a,\(\frac{x}{y}=\frac{7}{3}\Rightarrow\frac{x}{7}=\frac{y}{3}\Rightarrow\frac{5x}{35}=\frac{2y}{6}=\frac{5x-2y}{35-6}=\frac{87}{29}=3\)

=> x = 21; y = 9

b, \(\frac{x}{19}=\frac{y}{21}\Rightarrow\frac{2x}{38}=\frac{y}{21}=\frac{2x-y}{38-21}=\frac{34}{17}=2\)

=> x = 38; y = 42

Dễ lém sao đăng z?

​

giup mk voi

a) \(\frac{x}{y}=\frac{7}{3}\Rightarrow\frac{x}{7}=\frac{y}{3}\)

Theo tính chất dãy tỉ số bằng nhau:

  \(\frac{x}{7}=\frac{y}{3}=\frac{5x-2y}{5.7-2.3}=\frac{87}{29}=3\)

=> x = 7 x 3 = 21 ; y = 3x3 =9

b) \(\frac{x}{19}=\frac{y}{21}=\frac{2x-y}{2.19-21}=\frac{34}{17}=2\)

=> \(x=19.2=38\) ; \(y=21.2=42\)

1. Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y-z}{6+5-3}=\dfrac{54}{8}=\dfrac{27}{4}\)

+\(\dfrac{x}{6}=\dfrac{27}{4}\Rightarrow x=\dfrac{27.6}{4}=\dfrac{81}{2}\)

+\(\dfrac{y}{5}=\dfrac{27}{4}\Rightarrow y=\dfrac{27.5}{4}=\dfrac{135}{4}\)

+\(\dfrac{z}{3}=\dfrac{27}{4}\Rightarrow z=\dfrac{27.3}{4}=\dfrac{81}{4}\)

Vậy \(x=\dfrac{81}{2};y=\dfrac{135}{4};z=\dfrac{81}{4}\)

2,Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}=\dfrac{x+2y-3c}{2+2.3+3.4}=\dfrac{-20}{20}=-1\)

+\(\dfrac{x}{2}=-1\Rightarrow x=-1.2=-2\)

+\(\dfrac{y}{3}=-1\Rightarrow y=-1.3=-3\)

+\(\dfrac{c}{4}=-1\Rightarrow c=-1.4=-4\)

Vậy \(x=-2;y=-3;c=-4\)

3,Từ \(5x=8y=20z\Rightarrow\dfrac{5x}{160}=\dfrac{8y}{160}=\dfrac{20z}{160}\)

\(\Rightarrow\dfrac{x}{32}=\dfrac{y}{20}=\dfrac{z}{8}\)

Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{32}=\dfrac{y}{20}=\dfrac{z}{8}=\dfrac{x-y-z}{32-20-8}=\dfrac{3}{4}\)

+\(\dfrac{x}{32}=\dfrac{3}{4}\Rightarrow x=\dfrac{32.3}{4}=24\)

+\(\dfrac{y}{20}=\dfrac{3}{4}\Rightarrow y=\dfrac{20.3}{4}=15\)

+\(\dfrac{z}{8}=\dfrac{3}{4}\Rightarrow z=\dfrac{3.8}{4}=6\)

Vậy \(x=24;y=15;z=6\)

Theo đề bài, ta có:

\(\frac{x}{19}=\frac{y}{21}\) và 5x-2y=87

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{x}{19}=\frac{y}{21}=\frac{5x-2y}{5.19-2.21}=\frac{87}{53}\)

Vậy \(x=1007,y=\frac{1827}{53}\)

(Bài làm có gì ko hiueer cứ hỏi mk nhé  ^...^ )

\(\frac{x}{y}=\frac{7}{3}\Rightarrow\frac{x}{7}=\frac{y}{3}\Rightarrow\frac{5x}{35}=\frac{2y}{6}=\frac{5x-2y}{35-6}=\frac{87}{29}=3\)

=> x/7 = 3 => x=21

y/3 = 3 => y=9

\(\frac{x}{y}=\frac{7}{3}=3x=7y=\frac{x}{7}=\frac{y}{3}\)

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

\(\frac{x}{7}=\frac{y}{3}=\frac{5x}{5.7}=\frac{2y}{2.3}=\frac{2x-2y}{35-6}=\frac{87}{29}=3\)

\(\frac{x}{7}=3\Rightarrow x=3.7=21\)

\(\frac{y}{3}=3\Rightarrow y=3.3=9\)

Vậy x=21 ; y=9

Ta có: \(\frac{x}{y}\)= \(\frac{7}{3}\)\(\Rightarrow\)\(\frac{x}{7}\)=\(\frac{y}{3}\)

Ta lại có: \(\frac{x}{7}\)=\(\frac{y}{3}\)=\(\frac{5x-2y}{5.7-2.3}\)=\(\frac{87}{29}=3\)

Từ đó tính x và y nha bn.

Từ \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\)

Và \(\dfrac{y}{6}=\dfrac{z}{8}\Rightarrow\)\(\dfrac{y}{12}=\dfrac{z}{16}\)

Suy ra \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)\(\Rightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\)

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

\(\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=-1\Rightarrow x=-1\cdot9=-9\\\dfrac{y}{12}=-1\Rightarrow y=-1\cdot12=-12\\\dfrac{z}{16}=-1\Rightarrow z=-1\cdot16=-16\end{matrix}\right.\)

Ta có :

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x}{9}=\dfrac{y}{12}\)(1)

\(\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{y}{12}=\dfrac{z}{16}\)(2)

Từ (1) và (2) , suy ra \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)

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

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}=\dfrac{3x-2y-z}{27-24-16}=\dfrac{13}{-13}=-1\)

Do đó :

\(\dfrac{x}{9}=-1\Rightarrow x=-1.9=-9\)

\(\dfrac{y}{12}=-1\Rightarrow y=-1.12=-12\)

\(\dfrac{z}{16}=-1\Rightarrow z=-1.16=-16\)

Vậy x = -9 ; y = -12 ; z = -16

\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{6}=\dfrac{z}{8}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{16}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)

\(\Rightarrow\dfrac{3x}{27}=\dfrac{2y}{24}=\dfrac{z}{16}\)

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

\(\dfrac{3x}{54}=\dfrac{y}{24}=\dfrac{z}{32}\)

\(=\dfrac{3x-2y-z}{27-24-16}\)

\(=\dfrac{13}{-13}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=9.-1=-9\\y=12.-1=-12\\z=16.-1=-16\end{matrix}\right.\)

Đề sai sao á :))

Bạn ơi đề có sai ko

Sao lại \(\dfrac{y}{y}\)

- Theo đề bài ta có:

\(\dfrac{x}{3}=\dfrac{y}{4},\dfrac{y}{6}=\dfrac{z}{8}\)

=> \(\dfrac{x}{18}=\dfrac{y}{24}=\dfrac{z}{32}\)

=> \(\dfrac{3x}{54}=\dfrac{2y}{48}=\dfrac{z}{32}\)

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

\(\dfrac{3x}{54}=\dfrac{2y}{48}=\dfrac{z}{32}\)=\(\dfrac{3x-2y-z}{54-48-32}\)=\(\dfrac{13}{-26}=-2\)

- Suy ra:

x = \(\dfrac{-2.54}{3}=-36\)

y = \(\dfrac{-2.48}{2}=-48\)

z = \(-2.32=-64\)

- Vậy x = -36; y = -48; z = -64

\(\dfrac{2a\cdot x^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{-2a\left(x-1\right)^2}{5b\left(x-1\right)\left(x+1\right)}=\dfrac{-2a\left(x-1\right)}{5b\left(x+1\right)}\)

\(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)

\(=\dfrac{4x\cdot x-4x\cdot y}{5x^2\cdot x-5x^2\cdot y}\)

\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)

\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}\)

=x+y-z

\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)

\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)

\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3+y^3\right)\left(x^3-y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)