(16x8x10)-(12x8x5)=800cm3

2916002

bằng 2916002

=> 1 - 3 . X=x - 7 hoặc 1 - 3 . X =-(x-7)

*1 - 3x =x - 7                 *1 - 3x = -(x - 7 )

   8         =x + 3x             1 - 3x = -x + 7     

   8         =4x                   -3x+x =7-1

   8 : 4     =x                    -2x    =6

    2          = x                      x   = 6:(-2)

  =>x = 2                            x    = -3

vậy x \(\in\){2; -3}

đúng + x  =1

           x   =1 -đúng 

           x    = thích

Ta có: \(\frac{x+2}{y+10}\)\(=\)\(\frac{1}{5}\)\(\Rightarrow\)\(5\left(x+2\right)=y+10\)(1)

             \(y-3x=2\)\(\Rightarrow\)\(y+2=3x\)                              (2)

Thay (2) vào (1) ta có:

\(5\left(x+2\right)=\left(y+2\right)+8\)

\(5x+10=3x+8\)

\(5x-3x=8-10\)

\(2x=-2\)

\(x=-2:2\)

\(x=-1\)

Vậy: x=-1

Chúc bạn làm bài tốt!

13 - (2x + 5) = 21 + (22-x) 
<=>  13 - 21 = (22-x) + (2x + 5) 
=> -8 = 22 - x +2x +5 
=> -8 = x + 27 
=> x = -8 - 27 
=> x = -35.

13-(2x +5) = 21 + (22-x)

=>13 - 2x - 5 = 21 + 22 - x

=>8 - 2x = 43 - x

=>43 - x - (8 - 2x) = 0

=>43 - x - 8 + 2x = 0

=>35 + 2x - x = 0

=>35 + x = 0

=> x = 0 - 35

=> x = - 35

ta có : Do NB song song với MA nên

\(\hept{\begin{cases}\widehat{ABN}+\widehat{MAB}=180^0\\\widehat{ABN}-\widehat{MAB}=40^0\end{cases}}\Rightarrow2\widehat{MAB}=180^0-40^0=140^0\)

Nên \(\widehat{MAB}=70^0\)

30 D B C A E

BN Ê ĐỀ SAI NHÉ BN

Bài 1:

a: \(\frac17-\left(-\frac35\right)+\frac67=\frac77+\frac35=1+\frac35=\frac85\)

b: \(\frac{20}{13}\cdot\frac{-8}{9}+\frac{7}{18}\cdot\frac{20}{13}\)

\(=\frac{20}{13}\left(-\frac89+\frac{7}{18}\right)\)

\(=\frac{20}{13}\left(-\frac{16}{18}+\frac{7}{18}\right)=\frac{20}{13}\cdot\frac{-9}{18}=\frac{20}{13}\cdot\frac{-1}{2}=-\frac{20}{26}=-\frac{10}{13}\)

c: \(0,75+\frac{15}{6}:5-\frac{1}{36}\cdot\left(-3\right)^2+\left(-\frac{2020}{2021}\right)^0\)

\(=\frac34+\frac{15}{6\cdot5}-\frac{1}{36}\cdot9+1\)

\(=\frac34-\frac14+\frac12+1=\frac12+\frac12+1=2\)

Bài 2:

a: \(3x-\frac23=-\frac59\)

=>\(3x=-\frac59+\frac23=-\frac59+\frac69=\frac19\)

=>\(x=\frac19:3=\frac{1}{27}\)

b: \(\frac45-\left|x+\frac12\right|=\frac14\)

=>\(\left|x+\frac12\right|=\frac45-\frac14=\frac{16-5}{20}=\frac{11}{20}\)

=>\(\left[\begin{array}{l}x+\frac12=\frac{11}{20}\\ x+\frac12=-\frac{11}{20}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{20}-\frac12=\frac{1}{20}\\ x=-\frac{11}{20}-\frac12=-\frac{21}{20}\end{array}\right.\)

c: \(\left(\frac32x-\frac15\right)^2\cdot\left(\frac{51}{67}x^2+\frac12\right)=0\)

=>\(\left(\frac32x-\frac15\right)^2=0\)

=>\(\frac32x-\frac15=0\)

=>\(\frac32x=\frac15\)

=>\(x=\frac15:\frac32=\frac15\cdot\frac23=\frac{2}{15}\)