a)Vì \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)nên \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{x}{28}\).

Áp dụng t/c dãy tỉ số = nhau, ta có :

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)

⇒2x = 3.30 = 90 ⇒ x = 45

3y = 3.60 = 180 ⇒ y = 60

z = 3.28 = 84

Ý b) có gì đó sai sai ?

c)Ta có :

\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Áp dụng t/c dãy tỉ số = nhau, ta có :

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)

⇒x = 5.15 = 75

y = 5.10 = 50

z = 5.6 = 30

d)Ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\left(k\in Z\right)\)

⇒ x = 2k ; y = 3k ; z = 5k

⇒ xyz = 2k.3k.5k = 30k³ = 810

Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\left(\dfrac{a}{2}\right)^2=\left(\dfrac{b}{3}\right)^2=\left(\dfrac{c}{4}\right)^2\)

\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{z^2}{16}\)\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}\)

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

\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}=\dfrac{a^2+b^2-2c^2}{4+9-32}=\dfrac{-76}{-19}=4\)

\(\Rightarrow\dfrac{a^2}{4}=4\Rightarrow a=4\)

\(\dfrac{b^2}{9}=4\Rightarrow b=6\)

\(\dfrac{2c^2}{32}=4\Rightarrow2c^c=128\Rightarrow c=8\)

Vậy \(\left\{{}\begin{matrix}a=4\\b=6\\c=8\end{matrix}\right.\)

a = 4 hoặc a = -4
b = 6 hoặc b = -6
c = 8 hoặc c = -8

Bài 2:

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

\(\dfrac{a}{2}=\dfrac{b}{-5}=\dfrac{a+b}{2+\left(-5\right)}=\dfrac{21}{-3}=-7\)

(do \(a+b=21\))

\(\Rightarrow\left\{{}\begin{matrix}a=-7.2=-14\\b=-7.\left(-5\right)=35\end{matrix}\right.\)

Vậy \(a=-14;b=35\)

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

\(\dfrac{-10}{a}=\dfrac{-15}{b}=\dfrac{-10-\left(-15\right)}{a-b}=\dfrac{5}{-5}=-1\)

(do \(a-b=-5\))

\(\Rightarrow\left\{{}\begin{matrix}a=-10:\left(-1\right)=10\\b=-15:\left(-1\right)=15\end{matrix}\right.\)

Vậy \(a=10;b=15\)

Chúc bạn học tốt!!!

c, Ta có:

\(3x=2y\Rightarrow21x=14y\)

\(7y=5z\Rightarrow14y=10z\)

\(\Rightarrow21x=14y=10z\Rightarrow\dfrac{21x}{210}=\dfrac{14y}{210}=\dfrac{10z}{210}\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)

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

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x-y+z}{10-15+21}=\dfrac{32}{16}=2\)

(do \(x-y+z=32\))

\(\Rightarrow\left\{{}\begin{matrix}x=2.10=20\\y=2.15=30\\z=2.21=42\end{matrix}\right.\)

Vậy \(x=20;y=30;z=42\)

Chúc bạn học tốt!!!

a) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\)

\(\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{12}\)(1)

Ta có: \(\dfrac{b}{4}=\dfrac{c}{5}\)

nên \(\dfrac{b}{12}=\dfrac{c}{15}\)(2)

Từ (1) và (2) suy ra \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\)

mà a+b+c=2 

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

\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{2}{35}\)

Do đó: 

\(\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{2}{35}\\\dfrac{b}{12}=\dfrac{2}{35}\\\dfrac{c}{15}=\dfrac{2}{35}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{16}{35}\\b=\dfrac{24}{35}\\c=\dfrac{30}{35}=\dfrac{6}{7}\end{matrix}\right.\)

Vậy: \(a=\dfrac{16}{35}\); \(b=\dfrac{24}{35}\); \(c=\dfrac{6}{7}\)

b) Ta có: 2a=3b=5c

nên \(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}\)

mà a+b-c=3

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

\(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b-c}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{19}{30}}=\dfrac{90}{19}\)

Do đó: 

\(\left\{{}\begin{matrix}2a=\dfrac{90}{19}\\3b=\dfrac{90}{19}\\5c=\dfrac{90}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{45}{19}\\b=\dfrac{30}{19}\\c=\dfrac{18}{19}\end{matrix}\right.\)

Vậy: \(a=\dfrac{45}{19}\); \(b=\dfrac{30}{19}\); \(c=\dfrac{18}{19}\)

a,\(x-\dfrac{3}{5}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}+\dfrac{3}{5}\)

\(x=\dfrac{6}{5}\)

b,\(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)

\(\left|x\right|=\dfrac{2}{5}+\dfrac{4}{5}\)

\(\left|x\right|=\dfrac{6}{5}\)

\(\Rightarrow x=\pm\dfrac{6}{5}\)

c,\(\dfrac{x}{-5}=\dfrac{24}{15}\)

\(x=\dfrac{-5.24}{15}\)

\(x=\dfrac{-24}{5}\)

d,Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)

+\(\dfrac{x}{4}=-21\Rightarrow x=-21.4=-84\)

+\(\dfrac{y}{5}=-21\Rightarrow y=-21.5=-105\)

Vậy x=-84 ; y=-105

a/ \(x-\dfrac{3}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow x=\dfrac{3}{5}+\dfrac{3}{5}\)

\(\Leftrightarrow x=\dfrac{6}{5}\)

Vậy...

b/ \(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{2}{5}+\dfrac{4}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{6}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-\dfrac{6}{5}\end{matrix}\right.\)

Vậy...

c/ \(\dfrac{x}{-5}=\dfrac{24}{15}\)

\(\Leftrightarrow15x=-120\)

\(\Leftrightarrow x=-8\)

Vậy...

c/ Theo t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=-21\\\dfrac{y}{5}=-21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-84\\y=-105\end{matrix}\right.\)

Vậy..

e: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x+5}{2}=\dfrac{y-2}{3}=\dfrac{x-y+5+2}{2-3}=\dfrac{10+7}{-1}=-17\)

=>x+5=-34; y-2=-51

=>x=-39; y=-49

g: Áp dụng tính chất của DTSBN, ta được

\(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}=\dfrac{5a-3b-4c-5-9+20}{5\cdot2-3\cdot4-6\cdot4}=\dfrac{-253}{13}\)

=>a-1=-506/13; b+3=-1012/13; c-5=-1518/13

=>a=-493/13; b=-1051/13; c=-1453/13

Lời giải:
e. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-(y-2)}{2-3}=\frac{(x-y)+5+2}{2-3}=\frac{10+5+2}{-1}=-17$

Suy ra:

$x+5=2(-17)=-34\Rightarrow x=-39$

$y-2=3(-17)=-51\Rightarrow y=-49$

f. Đề thiếu. Bạn xem lại

h. Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}$

$=\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}$

$=\frac{5a-5-(3b+9)-(4c-20)}{10-12-24}$

$=\frac{5a-3b-4c-5-9+20}{-26}=\frac{500-5-9+20}{-26}=\frac{-253}{13}$

Suy ra:
$a-1=2.\frac{-253}{13}\Rightarrow a=\frac{-493}{13}$

$b+3=4.\frac{-253}{13}\Rightarrow b=\frac{-1051}{13}$

$c-5=6.\frac{-253}{13}\Rightarrow c=\frac{-1453}{13}$

a. Đặt \(\dfrac{x}{-3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=-3k\\y=5k\end{matrix}\right.\)

mà \(x.y=\dfrac{-5}{27}\)

hay \(-3k.5k=\dfrac{-5}{27}\)

\(\Rightarrow-15.k^2=\dfrac{-5}{27}\)

\(\Rightarrow k^2=\dfrac{1}{81}=\left(\pm\dfrac{1}{9}\right)^2\)

Với \(k=\dfrac{1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{5}{9}\end{matrix}\right.\)

Với \(k=\dfrac{-1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{-5}{9}\end{matrix}\right.\)

Vậy.......

b. Từ \(\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{3}=\dfrac{z}{5}\end{matrix}\) \(\Rightarrow\begin{matrix}\dfrac{x}{9}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{20}\end{matrix}\) \(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

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

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{x-y+z}{9-12+20}=\dfrac{32}{17}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{32}{17}\Rightarrow x=\dfrac{32.9}{17}=\dfrac{288}{17}\\\dfrac{y}{12}=\dfrac{32}{17}\Rightarrow y=\dfrac{32.12}{17}=\dfrac{384}{17}\\\dfrac{z}{20}=\dfrac{32}{17}\Rightarrow z=\dfrac{32.20}{17}=\dfrac{640}{17}\end{matrix}\right.\)

Vậy.........

1.a)\(2.x-\dfrac{5}{4}=\dfrac{20}{15}\)

\(\Leftrightarrow2.x=\dfrac{20}{15}+\dfrac{5}{4}=\dfrac{4}{3}+\dfrac{5}{4}=\dfrac{16+15}{12}=\dfrac{31}{12}\)

\(\Leftrightarrow x=\dfrac{31}{12}:2=\dfrac{31}{12}.\dfrac{1}{2}=\dfrac{31}{24}\)

b)\(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{8}\right)\)

\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}=-\dfrac{5}{6}\)

2.Theo đề bài, ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\) và \(a+b=-15\)

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

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{-15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=-3\Rightarrow a=-6\\\dfrac{b}{3}=-3\Rightarrow b=-9\end{matrix}\right.\)

3.Ta xét từng trường hợp:

-TH1:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow x\in\left\{0;1\right\}\)

-TH2:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)

Vậy \(x\in\left\{0;1\right\}\)

4.\(B=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^9=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^9=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{18}=\left(\dfrac{3}{7}\right)^3=\dfrac{27}{343}\)

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 \(x=-2;y=-3;c=-4\)