(4x-2).x+5=0

4x-2x+5=0

2x+5=0

x+5=0:2

x+5 =0

x=0-5

x=  -5

Vậy x= -5

a, x - 34 - ( 15 + x ) - ( 23 - x )

= x - 34 - 15 - x - 23 + x

= x - 72

b, - ( -a + b - 2c ) + ( 2b + c - 1 ) - ( 6 - 26 )

= a - b + 2c + 2b + c - 1 - 6 + 26

= a + b + 3c + 19

\(2^x+2^{x+3}=288\)

\(\Rightarrow2^x+2^x.2^3=288\)

\(\Rightarrow2^x\left(1+2^3\right)=288\)

\(\Rightarrow2^x.9=288\)

\(\Rightarrow2^x=288:9\)

\(\Rightarrow2^x=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

Vậy \(x=5\)

\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+30\right)=1240\)

\(\Rightarrow x+x+1+x+2+x+3+...+x+30=1240\)

\(\Rightarrow x+x+x+...+x+1+2+3+...+30=1240\)

Từ \(1\rightarrow30\)có: \(\left(30-1\right):1+1=30\)( số )

\(\Rightarrow31.x+\left(30+1\right).30:2=1240\)

\(\Rightarrow31.x+31.15=1240\)

\(\Rightarrow31.x+465=1240\)

\(\Rightarrow31.x=1240-465\)

\(\Rightarrow31.x=775\)

\(\Rightarrow x=775:31\)

\(\Rightarrow x=25\)

Chúc bạn học tốt !!!

x+(x+1)+(x+2)+....+(x+30)=1240
31 . x + (1 + 2 + 3 + 4 +...+ 29 + 30) = 1240
31 . x + 31.15 = 1240
31 . x = 1240 - 31.15
31 . x = 775
x = 775 : 31
x = 25

      Vậy x=25 

(2x + 1) (3y - 2) = -55

<=> 2x+1 ; 3y-2 \(\in\)Ư(-55)={-1,-5,-11,-55,1,5,11,55}

Ta có bảng :

.....x,y bạn tự tính nhé

4.5\(^2\)-3.(24-9)=4.25-3.15

                         =100-45

                         =55

~ Chúc bạn học tốt~

4.5² -3.(24-9)

\(=4.25-3.15\)

\(=100-45\)

= 55 

Học tốt

# Chiyuki Fujito

a) 100+-570+(-430)

=-470+(-430)

=-900

b)-2005+-596+-201+496+301

=-2005+(-596+496)+(-201+301)

=-2005+(-100)+100

=-2005+[-100+100]

=-2005+0

=-2005

a) 100+(-570)+(-430)

=(-470)+(-430)

=-900

b)-2005+(-596)+(-201)+496+301

=-2005+(-596+496)+(-201+301)

=-2005+(-100)+100

=-2005

c)-5+11+(-15)+21+(-25)+31+...+(-95)+101

= [(-5)+11]+[(-15)+21]+...[(-95)+101]

=6+6+6...+6

dãy số trên có 10 số hạng ta có:

6.10=60

Câu 4: \(\left|x+3\right|+\left|x+4\right|=1\)

Ta có: \(\left|x+3\right|\ge0\forall x\) và \(\left|x+4\right|\ge0\forall x\)

Nên: \(\left|x+3\right|+\left|x+4\right|=1\)

\(\Leftrightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left|x+4\right|=1\end{cases}}\)

Ta có: \(\left|x+3\right|=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=0-3\)

\(\Leftrightarrow x=-3\) \(\left(1\right)\)

Lại có: \(\left|x+4\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1-4\\x=-1-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\) \(\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra: \(x=-3\)

Vậy: \(x=-3\)

Câu 7:

 \(11-x+\left|x+2\right|=0\)

\(\Leftrightarrow11-x=-\left|x+2\right|\)

\(\Leftrightarrow-\left(11-x\right)=\left|x+2\right|\)

\(\Leftrightarrow-11+x=\left|x+2\right|\)

\(\Leftrightarrow\orbr{\begin{cases}-11+x=x+2\\-11+x=-\left(x+2\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-11-2=x-x\\-11+x=-x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-13=0\\x+x=-2+11\end{cases}}\)( T/h 1 vô lí )

\(\Leftrightarrow2x=9\)

\(\Leftrightarrow x=9:2\)

\(\Leftrightarrow x=\frac{9}{2}\)

P/s: Chắc sai =))