đây là bất đẳng thức

giải cho 1 bài thôi:

(x-2)(x+2)<0

=>x-2 và x+2 trái dấu

mà x-2<x+2

=>x-2<0 và x+2>0

=>x<-2 và x>-2

=>-2<x<2=>x E {-1;0;1}

còn lại tương tự

lớp 6 hok bài nek rùi ak, như bài lớp 9 í

Bài 15:

a: 2x+x=45

=>3x=45

=>\(x=\frac{45}{3}=15\)

b: 2x+7x=918

=>\(x\cdot\left(7+2\right)=918\)

=>9x=918

=>\(x=\frac{918}{9}=102\)

c: 2x+3x=60+5

=>5x=65

=>\(x=\frac{65}{5}=13\)

d: \(11x+22x=33\cdot2\)

=>33x=66

=>\(x=\frac{66}{33}=2\)

Bài 14:

a: (12-x)(2-x)=0

=>\(\left[\begin{array}{l}12-x=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=12\\ x=2\end{array}\right.\)

b: (x-33)(11-x)=0

=>\(\left[\begin{array}{l}x-33=0\\ 11-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=33\\ x=11\end{array}\right.\)

c: (21-x)(12-x)=0

=>\(\left[\begin{array}{l}21-x=0\\ 12-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=21\\ x=12\end{array}\right.\)

d: (50-x)(x-150)=0

=>\(\left[\begin{array}{l}50-x=0\\ x-150=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=50\\ x=150\end{array}\right.\)

Bài 13:

a: (x-2)(x-3)=0

=>\(\left[\begin{array}{l}x-2=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=3\end{array}\right.\)

b: (x-3)(x-4)=0

=>\(\left[\begin{array}{l}x-3=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=4\end{array}\right.\)

c: (x-7)(6-x)=0

=>\(\left[\begin{array}{l}x-7=0\\ 6-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=6\end{array}\right.\)

d: (x-3)(x-13)=0

=>\(\left[\begin{array}{l}x-3=0\\ x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=13\end{array}\right.\)

a) x² + 5x = 0 <=> x(x + 5) = 0

=. X = 0 hoặc x = - 5

b) (x - 1).(x + 2) > 0

x - 1 > 0 hoặc x + 2 > 0

=> x > 1 hoặc x > - 2

c) (x - 1).(x + 2) < 0

x - 1 < 0 hoặc x + 2 < 0

=> x < 1 hoặc x < - 2

chọn x = 0

d) 3(2x + 3).(3x - 5) < 0

2x + 3 < 0 hoặc 3x - 5 < 0

=> x < -3/2 hoặc x < 5/3

chọn x <5/3

a) Ta có \(x^2+y^2+2x-4y+5=0\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)=0\Leftrightarrow\left(x+1\right)^2+\left(y-2\right)^2=0\)

<=> x=-1;y=2

b)Ta có:\(x^2+4y^2-x+4y+\frac{5}{4}=0\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)+\left(4y^2+4y+1\right)=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\left(2y+1\right)^2=0\)

<=> x=1/2 ;y=-1/2

a, \(x^2+y^2+2x-4y+5=0\Rightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)=0.\)

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

   \(\Rightarrow x+1=0\)và \(y-2=0\)

\(\left(+\right)x+1=0\Rightarrow x=-1\)

\(\left(+\right)y-2=0\Rightarrow y=2\)

Vậy x=-1 ; y=2 

b, \(x^2+4y^2-x+4y+\frac{5}{4}=0\)

\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(4y^2+4y+\frac{4}{4}\right)=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(2y+1\right)^2=0\)

\(\Rightarrow x-\frac{1}{2}=0\) và \(2y+1=0\)

\(\left(+\right)x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

\(\left(+\right)2y+1=0\Rightarrow2y=-1\Rightarrow y=-\frac{1}{2}\)

Vậy \(x=\frac{1}{2};y=-\frac{1}{2}\)