\(a.3,2.x+\left(-1,2\right).x+2,7=-4,9\)

    \(x.\left[3,2+\left(-1,2\right)+2,7\right]=-4,9\)

    \(x.4,7=-4,9\)

    \(x=-4,9+4,7\)

    \(x=-0,2\)

a, 

3,2x - 1,2x = -4,9 - 2,7 

2x = -7,6 

x= -3,8 

b, 

5,6x + 2,9x = -9,8 + 3,38 

8,5x = -6,42 

\(x=\frac{-321}{425}\)    

A = -5,13 : (25/28 - 8/9 . 1,25 + 16/63)

   = -5,13 : (25/28 - 10/9 + 16/63)

   = -5,13 : 1/28 = -3591/25 (-143,64)

B = (1 . 1,9 + 19,5 : 4/3) . (62/75 . 4/25)

   = ( 1,9 + 117/8 ) . 248/1875

  =  661/40 . 248/1875 = 2,185...

câu trl là do bấm máy tính đó nheaaaa

a) \(\left(\frac{5}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{17}\right]\)

\(=\left(\frac{1}{5}-\frac{126}{125}\right):\frac{4}{7}:\left[\left(\frac{13}{4}-\frac{59}{9}\right)\cdot\frac{36}{17}\right]\)

\(=\left(\frac{25}{125}-\frac{126}{125}\right):\frac{4}{7}:\left[-\frac{119}{36}\cdot\frac{36}{17}\right]\)

\(=-\frac{101}{125}:\frac{4}{7}:\left(-7\right)=-\frac{101}{125}\cdot\frac{7}{4}\cdot\left(-\frac{1}{7}\right)=\frac{101}{500}\)

b) \(\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)

\(=\left(-\frac{1}{2}-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right)\cdot\left(-\frac{1}{2}\right)\)

\(=-\frac{11}{10}:\left(-3\right)+\frac{1}{3}-\frac{1}{12}\)

\(=\frac{11}{30}+\frac{1}{3}-\frac{1}{12}=\frac{37}{60}\)

Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)\(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)( đpcm )

a, 3y(x2-xy)-7x2(y+xy)

= 6xy - 3xy² - 14xy - 14x²y

=-8xy-3xy²-14x²y

Bậc: 2

Ta có \(\frac{y+6}{y-2}=\frac{y}{y-4}\)

=> (y + 6)(y - 4) = y(y - 2)

=> y² - 4y + 6y - 24 = y² - 2y

=> y² + 2y - 24 = y² - 2y

=> 2y - 24 = -2y

=> 2y + 2y = 24

=> 4y = 24

=> y = 6

Vậy y = 6

\(\frac{y+6}{y-2}=\frac{y}{y-4}\)

=> \(\frac{y-2+8}{y-2}=\frac{y}{y-4}\)

=> \(1+\frac{8}{y-2}=\frac{y}{y-4}\)

=> \(\frac{8}{y-2}=\frac{y}{y-4}-1=\frac{4}{y-4}\)

=> \(\frac{8}{y-2}=\frac{4}{y-4}\)

=> 8(y - 4) = 4(y - 2)

=> 8y - 32 = 4y - 8

=> 8y - 32 - 4y + 8 = 0

=> 4y - 24 = 0

=> 4y = 24

=> y = 6

Vậy y = 6

\(\frac{2x-6}{4}=\frac{x+1}{3}\)

=> \(\frac{2\left(x-3\right)}{4}=\frac{x+1}{3}\)

=> \(\frac{x-3}{2}=\frac{x+1}{3}\)

=> \(\frac{3\left(x-3\right)}{6}=\frac{2\left(x+1\right)}{6}\)

=> 3(x - 3) = 2(x + 1)

=> 3x - 9 = 2x + 2

=> 3x - 9 - 2x - 2 = 0

=> x - 11 = 0

=> x = 11

Vậy x = 11

Ta có: (2x - 6 ) : 4 = ( x + 1 ) : 3

      =>  3( 2x - 6 ) = 4 ( x + 1 )

      =>  6x - 18 = 4x + 4

      =>  2x = 22

      => x = 11

Bài 1:

Vẽ hình
O x y t z

Ta có: \(\widehat{xOy}+\widehat{yOt}+\widehat{zOt}+\widehat{xOz}=360^o\)(Tổng các góc trong không có điểm trong chung ) 

\(\Rightarrow\widehat{xOy}+90^o+\widehat{zOt}+90^o=360^o\)

\(\Rightarrow\widehat{xOy}+\widehat{zOt}=360^o-90^o-90^o\)

\(\Rightarrow\widehat{xOy}+\widehat{zOt}=180^o\)

Vậy \(\widehat{xOy}+\widehat{zOt}=180^o\)

Bài 2:

O B A C D x 100 độ y

A) Ta có: \(\widehat{AOB}=100^o,\widehat{AOC}=90^o,\widehat{BOD}=90^o\)

\(\Rightarrow\widehat{COD}=360^o-\left(\widehat{AOB}+\widehat{AOC}+\widehat{BOD}\right)\)

\(=360^o-\left(100^o+90^o+90^o\right)=360^o-280^o=80^o\)

Ox là tia phân giác của \(\widehat{AOB}\)nên \(\widehat{xOA}=\frac{\widehat{AOB}}{2}=\frac{100^o}{2}=50^o\)

Oy là tia phân giác của \(\widehat{COD}\)nên \(\widehat{COy}=\frac{\widehat{COD}}{2}=\frac{80^o}{2}=40^o\)

Do đó: \(\widehat{xOy}=\widehat{xOA}+\widehat{AOC}+\widehat{COy}=50^o+90^o+40^o\)

Hay \(\widehat{xOy}=180^o\)

=> Ox và Oy là hai tia đối nhau ( đpcm ) 

b) Ta có: \(\widehat{xOC}=\widehat{xOA}+\widehat{AOC}=50^o+90^o=140^o\)

\(\widehat{BOy}=\widehat{BOD}+\widehat{DOy}=90^o+40^o=130^o\)