thế thì chúc bạn may mắn

trong sách có đấy               

a, 5x = 8y => \(\frac{x}{8}=\frac{y}{5}\)

8y = 20z => 2y = 5z => \(\frac{y}{5}=\frac{z}{2}\)

=> \(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}\)

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

\(\frac{x}{8}=\frac{y}{5}=\frac{z}{2}=\frac{x-y-z}{8-5-2}=\frac{3}{1}=3\)

=> x = 24,y = 15,z = 6

b, \(\frac{6}{11}x=\frac{9}{2}y\)=> \(\frac{12x}{22}=\frac{99y}{22}\)=> 12x = 99y => 4x = 33y => \(\frac{x}{33}=\frac{y}{4}\)

\(\frac{9}{2}y=\frac{18}{5}z\)=> \(\frac{45y}{10}=\frac{36z}{10}\)=> 45y = 36z => 5y = 4z => \(\frac{y}{4}=\frac{z}{5}\)

=> \(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)

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

\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{120}{-24}=-5\)

=> x = -165 , y = -20 , z = -25

c, Đặt : \(\frac{x}{12}=\frac{y}{9}=\frac{z}{5}=k\)=> x = 12k , y = 9k , z = 5k

=> xyz = 12k . 9k . 5k

=> xyz = 540k³

=> 540k³ =20

=> k³ = 20/540

=> k³ = 1/27

=> k = 1/3

Do đó : x= 4 , y = 3 , z = 5/3

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

\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{-24}{8}=-3\)

\(\frac{x}{3}=-3\Rightarrow x=\left(-3\right).3=-9\)

\(\frac{y}{5}=-3\Rightarrow y=\left(-3\right).5=-15\)

b) \(\frac{x}{5}=\frac{y}{8}=\frac{x-y}{5-8}=\frac{15}{-3}=-5\)

\(\frac{x}{5}=-5\Rightarrow x=\left(-5\right).5=-25\)

\(\frac{y}{8}=-5\Rightarrow y=\left(-5\right).8=-40\)

c) 7x=4y <=> x/4=y/7

\(\frac{x}{4}=\frac{y}{7}=\frac{x+y}{4+7}=\frac{12}{11}\)

\(\frac{x}{4}=\frac{12}{11}\Rightarrow x=\frac{12}{11}.4=\frac{48}{11}\)

\(\frac{y}{7}=\frac{12}{11}\Rightarrow y=\frac{12}{11}.7=\frac{84}{11}\)

d) tt câu c

e) x/5=y/8;z/3=y/12 <=> x/60=y/96=z/24

\(\frac{x}{60}=\frac{y}{96}=\frac{z}{24}=\frac{4x}{4.60}=\frac{2y}{2.96}=\frac{z}{24}=\frac{2y+z-4x}{192+24-240}=\frac{30}{-24}=\frac{-5}{4}\)

\(\frac{x}{60}=\frac{-5}{4}\) => x=-5/4.60=-75

y/96=-5/4 => y=-5/4.96=-120

z/24=-5/4 => z=-5/4.24=-30

chắc bạn đang học lớp 7 nên mik sẽ giải kiểu lớp 7 nha
mỗi câu mik chia làm 2 bài nhé!
Bài 1. Tìm \(\left(\right. x , y \left.\right) \in \mathbb{Q}^{2}\)

(a) \(x + 3 y - x \sqrt{5} = y \sqrt{5} + 7\)

\(\Rightarrow - \left(\right. x + y \left.\right) \sqrt{5} = 7 - x - 3 y\).

Vế trái vô tỉ (nếu \(x + y \neq 0\)), vế phải hữu tỉ.
\(\Rightarrow x + y = 0 , \textrm{ }\textrm{ } 7 - x - 3 y = 0\).

\(\Rightarrow x = - y , \textrm{ }\textrm{ } 7 + y - 3 y = 0 \Rightarrow y = \frac{7}{2} , x = - \frac{7}{2}\).

Đáp số: \(\left(\right. - \frac{7}{2} , \frac{7}{2} \left.\right)\).

(b) \(5 x + y - \left(\right. 2 x - 1 \left.\right) \sqrt{7} = y \sqrt{7} + 2\).

\(\Rightarrow - \left(\right. 2 x + y - 1 \left.\right) \sqrt{7} = 2 - 5 x - y\).

\(\Rightarrow 2 x + y - 1 = 0 , \textrm{ }\textrm{ } 2 - 5 x - y = 0\).

Giải hệ:

\(\left{\right. 2 x + y = 1 \\ 5 x + y = 2 \Rightarrow x = \frac{1}{3} , y = \frac{1}{3} .\)

Đáp số: \(\left(\right. \frac{1}{3} , \frac{1}{3} \left.\right)\).

Bài 2. Tìm \(\left(\right. x , y \left.\right) \in \mathbb{Q}^{2}\)

(a) \(x + y + 61 = 10 \sqrt{x} + 12 \sqrt{y}\).

Đặt \(x = a^{2} , y = b^{2}\).

\(\Rightarrow a^{2} + b^{2} + 61 = 10 a + 12 b\).

Thử \(a = 5 , b = 6\): \(25 + 36 + 61 = 122 , \textrm{ }\textrm{ } 10 \cdot 5 + 12 \cdot 6 = 122\).

Đáp số: \(\left(\right. 25 , 36 \left.\right)\).

(b) \(2 x + y + 4 = 2 \sqrt{x} \left(\right. \sqrt{y} + 2 \left.\right)\).

Đặt \(x = a^{2} , y = b^{2}\).

\(\Rightarrow 2 a^{2} + b^{2} + 4 = 2 a b + 4 a\).

\(\Rightarrow \left(\right. a - b \left.\right)^{2} + 2 \left(\right. a - 2 \left.\right) = 0\).

\(\Rightarrow a = 2 , b = 2\).

Đáp số: \(\left(\right. 4 , 4 \left.\right)\).

👉 Vậy:

• Bài 1(a): \(\left(\right. - 7 / 2 , 7 / 2 \left.\right)\).
• Bài 1(b): \(\left(\right. 1 / 3 , 1 / 3 \left.\right)\).
• Bài 2(a): \(\left(\right. 25 , 36 \left.\right)\).
• Bài 2(b): \(\left(\right. 4 , 4 \left.\right)\).
  cho mik xin tick nha. Cảm ơn cậu !

\(1)\)

\(VT=\left(\left|x-6\right|+\left|2022-x\right|\right)+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(\ge\left|x-6+2022-x\right|+\left|0\right|+\left|0\right|+\left|0\right|=2016\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-6\right)\left(2022-x\right)\ge0\left(1\right)\\x-10=y-2014=z-2015=0\left(2\right)\end{cases}}\)

\(\left(2\right)\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=10\\y=2014\\z=2015\end{cases}}\)

\(\left(1\right)\)

TH1 : \(\hept{\begin{cases}x-6\ge0\\2022-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge6\\x\le2022\end{cases}\Leftrightarrow}6\le x\le2022}\) ( nhận ) 

TH2 : \(\hept{\begin{cases}x-6\le0\\2022-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le6\\x\ge2022\end{cases}}}\) ( loại ) 

Vậy \(x=10\)\(;\)\(y=2014\) và \(z=2015\)

\(2)\)

\(VT=\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=\left|-4\right|=4\)

\(VP=\frac{12}{\left|y+1\right|+3}\le\frac{12}{3}=4\)

\(\Rightarrow\)\(VT\ge VP\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-5\right)\left(1-x\right)\ge0\left(1\right)\\\left|y+1\right|=0\left(2\right)\end{cases}}\)

\(\left(1\right)\)

TH1 : \(\hept{\begin{cases}x-5\ge0\\1-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge5\\x\le1\end{cases}}}\) ( loại ) 

TH2 : \(\hept{\begin{cases}x-5\le0\\1-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le5\\x\ge1\end{cases}\Leftrightarrow}1\le x\le5}\) ( nhận ) 

\(\left(2\right)\)\(\Leftrightarrow\)\(y=-1\)

Vậy \(1\le x\le5\) và \(y=-1\)

a ) 

Ta có : 

\(\frac{1+5y}{5x}=\frac{1+7y}{4x}\)

\(\Rightarrow\frac{4\left(1+5y\right)}{20x}=\frac{5\left(1+7y\right)}{20x}\)

\(\Rightarrow\frac{4+20y}{20x}=\frac{5+35y}{20x}\)

\(\Rightarrow4+20y=5+35y\)

\(\Rightarrow35y-20y=4-5\)

\(\Rightarrow15y=4-5\)

\(\Rightarrow15y=-1\)

\(\Rightarrow y=-\frac{1}{15}\)

Lại có : 

\(\frac{1+3y}{12}=\frac{1+5y}{5x}\)

\(\Rightarrow\frac{1+3.-\frac{1}{15}}{12}=\frac{1+5.-\frac{1}{15}}{5x}\)

\(\Rightarrow\frac{1-\frac{1}{5}}{12}=\frac{1-\frac{1}{3}}{5x}\)

\(\Rightarrow\frac{4}{5}:12=\frac{4}{3}:5x\)

\(\Rightarrow\frac{1}{15}=\frac{4}{3}:5x\)

\(\Rightarrow5x=\frac{4}{3}:\frac{1}{15}\)

\(\Rightarrow5x=20\)

\(\Rightarrow x=4\)

Vậy \(x=4;y=-\frac{1}{15}\)

a) Xét \(\frac{1+5y}{5x}=\frac{1+7y}{4x}\)

\(\Rightarrow\frac{4x\left(1+5y\right)}{20x}=\frac{5\left(1+7y\right)}{20x}\)

\(\Rightarrow4x\left(1+5y\right)=5\left(1+7y\right)\)

\(\Rightarrow4+20y=5+35y\)

\(\Rightarrow35y-20y=4-5\)

\(\Rightarrow15y=-1\)

\(\Rightarrow y=\frac{-1}{15}\)

Xét \(\frac{1+3y}{12}=\frac{1+5y}{5x}\)

\(\Rightarrow\frac{1+3.\frac{-1}{15}}{12}=\frac{1+5.\frac{-1}{15}}{5x}\)

\(\Rightarrow\frac{1+\frac{-1}{5}}{12}=\frac{1+\frac{-1}{3}}{5x}\)

\(\Rightarrow\frac{\frac{4}{5}}{12}=\frac{\frac{2}{3}}{5x}\)

\(\Rightarrow\frac{4}{5}:12=\frac{2}{3}:5x\)

\(\Rightarrow\frac{1}{15}=\frac{2}{3}:5x\)

\(\Rightarrow5x=\frac{2}{3}:\frac{1}{15}\)

\(\Rightarrow5x=\frac{30}{3}\)

\(\Rightarrow x=\frac{30}{3}:5\)

\(\Rightarrow x=\frac{30}{3}.\frac{1}{5}\)

\(\Rightarrow x=2\)

Vậy x = 2 ; y = \(\frac{-1}{15}\)