\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7x}{x-6}+\dfrac{4}{y+10}=3\\\dfrac{-x}{x-6}+\dfrac{3y}{y+10}=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7x+42-42}{x-6}+\dfrac{4}{y+10}=3\\\dfrac{-x+6-6}{x-6}+\dfrac{3y+30-30}{y+10}=-11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{42}{x-6}-\dfrac{4}{y+10}=-10\\\dfrac{-6}{x-6}+\dfrac{-30}{y+10}=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{42}{x-6}-\dfrac{4}{y+10}=-10\\\dfrac{-42}{x-6}+\dfrac{-210}{y+10}=-70\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-214}{y+10}=-80\\\dfrac{-6}{x-6}+\dfrac{-30}{y+10}=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-293}{40}\\x=\dfrac{69}{65}\end{matrix}\right.\)

a/ 2x = 5y và x - 2y = -12

Ta có: 2x = 5y => \(\frac{x}{5}=\frac{y}{2}\)

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

\(\frac{x}{5}=\frac{y}{2}=\frac{x-y}{5+2}=\frac{x-2y}{5+2.2}=\frac{-12}{9}=-\frac{4}{3}\)

\(\frac{x}{5}=-\frac{4}{3}\Rightarrow x=\frac{-4}{3}.5=-\frac{20}{3}\)

\(\frac{y}{2}=-\frac{4}{3}\Rightarrow y=-\frac{4}{3}.2=-\frac{8}{3}\)

Vậy:.................

b/ 2x = 3y = 4z và x + y + z =21

Ta có: 2x = 3y = 4z

=> \(\frac{2x}{12}=\frac{3y}{12}=\frac{4z}{12}\)

=> \(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)

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

\(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{6+4+3}=\frac{21}{13}\)

\(\frac{x}{6}=\frac{21}{13}\Rightarrow x=\frac{21}{13}.6=\frac{126}{13}\)

\(\frac{y}{4}=\frac{21}{13}\Rightarrow y=\frac{21}{13}.4=\frac{84}{13}\)

\(\frac{z}{3}=\frac{21}{13}\Rightarrow z=\frac{21}{13}.3=\frac{63}{13}\)

Vậy:...............

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

\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{32}{8}=4\)

\(\frac{x}{3}=4\Rightarrow x=4.3=12\)

\(\frac{y}{5}=4\Rightarrow y=4.5=20\)

Vậy:................

d/ Ta có: 7x = 3y

=> \(\frac{7x}{21}=\frac{3y}{21}\)

=> \(\frac{x}{3}=\frac{y}{7}\)

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

\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\)

\(\frac{x}{4}=-4\Rightarrow x=\left(-4\right).4=-16\)

\(\frac{y}{7}=-4\Rightarrow y=\left(-4\right).7=-28\)

Vậy:................

1,\(2x=5y\Leftrightarrow\frac{x}{5}=\frac{y}{2}\Rightarrow\frac{x}{5}=\frac{2y}{4}\)

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

\(\frac{x}{5}=\frac{2y}{4}=\frac{x-2y}{5-4}=\frac{-12}{1}=-12\)

Do đó:

\(\frac{x}{5}=-12\Rightarrow x=-60\)

\(\frac{2y}{4}=-12\Leftrightarrow\frac{y}{2}=-12\Rightarrow x=-24\)

Vây x = -60,y = -24

2, 2x = 3y = 4z \(\Rightarrow BCNN\left(2;3;4\right)=12\)

nên \(\frac{2x}{12}=\frac{3y}{12}=\frac{4z}{12}\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)

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

\(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{6+4+3}=\frac{21}{13}\)

Do đó

\(\frac{x}{6}=\frac{21}{13}\Rightarrow x=\frac{6.21}{13}=\frac{126}{13}\)

\(\frac{y}{4}=\frac{21}{13}\Rightarrow y=\frac{4.21}{13}=\frac{84}{13}\)

\(\frac{z}{3}=\frac{21}{13}\Rightarrow z=\frac{3.21}{13}=\frac{63}{13}\)

f/ \(\frac{x}{3}=\frac{y}{5};\frac{y}{2}=\frac{z}{7}\)

=> \(\frac{x}{6}=\frac{y}{10};\frac{y}{10}=\frac{z}{35}\)

=> \(\frac{x}{6}=\frac{y}{10}=\frac{z}{35}\)

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

\(\frac{x}{6}=\frac{y}{10}=\frac{z}{35}=\frac{x+y+z}{6+10+35}=\frac{102}{51}=2\)

\(\frac{x}{6}=2\Rightarrow x=2.6=12\)

\(\frac{y}{10}=2\Rightarrow y=2.10=20\)

\(\frac{z}{35}=2\Rightarrow z=2.35=70\)

Vậy:.................

h/ Đăt: \(\frac{x}{3}=\frac{y}{16}=k\)

\(\frac{x}{3}=k\Rightarrow x=3k\)

\(\frac{y}{16}=k\Rightarrow y=16k\)

Ta có: x. y = 192

=> 3k. 16k = 192

=> k². (3. 16) = 192

=> k². 48 = 192

=> k² = 192 : 48 = 4

=> k = \(\pm\) 2

*Với k = 2

\(\frac{x}{3}=k\Rightarrow x=3.k=3.2=6\)

\(\frac{y}{16}=k\Rightarrow y=16.k=16.2=32\)

*Với k = -2

\(\frac{x}{3}=k\Rightarrow x=3.k=3.\left(-2\right)=-6\)

\(\frac{y}{16}=k\Rightarrow y=16.k=16.\left(-2\right)=-32\)

Vậy:..........

\(\left\{ \begin{array}{l} \dfrac{6}{{x + y}} + \dfrac{{11}}{{x - y}} = 21\\ \dfrac{6}{{x + y}} + \dfrac{5}{{x - y}} = 9 \end{array} \right.\)

Đặt \(\left\{ \begin{array}{l} t = \dfrac{1}{{x + y}}\\ r = \dfrac{1}{{x - y}} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 6t - 11r = 21\\ 6t + 5r = 9 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} t = \dfrac{{17}}{8}\\ r = - \dfrac{3}{4} \end{array} \right.\)

Với \(\left\{ \begin{array}{l} t = \dfrac{{17}}{8}\\ r = - \dfrac{3}{4} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \dfrac{1}{{x + y}} = \dfrac{{17}}{8}\\ \dfrac{1}{{x - y}} = - \dfrac{3}{4} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = - \dfrac{{22}}{{51}}\\ y = \dfrac{{46}}{{51}} \end{array} \right.\)

\(\left\{{}\begin{matrix}\frac{6}{x+y}+\frac{11}{x-y}=21\\\frac{6}{x+y}+\frac{5}{x-y}=9\end{matrix}\right.\) (*)

Đặt \(\frac{1}{x+y}\) là a; \(\frac{1}{x-y}\) là b.

Phương trình (*) trở thành:

\(\left\{{}\begin{matrix}6a+11b=21\\6a+5b=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6b=12\\6a+5b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=-\frac{1}{6}\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}\frac{1}{x+y}=-\frac{1}{6}\\\frac{1}{x-y}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{6}\left(x+y\right)=1\\6\left(x-y\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{6}x-\frac{1}{6}y=1\\6x-6y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{6}\left(\frac{1+6y}{6}\right)-\frac{1}{6}y=1\\x=\frac{1+6y}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\frac{37}{12}\\x=-\frac{35}{12}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\frac{6}{x+y}+\frac{11}{x-y}=21\\\frac{6}{x+y}+\frac{5}{x-y}=9\end{matrix}\right.\)

Đặt u = \(\frac{1}{x+y}\)

v = \(\frac{1}{x-y}\)

Ta có:

<=>\(\left\{{}\begin{matrix}6u+11v=21\\6u+5v=9\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}6v=12\\6u+5v=9\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}v=2\\6u+5.2=9\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}v=2\\6u=-1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}v=2\\u=-\frac{1}{6}\end{matrix}\right.\)

Câu 2/

Điều kiện xác định b tự làm nhé:

\(\frac{6}{x^2-9}+\frac{4}{x^2-11}-\frac{7}{x^2-8}-\frac{3}{x^2-12}=0\)

\(\Leftrightarrow x^4-25x^2+150=0\)

\(\Leftrightarrow\left(x^2-10\right)\left(x^2-15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=10\\x^2=15\end{cases}}\)

Tới đây b làm tiếp nhé.

a. ĐK: \(\frac{2x-1}{y+2}\ge0\)

Áp dụng bđt Cô-si ta có: \(\sqrt{\frac{y+2}{2x-1}}+\sqrt{\frac{2x-1}{y+2}}\ge2\)

\(\)Dấu bằng xảy ra khi  \(\frac{y+2}{2x-1}=1\Rightarrow y+2=2x-1\Rightarrow y=2x-3\) 

Kết hợp với pt (1) ta tìm được x = -1, y = -5 (tmđk)

b. \(pt\Leftrightarrow\left(\frac{6}{x^2-9}-1\right)+\left(\frac{4}{x^2-11}-1\right)-\left(\frac{7}{x^2-8}-1\right)-\left(\frac{3}{x^2-12}-1\right)=0\)

\(\Leftrightarrow\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}+\frac{1}{x^2-8}+\frac{1}{x^2-12}\right)=0\)

\(\Leftrightarrow x^2-15=0\Leftrightarrow\orbr{\begin{cases}x=\sqrt{15}\\x=-\sqrt{15}\end{cases}}\)

BALABOLO

TK NHA

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

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}=\dfrac{x-3y+4z}{4-3\cdot3+4\cdot9}=\dfrac{62}{31}=2\)

Do đó: x=8; y=6; z=18

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

\(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=\dfrac{2x+5y-2z}{2\cdot7+5\cdot20-2\cdot32}=\dfrac{100}{50}=2\)

Do đó: x=14; y=40; z=64

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

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

DO đó: x=-27; y=-21; z=-9

a/ \(\Leftrightarrow\left\{{}\begin{matrix}3x-4y=11\\-x-10y=-15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\)

b/ \(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\\frac{2x}{3}+\frac{x}{4}-\frac{y}{6}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\\frac{11}{12}x-\frac{y}{6}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\11x-2y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{28}{13}\\y=\frac{76}{13}\end{matrix}\right.\)