Bài 1:

a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)

\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{6}{7}\)

b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)

\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)

\(\Rightarrow x=\dfrac{1}{6}\)

help me

\(A=\left(a\text{x}7+a\text{x}8-a\text{x}15\right):\left(1+2+3+...+10\right)\)

\(A=\left(a\text{x}\left(7+8-15\right)\right):\left(1+2+3+...+10\right)\)

\(A=\left(a\text{x}0\right):\left(1+2+3+..+10\right)\)

\(A=0:\left(1+2+3+...+10\right)\)

\(A=0\)

\(B=\left(18-9\text{x}2\right)\text{x}\left(2+4+6+8+10\right)\)

\(B=\left(18-18\right)\text{x}\left(2+4+6+8+10\right)\)

\(B=0\text{x}\left(2+4+6+8+10\right)\)

\(B=0\)

Câu trả lời B=0

a)456+[1263-x]=1362

          1263-x=1362-456

          1263-x=906

                    x=1263-906

                   x=357

b)22.x+4=20

22.x    =20-4

22.x    =16

     x     =16:20

     x      =0.8

c)34.x+1=41

   34.x    =41-1

   34.x     =40

        x    =40:34

        x     =20/17

chắc chắn 100% k cho mình nha nguyenthuyngan

456 + [ 1263 - x] = 1326

1263 - x = 1326 - 456

1263 - x =  870

x = 1326 - 870

x= 456

22x + 4 = 20

22x = 24

x = 24/22

34x + 1 = 41

=> 34x = 40

=> x = 40/34

a) 3x4-3x-2x+2=x+20   ​​​b) 10+13-(x²- 6x+9)=0

    12-5x=x+20                 -x²+ 6x+14=0... 

    6x=-8

    x=-3/2

Ta có: 34 chia hết cho x 

=> x E Ư(34) = { -1 ; 1 ; -2;2;-17;17;-34;34}

7 chia hết x - 1

=> x - 1 = Ư(7) = { -1;1;-7;7}

=> x = {0;2;-6;8}

Ta có: 34 chia hết cho x 

=> x E Ư(34) = { -1 ; 1 ; -2;2;-17;17;-34;34}

7 chia hết x - 1

=> x - 1 = Ư(7) = { -1;1;-7;7}

=> x = {0;2;-6;8}

=1

=4

=9

=16

=25

=36

=42

=64

=81

=100

1

4

9

16

25

36

49

64

81

100

bài 1

a)\(\frac{x}{y}=\frac{7}{3}\Rightarrow\frac{x}{7}=\frac{y}{3}\)
\(\Rightarrow\frac{5x}{5.7}=\frac{2y}{2.3}=\frac{5x-2y}{35-6}=\frac{87}{29}=3\)
\(\Rightarrow x=3.7=21;y=3.3=9\)

Bài dưới tướng tự nhé

-2x - (x - 7) = 34 - (-x + 25)

-3x+7=34+x-25

-4x=2

x=-1/2

-2x-(x-7)=34-(-x+25)

-2x-x+7=34+x-25

-2x-x+7-34-x+25=0

-4x-2=0

-4x=2

x=\(-\frac{1}{2}\)

#H

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