\(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\\ \dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{9}{10}\\ y=\dfrac{9}{10}\times\dfrac{11}{4}=\dfrac{99}{40}\\ b,1\dfrac{1}{4}+2\dfrac{1}{5}\times y=2\dfrac{3}{5}\\ \dfrac{5}{4}+\dfrac{11}{5}\times y=\dfrac{13}{5}\\ \dfrac{11}{5}\times y=\dfrac{13}{5}-\dfrac{5}{4}\\ \dfrac{11}{5}\times y=\dfrac{27}{20}\\ y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{44}\)

\(c,2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\\ \dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{41}{20}\\ y=\dfrac{9}{4}:\dfrac{41}{20}=\dfrac{45}{41}\\ c2,x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{31}{10}\\ x=\dfrac{31}{10}\times\dfrac{10}{3}=\dfrac{31}{3}\)

a) \(...\Rightarrow\dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\)

\(\Rightarrow y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\Rightarrow y:\dfrac{11}{4}=\dfrac{24}{10}-\dfrac{15}{10}\)

\(\Rightarrow y:\dfrac{11}{4}=\dfrac{9}{10}\Rightarrow y=\dfrac{9}{10}x\dfrac{11}{4}=\dfrac{99}{40}\)

b) \(...\Rightarrow\dfrac{5}{4}+\dfrac{11}{5}xy=\dfrac{13}{5}\Rightarrow\dfrac{11}{5}xy=\dfrac{13}{5}-\dfrac{5}{4}\)

\(\Rightarrow\dfrac{11}{5}xy=\dfrac{52}{20}-\dfrac{25}{20}\Rightarrow\dfrac{11}{5}xy=\dfrac{27}{20}\)

\(\Rightarrow y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{20}x\dfrac{5}{11}=\dfrac{27}{44}\)

c) \(...\Rightarrow\dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\Rightarrow\dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{9}{4}:y=\dfrac{56}{20}-\dfrac{15}{20}\Rightarrow\dfrac{9}{4}:y=\dfrac{39}{20}\)

\(\Rightarrow y=\dfrac{9}{4}:\dfrac{39}{20}\Rightarrow y=\dfrac{9}{4}x\dfrac{20}{39}=\dfrac{15}{13}\)

d) \(...\Rightarrow x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\Rightarrow x:\dfrac{10}{3}=\dfrac{24}{10}+\dfrac{7}{10}\)

\(\Rightarrow x:\dfrac{10}{3}=\dfrac{31}{10}\Rightarrow x=\dfrac{31}{10}x\dfrac{10}{3}=\dfrac{31}{3}\)

a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)

    y         =    \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)

    y         = \(\dfrac{4}{3}\)

b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)

    y - 0,5 + 0,5 = \(\dfrac{3}{4}\)

   y                   = \(\dfrac{3}{4}\)

c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2

   0,8 - 0,4y = 0,2

           0,4y = 0,8 - 0,2

           0,4y  = 0,6

               y = 1,5

d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)

    y + \(\dfrac{3}{4}\)           = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)

   y + \(\dfrac{3}{4}\)            = \(\dfrac{14}{9}\)

y                    = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)

 y                   =   \(\dfrac{29}{36}\)

e, y : \(\dfrac{5}{4}\)         = \(\dfrac{9}{5}\)  + \(\dfrac{1}{2}\)

   y : \(\dfrac{5}{4}\)         =   \(\dfrac{23}{10}\)

  y                =      \(\dfrac{23}{10}\)

  y               =   \(\dfrac{23}{8}\)

f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y   = \(\dfrac{4}{5}\)

   y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\))      =  \(\dfrac{4}{5}\)

   2y                       = \(\dfrac{4}{5}\)

    y                        = \(\dfrac{2}{5}\)

vd câu 1:
ta có x-y=4 =>x=4+y
ta có pt:
4+y/y-2=3/2
=>8+2y=3y-6
=>-y=-14
=>y=14
=>x=4+y=4+14=18
các bài khác cũng tương tự thôi bạn

dấu chéo có nghĩa là phân số hí

Vậy đề bài yêu cầu gì vậy bạn?

\(C=5x^3y^2-4x^3y^2+3x^2y^3+\dfrac{1}{2}x^2y^3+\dfrac{1}{3}x^4y^5-3x^4y^5-\dfrac{1}{7}\)

    \(=x^3y^2+\dfrac{7}{2}x^2y^3-\dfrac{8}{3}x^4y^5-\dfrac{1}{7}\)

gg

\(a,2\dfrac{2}{5}:y\times1\dfrac{3}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y\times\dfrac{7}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y=\dfrac{7}{8}:\dfrac{7}{4}\\ \dfrac{12}{5}:y=\dfrac{1}{2}\\ y=\dfrac{12}{5}:\dfrac{1}{2}=\dfrac{24}{5}\\ b,3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\\ \dfrac{17}{5}:y:\dfrac{5}{4}=\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{5}:\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{13}\\ y=\dfrac{17}{13}\times\dfrac{5}{4}=\dfrac{85}{52}\)

\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)

a, 2\(\dfrac{2}{5}\): y \(\times\)1\(\dfrac{3}{4}\) = \(\dfrac{7}{8}\)

     \(\dfrac{12}{5}\) : y   \(\times\dfrac{7}{4}\)    = \(\dfrac{7}{8}\)

   \(\dfrac{12}{5}\)  : y         = \(\dfrac{7}{8}\) : \(\dfrac{7}{4}\)

          \(\dfrac{12}{5}\) : y        = \(\dfrac{1}{2}\)       

                  y         = \(\dfrac{12}{5}\) : \(\dfrac{1}{2}\)

                  y         =    \(\dfrac{24}{5}\)

b, 3\(\dfrac{2}{5}\): y : 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\) 

     \(\dfrac{17}{5}\): y: \(\dfrac{5}{4}\)    = \(\dfrac{13}{5}\)

       \(\dfrac{17}{5}\):y         = \(\dfrac{13}{5}\times\dfrac{5}{4}\)

       \(\dfrac{17}{5}\) : y       = \(\dfrac{13}{4}\)

               y        =   \(\dfrac{17}{5}\) : \(\dfrac{13}{4}\)

               y        =  \(\dfrac{68}{65}\)

c, \(\dfrac{12}{5}\) - 2\(\dfrac{2}{5}\)\(\times y\) = 1\(\dfrac{1}{4}\)

     \(\dfrac{12}{5}\) - \(\dfrac{12}{5}\)\(\times\)y = \(\dfrac{5}{4}\)

            \(\dfrac{12}{5}\times y\) =  \(\dfrac{12}{5}\) - \(\dfrac{5}{4}\)

           \(\dfrac{12}{5}\) \(\times\) y =    \(\dfrac{23}{20}\)

                     \(y\)  = \(\dfrac{23}{20}\): \(\dfrac{12}{5}\)

                     y   = \(\dfrac{23}{48}\)

\(\dfrac{8}{9}\) : ( 2 - 3 \(\times\) y) = \(\dfrac{5}{3}\) 

        2 - 3 \(\times\) y = \(\dfrac{8}{9}\) : \(\dfrac{5}{3}\)

        2 - 3 \(\times\) y = \(\dfrac{8}{15}\)

             3 \(\times\) y = 2 - \(\dfrac{8}{15}\)

             3 \(\times\) y = \(\dfrac{22}{15}\)

                   y  = \(\dfrac{22}{15}\) : 3 

                   y = \(\dfrac{22}{45}\)

e: \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{3}{y}=3\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7}{y}=-2\\\dfrac{1}{x}-\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\\dfrac{1}{x}=1+\dfrac{2}{7}=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\x=\dfrac{7}{9}\end{matrix}\right.\)

\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)

\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)

\(=6x^2y\)

\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)

\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)

\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)

1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy

2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

3: =(x+y-x+y)^2=(2y)^2=4y^2

4: =(2x+3-2x-5)^2=(-2)^2=4

5: =18^8-18^8+1=1

bài 1

A)37/60

B)13/45

C)1/4

bài 2

a)42/25:y/5=5/6

y=10,08

b)27/y:9/4=3/7

y=28

1) 2^{2x + 1} = 32

=> 2^{2x + 1} = 2⁵

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 2

(2) 3.x³ - 100 = 275

=> 3x³ = 275 + 100

=> 3x³ = 375

=> x³ = 375 : 3

=> x³ = 125

=> x³ = 5³

=> x = 5

(4) (x - 1)³ - 2⁵ = 7²

=> (x - 1)³ = 49 + 32

=> (x - 1)³ = 81

(xem lại đề)

5) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

  \(\frac{x}{3}=\frac{y}{5}=\frac{x-y}{3-5}=\frac{-4}{-2}=2\)

=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.5=10\end{cases}}\)

Vậy ...

6) Ta có: \(\frac{x}{2}=\frac{y}{3}\) => \(\frac{x}{10}=\frac{y}{15}\)

       \(\frac{y}{5}=\frac{z}{4}\) => \(\frac{y}{15}=\frac{z}{12}\)

=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=\frac{-49}{37}\)

=> \(\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}}\) => \(\hept{\begin{cases}x=-\frac{49}{37}\cdot10=\frac{-490}{37}\\y=-\frac{49}{37}\cdot15=-\frac{735}{37}\\z=-\frac{49}{37}\cdot12=-\frac{588}{37}\end{cases}}\)

Vậy ...

mk lm bài mà mk cho là ''khó'' nhất thôi nha 

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và \(x+y+z=-49\)

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{15}=\frac{z}{12}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

ADTC dãy tỉ số bằng nhau ta có 

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=-\frac{49}{37}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{49}{37}.10=-\frac{490}{37}\\y=-\frac{49}{37}.15=-\frac{735}{37}\\z=-\frac{49}{37}.12=-\frac{588}{37}\end{cases}}}\)