\(2x=3y\text{⇒}\dfrac{x}{3}=\dfrac{y}{2}\text{⇒}\dfrac{x}{21}=\dfrac{y}{14}\)

\(5y=7z\text{⇒}\dfrac{y}{7}=\dfrac{z}{5}\text{⇒}\dfrac{y}{14}=\dfrac{z}{10}\)

⇒\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)⇒\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\)

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

\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)

⇒x=42,y=28,z=20

\(\dfrac{x}{3}=\dfrac{y}{2}\)⇒\(\dfrac{x}{15}=\dfrac{y}{10}\)

\(\dfrac{x}{5}=\dfrac{z}{7}\text{⇒}\dfrac{x}{15}=\dfrac{z}{21}\)

⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{21}\)⇒\(\dfrac{x}{15}=\dfrac{2y}{20}\)

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

\(\dfrac{x}{15}=\dfrac{2y}{20}=\dfrac{x+2y}{15+20}=\dfrac{-112}{35}=\dfrac{-16}{5}\)

⇒x=48,y=32,z=336/5

Lời giải:

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

$\frac{x}{3}=\frac{y}{2}=\frac{2y}{4}=\frac{x+2y}{3+4}=\frac{-112}{7}=-16$

$\Rightarrow x=-16.3=-48; y=-16.2=-32$

Đoạn $\frac{x}{5}=\frac{x}{7}$ là sao em? Em xem lại đề.

2. 

$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}\Rightarrow \frac{x}{21}=\frac{y}{14}(1)$

$5y=7z\Rightarrow \frac{y}{7}=\frac{z}{5}\Rightarrow \frac{y}{14}=\frac{z}{10}(2)$

Từ $(1);(2)\Rightarrow \frac{x}{21}=\frac{y}{14}=\frac{z}{10}$

Áp dụng tính chất dãy tỷ số bằng nhau:

$\frac{x}{21}=\frac{y}{14}=\frac{z}{10}$

$=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2$

$\Rightarrow x=2.21=42; y=2.14=28; z=2.10=20$

a: \(A=-4x^5y^3-2x^2y^3z^2-2y^4\)

b: \(B=-4x^5y^3-2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3=-4x^5y^3+\dfrac{1}{5}x^4y^3-\dfrac{8}{3}y^4\)

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}< =>\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{15}\)

áp dụng tính chất dãy tỉ số = nhau

\(=>\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{3-8+15}=\dfrac{35}{10}=3,5\)

\(=>\dfrac{x}{3}=3.5=>x=10,5\)

\(\dfrac{2y}{8}=3,5=>y=14\)

\(\dfrac{3z}{15}=3,5=>z=17,5\)

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{3-8+15}=\dfrac{35}{10}\)

- \(\dfrac{35}{10}=\dfrac{x}{3}\Rightarrow x=\dfrac{21}{2}\)

- \(\dfrac{35}{10}=\dfrac{y}{4}\Rightarrow y=14\)

-\(\dfrac{35}{10}=\dfrac{z}{5}\Rightarrow z=\dfrac{35}{2}\)

Tick cho mình với.

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

nên \(\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{15}\)

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

\(\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{3-8+15}=\dfrac{35}{10}=\dfrac{7}{2}\)

Do đó: \(\left\{{}\begin{matrix}x=\dfrac{21}{2}\\y=14\\z=\dfrac{35}{2}\end{matrix}\right.\)

\(VT=\dfrac{2y+3z+5}{1+x}+1+\dfrac{3z+x+5}{2y+1}+1+\dfrac{x+2y+5}{1+3z}+1-3\)

\(VT=\dfrac{x+2y+3z+6}{1+x}+\dfrac{x+2y+3z+6}{1+2y}+\dfrac{x+2y+3z+6}{1+3z}-3\)

\(VT=24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)-3\ge\dfrac{24.9}{1+x+1+2y+1+3z}-3=\dfrac{216}{21}-3=\dfrac{51}{7}\)

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

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\\\Rightarrow x=-3.3=-9\\ \Rightarrow y=-3.5=-15\\ \Rightarrow z=-3.7=-21 \)

a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x}{9}=\dfrac{2z}{14}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\)  (Vì 3x-2z=15)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-3\\\dfrac{y}{5}=-3\\\dfrac{z}{7}=-3\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-21\end{matrix}\right.\)

Vậy ...
 

b) Ta có: \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x}{10}=\dfrac{3y}{9}=\dfrac{2x-3y}{10-9}=\dfrac{100}{1}=100\) (Vì 2x-3y=100)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=100\\\dfrac{y}{3}=100\\\dfrac{z}{2}=100\end{matrix}\right.\)    \(\Rightarrow\left\{{}\begin{matrix}x=500\\y=300\\z=200\end{matrix}\right.\)

Vậy ...

c) Ta có: \(\dfrac{x}{-3}=\dfrac{y}{-5}=\dfrac{z}{-4}=\dfrac{3z}{-12}=\dfrac{2x}{-6}=\dfrac{3z-2x}{\left(-12\right)-\left(-6\right)}=\dfrac{36}{-18}=-2\)                                                         (Vì 3z-2x=36)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-3}=-2\\\dfrac{y}{-5}=-2\\\dfrac{z}{-4}=-2\end{matrix}\right.\)     \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\\z=8\end{matrix}\right.\)

Vậy ... 

d: x/2=y/1=z/3

mà 3x+4z=16+2=18

nên Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{2}=\dfrac{y}{1}=\dfrac{z}{3}=\dfrac{3x+4z}{3\cdot2+4\cdot3}=\dfrac{18}{18}=1\)

=>x=2; y=1; z=3

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

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x-2z}{3\cdot3-2\cdot7}=\dfrac{15}{-5}=-3\)

=>x=-9; y=-15; z=-21

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

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x-3y}{2\cdot5-3\cdot3}=\dfrac{100}{1}=100\)

=>x=500; y=300; z=200

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

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{4}=\dfrac{3z-2x}{3\cdot4-2\cdot3}=\dfrac{36}{6}=6\)

=>x=18; y=30; z=24

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

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

\(=\dfrac{2y^2+8y+12}{y-1}\)

\(VT+3=\left(x+2y+3z+6\right)\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)\)

= \(24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)\)

Áp dụng BĐT cauchy-schwarz:

\(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\ge\dfrac{9}{3+x+2y+3z}=\dfrac{9}{21}\)

\(\Rightarrow VT\ge\dfrac{24.9}{21}-3=\dfrac{51}{7}\)

dấu = xảy ra khi x=2y=3z=6 hay x=6,y=3,z=2

cộng 3 vào rồi b-c-s