Question

1. Tìm x, y∈ Q

a) (x+5)(3x-y) =0

b) (x-4)(y²-x) =0

c) (x+y)² + (2x-y)² =0

2. Tìm x ∈Q

a) x+2x−3 >0

b) 4x² - x <0

 

                             Giải chi tiết giúp mình nhé!!!!!!!

Answer 1

mk ko chắc là đúng, bn coi sai chỗ nào thì nói mk nha

Answer 2

Answer 3

Bài 1:

a)Ta có: (x+5)(3x-y)=0

=> \(\left[\begin{array}{nghiempt}x+5=0\\3x-y=0\end{cases}< =>}\left[\begin{array}{nghiempt}x=-5\\x=-15\end{array}\right.\)

Vậy x=-5; y=-15

b)Ta có:(x-4)(y²-x)=0

=>\(\left[\begin{array}{nghiempt}x-4=0\\y^2-x=0\end{cases}< =>\left[\begin{array}{nghiempt}x=4\\y=\pm2\end{array}\right.}\)

Vậy x=4;y=\(\pm2\)

c)(x+y)²+(2y-x)²=0

Vì \(\left(x+y\right)^2\ge0;\left(2x-y\right)^2\ge0\)=>\(\left[\begin{array}{nghiempt}\left(x+y\right)^2=0\\\left(2x-y\right)^2=0\end{array}\right.\)

                                                   <=>  \(\left[\begin{array}{nghiempt}x=-y\\2x=y\end{cases}< =>\left[\begin{array}{nghiempt}2x=-2y\\2x=y\end{array}\right.< =>x=y=0}\)

Bài 2:

a) Ta thấy :\(\frac{x+2}{x-3}>0< =>\left[\begin{array}{nghiempt}\hept{\begin{cases}x+2>0\\x-3>0\end{array}\right.\\\hept{\begin{cases}x+2< 0\\x-3< 0\end{array}\right.\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}\hept{\begin{cases}x>-2\\x>3\end{array}\right.\\\hept{\begin{cases}x< -2\\x< 3\end{array}\right.\end{cases}< =>\left[\begin{array}{nghiempt}x>3\\x< -2\end{array}\right.}\)Vậy x>3 hoặc x<-2 (x\(\in Q\))

b) Ta có: 4x²-x<0

<=> x(4x-1)<0

<=>\(\left[\begin{array}{nghiempt}\hept{\begin{cases}x< 0\\4x-1>0\end{array}\right.\\\hept{\begin{cases}x>0\\4x-1< 0\end{array}\right.\end{cases}< =>\left[\begin{array}{nghiempt}\hept{\begin{cases}x< 0\\x>\frac{1}{4}\end{cases}\left(loai\right)}\\\hept{\begin{cases}x>0\\x< \frac{1}{4}\end{array}\right.\end{cases}=>}0< x< \frac{1}{4}}\) Vậy 0<x<1/4( x thuộc Q)

Answer 4

Answer 5

Chỉ ghi lại những cái bị lỗi thôi cho rõ nhé

1) a)=> \(\left[\begin{array}{nghiempt}x+5=0\\3x-y=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=-5\\y=-15\end{array}\right.\)

b)\(\left[\begin{array}{nghiempt}x-4=0\\y^2-x=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=4\\y=\pm2\end{array}\right.\)

c)\(\left[\begin{array}{nghiempt}x=-y\\2x=y\end{array}\right.\)<=> x=y=0

 

Answer 6

2)a)

\(\frac{x+2}{x-3}>0\)

TH1:\(\begin{cases}x+2>0\\x-3>0\end{cases}\)<=> \(\begin{cases}x>-2\\x>3\end{cases}\)<=> x>3

TH2:\(\begin{cases}x+2< 0\\x-3< 0\end{cases}\)<=> \(\begin{cases}x< -2\\x< 3\end{cases}\)=> x<-2

Vậy x>3 hoặc x<-2(x thuộc Q)

b)4x²-x<0

<=>x(4x-1)<0

Th1 : \(\begin{cases}x< 0\\4x-1>0\end{cases}\)<=> \(\begin{cases}x< 0\\x>\frac{1}{4}\end{cases}\)(loại)

TH2: \(\begin{cases}x>0\\4x-1< 0\end{cases}\)<=> \(\begin{cases}x>0\\x< \frac{1}{4}\end{cases}\)

VẬy 0<x<1/4( x thuộc Q)