Giải:

Theo bài ra ta có:

\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{12}\)

\(\Rightarrow-3\le x\le\frac{23}{12}\)

\(\Rightarrow x\varepsilon\left\{-2;-1;0;1\right\}\)

\(\frac{-5}{6}+\frac{16}{6}+-\frac{29}{6}\le x\le\frac{-6}{12}+\frac{24}{12}+\frac{5}{12}\)

=>-3\(\le\) x\(\le\) 23/12

=> x thuộc{-2-1;0;1}

x thuộc {-2;-1;0;1}

a) Ta có :

\(\left|x\right|\le6\)

\(\Leftrightarrow\left|x\right|\in\left\{0;1;2;3;4;5;6\right\}\) (do \(\left|x\right|\ge0\))

\(\Leftrightarrow x\in\left\{-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6\right\}\)

Vậy \(x\in\left\{-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6\right\}\)

b) Ta có :

\(2\le\left|y-1\right|\le3\)

\(\Leftrightarrow\left|y-1\right|\in\left\{2;3\right\}\)

\(\Leftrightarrow y-1\in\left\{-2;-3;2;3\right\}\)

\(\Leftrightarrow y\in\left\{-1;-2;3;4\right\}\)

c) \(\left(x+1\right)\left(y+2\right)=-12\)

Vì \(x,y\in Z\Leftrightarrow x+1;y-2\in Z,x+1;y-2\inƯ\left(-12\right)\)

Bn tự lập bảng rồi tính tiếp nhé!

a) \(\left|x\right|\le6\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le6\\x\le-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\left\{...;-4;-3;-2;-1;0;1;2;3;4;5;6\right\}\\x=\left\{...;-10;-9;-8;-7;-6\right\}\end{matrix}\right.\)

đặt x/3=y/4=k

=>x=3k

y=4k

=>xy=3k.4k=12.k^2 =300

=>k^2 =25

=>k=5

=>x=5.3=15

y=5.4=20

b)chờ chút

a, ta co\(\frac{x}{3}=\frac{y}{4}=>\frac{x^2}{9}=\frac{x}{3}.\frac{y}{4}=\)\(\frac{300}{12}=25\)

=> x= 15=> y=10

\(b.\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{117}{9}=13\)

\(\Rightarrow x=13.2=26\)

\(y=13.3=39\)

\(z=13.4=52\)

L-I-K-E tớ nhé !!

22222222222222222

  Ta có :  

\(\frac{x^3}{8}\)= \(\frac{y^3}{64}\)= \(\frac{z^3}{216}\) \(\Rightarrow\)\(\frac{x^3}{2^3}\)= \(\frac{y^3}{4^3}\)= \(\frac{z^3}{6^3}\)\(\Rightarrow\)\(\frac{x^2}{2^2}\)=\(\frac{y^2}{4^2}\)=\(\frac{z^2}{6^2}\)

và có : \(^{x^2+y^2+z^2=224}\)

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

\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{224}{56}=4\)

=>    \(\frac{x^2}{4}=4\Rightarrow x^2=16\Rightarrow x\in4;-4\)​

\(\frac{y^2}{16}=4\Rightarrow y^2=64\Rightarrow y\in8:-8\)

\(\frac{z^2}{36}=4\Rightarrow z^2=144\Rightarrow z\in12:-12\)

Vì \(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\)nên x,y,z cùng dấu 

Vậy \(x,y,z\in\left(4;8;12\right);\left(-4;-8;-12\right)\)

\(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\\ 5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\\ \Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}=\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a-7b+5c}{63-98+50}=\dfrac{-30}{15}=-2\\ \Rightarrow\left\{{}\begin{matrix}a=-42\\b=-28\\c=-20\end{matrix}\right.\)

\(x:y:z=3:4:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow x=3k;y=4k;z=5k\)

\(2x^2+2y^2-3z^2=-100\\ \Rightarrow18k^2+32k^2-75k^2=-100\\ \Rightarrow-25k^2=-100\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=6;y=8;z=10\\x=-6;y=-8;z=-10\end{matrix}\right.\)

\(2a=3b\text{⇒}a=\dfrac{3b}{2}\) , \(5b=7c\text{⇒}c=\dfrac{5c}{7}\) 

\(3a-7b+5c\) \(=-30\) 

⇔ \(3.\dfrac{3b}{2}-7b+5.\dfrac{5b}{7}=-30\)  

⇔\(63b-98b+50b=-420\) 

⇔\(b=-28\) ⇒\(\left\{{}\begin{matrix}a=-42\\c=-20\end{matrix}\right.\)

b) |x + 1| + |x + 2| + |x + 3| = 4x

Có: \(\left\{{}\begin{matrix}\left|x+1\right|>0\\\left|x+2\right|>0\\\left|x+3\right|>0\end{matrix}\right.\) \(\forall x\)

Do đó, \(4x>0=>x>0\).

Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)

=> \(3x+6=4x\)

=> \(4x-3x=6\)

=> \(1x=6\)

=> \(x=6:1\)

=> \(x=6\)

Vậy \(x=6\).

Chúc bạn học tốt!

\(\left(x-\frac{1}{2}\right)\left(y+\frac{1}{3}\right)\left(z-2\right)=0\) và \(x+2=y+3=z+4\)

\(\Rightarrow x-\frac{1}{2}=0\) hoặc \(y+\frac{1}{3}=0\) hoặc \(z-2=0\)

\(\Rightarrow x=\frac{1}{2}\)            |         \(y=-\frac{1}{3}\)     |       \(z=2\)

Khi \(x=\frac{1}{2}\) thì:

\(\frac{1}{2}+2=\frac{5}{2}\)

\(y=\frac{5}{2}-3=-\frac{1}{2}\)

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

Khi \(y=\frac{-1}{3}\)  thì:

\(\frac{-1}{3}+3=\frac{8}{3}\)

\(x=\frac{8}{3}-2=\frac{2}{3}\)

\(z=\frac{8}{3}-4=-\frac{4}{3}\)

Khi \(z=2\) thì:

\(2+4=6\)

\(x=6-2=4\)

\(y=6-3=3\)

Vậy (x,y,z) = \(\left(\frac{1}{2};-\frac{1}{2};-\frac{3}{2}\right)\)    ;    \(\left(\frac{2}{3};-\frac{1}{3};-\frac{4}{3}\right)\)  ;    \(\left(4;3;2\right)\)