

\(=\left(2a+b\right)^2\)

\(=\left(2x^2+2xy+2zx+yz\right)^2\ge0\left(\forall x,y,z\right)\)

=> đpcm

Ta có:\(4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2=4x\left(x+y+z\right)\left(x+y\right)\left(x+z\right)+y^2z^2=4\left(x^2+xy+xz\right)\left(x^2+xy+yz+zx\right)+y^2z^2\)Đặt \(x^2+xy+xz=t\)thì biểu thức trên trở thành \(4t\left(t+yz\right)+y^2z^2=4t^2+4yzt+y^2z^2=\left(2t+yz\right)^2=\left(2x^2+2xy+2xz+yz\right)^2\ge0\forall x,y,z\left(đpcm\right)\)

x/7=22/77 hay x/7=2/7

=>x=2

9/36=1/x <=> 1/4=1/x

=>x=4

x/27=12/18 =>x/27=2/3

=>x/27=18/27 =>x=18

35/49=10/x hay 5/7=10/x

=>10/14=10/x =>x=14

Bn tick mk nha

(x+12345):5=9725

x+12345=9725x5

x+12345=48625

x=48625-12345

x=36280