(x-40)(x-7)=0

=>x-40=0 hoặc x-7=0

=>x=40   hoặc x=7

tíc mình nha

1.\(\left(x-40\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-40=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=40\\x=7\end{cases}}\)

2. \(\left(x-12\right)\left(x+13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-12=0\\x+13=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-13\end{cases}}\)

3. \(\left(2-x-4\right)\left(3x-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2-x-4=0\\3x-9=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

Đề sai rồi bạn

a: -5x-16=x+40

=>-6x=56

hay x=-28/3

b: \(4x-10=15-x\)

=>5x=25

hay x=5

c: \(-12\left(x-5\right)+7\left(3-x\right)=5\)

=>-12x+60+21-7x=5

=>-19x+71=5

=>-19x=-76

hay x=4

d: \(\left(x-2\right)\left(x+15\right)=0\)

=>x-2=0 hoặc x+15=0

=>x=2 hoặc x=-15

l) (x + 9) . (x² – 25) = 0  

<=> (x + 9) . (x – 5) . (x + 5) = 0   

<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy S = \(\left\{-9,5,-5\right\}\)

      e) |x - 4 |< 7         

<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)

Vậy S = \(\left\{11;-3\right\}\)

I,(x+9).(x^2-25)=0

tương đương:x+9=0

                       x^2-25=0

tương đương : x=-9

                       x=5

e,\(\left|x-4\right|\)=7

tương đương x-4=4

                       x-4=-4

tương đương :x=0

                        x=-8

Bài 1: 

l) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-9;5;-5\right\}\)

e) Ta có: |x-4|<7

mà \(\left|x-4\right|\ge0\forall x\)

nên \(\left|x-4\right|\in\left\{0;1;2;3;4;5;6\right\}\)

\(\Leftrightarrow x-4\in\left\{0;1;-1;2;-2;3;-3;4;-4;5;-5;6;-6\right\}\)

hay \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

Vậy: \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

f) Ta có: \(40< 31+\left|x\right|< 47\)

\(\Leftrightarrow\left|x\right|+31\in\left\{41;42;43;44;45;46\right\}\)

\(\Leftrightarrow\left|x\right|\in\left\{10;11;12;13;14;15\right\}\)

hay \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

Vậy: \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

g) Ta có: \(\left|x+3\right|\le2\)

\(\Leftrightarrow\left|x+3\right|\in\left\{0;1;2\right\}\)

\(\Leftrightarrow x+3\in\left\{0;1;-1;2;-2\right\}\)

hay \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

Vậy: \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

a) (x-2).(x+1)=6

(x+1-3).(x+1)=6

(x+1).(1-3)=6

(x+1).-2=6

x+1=6:(-2)

x+1=-3

x=-3-1

x=-4

a) Xét ước

b)

\(\left(x^2+7\right)\left(x^2-49\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+7< 0\Rightarrow x^2< -7\left(KTM\right)\\x^2-49>0\Rightarrow x^2>49\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+7>0\Rightarrow x^2>-7\\x^2-49< 0\Rightarrow x^2< 9=49\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-7< x^2< 49\)

\(\Rightarrow0\le x^2< 49\)

\(\Rightarrow x^2\in\left\{0;1;4;9;16;25;36\right\}\)

\(\Rightarrow x\in\left\{0;\pm1;\pm2;\pm3;\pm4;\pm5;\pm6\right\}\)

\((x^2-7)(x^2-49)<0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-7< 0\Rightarrow x^2< 7\\x^2-49>0\Rightarrow x^2>49\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-7>0\Rightarrow x^2>7\\x^2-49< 0\Rightarrow x^2< 49\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow7< x^2< 49\)

\(\Rightarrow x^2\in\left\{9;16;25;36\right\}\)

\(\Rightarrow x\in\left\{\pm3;\pm4;\pm5;\pm6\right\}\)

a, -154+(x-9-18)=40

x-27=40-(-154)

x-27=194

x=194+27

x=221

b,\(\left|9-x\right|\)=64+(-7)

\(\left|9-x\right|\)=57

TH1:

9-x=57

x=9-57

x=-48

TH2:

9-x=-57

x=9-(-57)

x=66

a, -154+(x-9-18)=40

x-27=40-(-154)

x-27=194

x=194+27

x=221

b,=64+(-7)

=57

TH1:

9-x=57

x=9-57

x=-48

TH2:

9-x=-57

x=9-(-57)

x=66

Bạn ơi bài 1 ô trống đâu ạ ?

a)|x+25|+|-y+5|=0

=>|x+25|=0=>x+25=0

=>x=0+25=25

=>|-y+5|=0=>-y+5=0

=>-y=5&y=-5

mk chỉ bít nhiu thui