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

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!!!

\(x.y.z=810\)(1)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)(2)

Thay (2) vào (1) ta có:

\(2k.3k.5k=810\)

\(\Rightarrow30k^3=810\Rightarrow k^3=17\Rightarrow k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=3.3=9\\z=5.3=15\end{matrix}\right.\)

Vậy \(x=6;y=9;z=15\)

Chúc bạn học tốt!!!

Bài 1:

\(a^2+b^2+c^2=16\Rightarrow\left(a+b+c\right)^2-2ab-2bc-2ac=16\)\(\Leftrightarrow-2\left(ab+bc+ac\right)=16\Rightarrow ab+bc+ac=-8\)\(\Rightarrow\left(ab+bc+ac\right)^2=64\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=64\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=64\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=64\)

Ta có:

\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)\(=16^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=256-2.64=128\)

1. Ta có $a + b + c = 0$

\(\Rightarrow\) $( a + b + c)^2 = 0$

\(\Leftrightarrow\) $a^2+b^2 +c^2 +2ab+2bc+2ac = 0
\(\Leftrightarrow\) $a^2 + b^2 + c^2 = -2(ab+bc+ac)$

Thay $a^2 + b^2 + c^2 = 2$

\(\Rightarrow\)$2 = -2(ab+bc+ac)$ \(\Rightarrow\) $ab + bc +ac = -1 $
Ta có: $(a^2+b^2+c^2) = 2$

\(\Leftrightarrow\) $(a^2+b^2+c^2)^2 = 4$

\(\Leftrightarrow\)$a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2 = 4$

\(\Leftrightarrow\) $a^4+b^4+c^4 + 2(a^b^2+b^2c^2+a^2c^2) = 4$ (1)
Do $2(ab+bc+ac)^2 = 2(a^2b^2+b^2c^2+a^2c^2 + 2a^2bc+2ab^2c+2abc^2)$ (2)
Từ (1)(2) => $a^4+b^4+c^4+2(ab+bc+ac)^2 - 4abc(a+b+c) = 4$(3)
Thay $(ab+bc+ac) = -1$ và $a+b+c = 0$ (4)
Từ (3)(4) => $a^4 + b^4 + c^4 +2(-1)^2 -4abc.(0) = 4 $
<=> $a^4 + b^4 + c^4 + 2 = 4 => a^4 + b^4 + c^4 = 2 $

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..

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

\(x\) \(=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)

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

\(\left|x\right|\) \(=\dfrac{2}{5}+\dfrac{4}{5}=\dfrac{6}{5}\)

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

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

=> \(x:\left(-5\right)=24:15\)

=> \(x\) \(=\left(24:15\right).\left(-5\right)=-8\)

d, \(\dfrac{x}{4}=\dfrac{y}{5}\) và \(x-y=21\)

Theo tính chất dãy tỉ số bằng nhau, ta có:

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

Suy ra:

\(\dfrac{x}{4}=-21\) => \(\left(-21\right).4=-84\)

\(\dfrac{y}{5}=-21\) => \(\left(-21\right).5=-105\)

Vậy x = -84 và y = -105.

\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

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: \(\Leftrightarrow-4< =x< =-3\)

hay \(x\in\varnothing\)

b: =>-9<x<=3

hay \(x\in\left\{0;1;2;3\right\}\)