Bài 2: 

Ta có: \(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)

\(\Leftrightarrow\left(\dfrac{x-1}{65}-1\right)+\left(\dfrac{x-3}{63}-1\right)=\left(\dfrac{x-5}{61}-1\right)+\left(\dfrac{x-7}{59}-1\right)\)

\(\Leftrightarrow\left(x-66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)

=>x-66=0

hay x=66

a) Áp dụng tc

\(\frac{x}{3}\) = \(\frac{y}{4}\) = \(\frac{x+y}{3+4}\) = \(\frac{28}{7}\) = 4

Do \(\left[\begin{matrix}\frac{x}{3}=4\\\frac{y}{4}=4\end{matrix}\right.\)

=> \(\left[\begin{matrix}x=3.4\\y=4.4\end{matrix}\right.\) => \(\left[\begin{matrix}x=12\\y=16\end{matrix}\right.\)

Vậy x = 12 và y = 16.

b) Áp dụng tc dãy tỉ số bằng nhau ta có:

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

Do \(\left[\begin{matrix}\frac{x}{2}=-1\\\frac{y}{-5}=-1\end{matrix}\right.\)

=> \(\left[\begin{matrix}x=-1.2\\y=-5.\left(-1\right)\end{matrix}\right.\) => \(\left[\begin{matrix}x=-2\\y=5\end{matrix}\right.\)

Vậy x = -2 và y = 5.

c) Ta có: \(\frac{x}{2}\) = \(\frac{y}{3}\) => \(\frac{x}{8}\) = \(\frac{y}{12}\) (1)

\(\frac{y}{4}\) = \(\frac{z}{5}\) => \(\frac{y}{12}\) = \(\frac{z}{15}\) (2)

Từ (1) và (2) suy ra \(\frac{x}{8}\) = \(\frac{y}{12}\) = \(\frac{z}{15}\)

Áp dụng tc dãy tỉ số bằng nhau ta có:

\(\frac{x}{8}\) = \(\frac{y}{12}\) = \(\frac{z}{15}\) = \(\frac{x+y-z}{8+12-15}\) = \(\frac{10}{5}\) = 2

Do \(\left[\begin{matrix}\frac{x}{8}=2\\\frac{y}{12}=2\\\frac{z}{15}=2\end{matrix}\right.\)

=> \(\left[\begin{matrix}x=8.2\\y=12.2\\z=15.2\end{matrix}\right.\) => \(\left[\begin{matrix}x=16\\y=24\\z=30\end{matrix}\right.\)

Vậy x = 16; y = 24 và z = 30.

1.

Ta có: \(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}.\)

=> \(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{25}.\)

=> \(\frac{2x^2}{8}=\frac{2y^2}{32}=\frac{3z^2}{75}\) và \(2x^2+2y^2-3z^2=-100.\)

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

\(\frac{2x^2}{8}=\frac{2y^2}{32}=\frac{3z^2}{75}=\frac{2x^2+2y^2-3z^2}{8+32-75}=\frac{-100}{-35}=\frac{20}{7}.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x^2}{4}=\frac{20}{7}\Rightarrow x^2=\frac{80}{7}\Rightarrow\left[{}\begin{matrix}x=\sqrt{\frac{80}{7}}\\x=-\sqrt{\frac{80}{7}}\end{matrix}\right.\\\frac{y^2}{16}=\frac{20}{7}\Rightarrow y^2=\frac{320}{7}\Rightarrow\left[{}\begin{matrix}y=\sqrt{\frac{320}{7}}\\y=-\sqrt{\frac{320}{7}}\end{matrix}\right.\\\frac{z^2}{25}=\frac{20}{7}\Rightarrow z^2=\frac{500}{7}\Rightarrow\left[{}\begin{matrix}z=\sqrt{\frac{500}{7}}\\z=-\sqrt{\frac{500}{7}}\end{matrix}\right.\end{matrix}\right.\)

Vậy.......

Chúc bạn học tốt!

1,

\(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{25}\)

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

\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{25}=\frac{2x^2+2y^2-3z^2}{2\cdot4+2\cdot16-3\cdot25}=\frac{-100}{-35}=\frac{20}{7}\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x^2}{4}=\frac{20}{7}\\\frac{y^2}{16}=\frac{20}{7}\\\frac{z^2}{25}=\frac{20}{7}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=\frac{20}{7}\cdot4=\frac{80}{7}\\y^2=\frac{20}{7}\cdot16=\frac{320}{7}\\z^2=\frac{20}{7}\cdot25=\frac{500}{7}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\frac{4\sqrt{35}}{7}\\x=\frac{-4\sqrt{35}}{7}\end{matrix}\right.\\\left[{}\begin{matrix}y=\frac{8\sqrt{35}}{7}\\y=\frac{-8\sqrt{35}}{7}\end{matrix}\right.\\\left[{}\begin{matrix}z=\frac{10\sqrt{35}}{7}\\z=\frac{-10\sqrt{35}}{7}\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)\in\left\{\left(\frac{4\sqrt{35}}{7};\frac{8\sqrt{35}}{7};\frac{10\sqrt{35}}{7}\right);\left(\frac{-4\sqrt{35}}{7};\frac{-8\sqrt{35}}{7};\frac{-10\sqrt{35}}{7}\right)\right\}\)

2,

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

\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{9}\\ \Rightarrow\frac{x^3}{64}=\frac{y^3}{216}=\frac{z^3}{729}\)

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

\(\frac{x^3}{64}=\frac{y^3}{216}=\frac{z^3}{729}=\frac{x^3+y^3+z^3}{64+216+729}=\frac{-1009}{1009}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x^3}{64}=-1\\\frac{y^3}{216}=-1\\\frac{z^3}{729}=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^3=-64\\y^3=-216\\z^3=-729\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-4\\y=-6\\z=-9\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(-4;-6;-9\right)\)

a, \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}vàx+y-10=26\)

=> x + y - 10 = 26

x + y = 26 + 10 = 36

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

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y}{3+4}=\dfrac{36}{7}\)

Suy ra:

x = \(\dfrac{36.3}{7}=\dfrac{108}{7}\)

\(y=\dfrac{36.4}{7}=\dfrac{144}{7}\)

\(z=\dfrac{36.5}{7}=\dfrac{180}{7}\)

Vậy\(x=\dfrac{108}{7},y=\dfrac{144}{7},z=\dfrac{180}{7}.\)

a) đặt \(\dfrac{3}{7x}=\dfrac{8}{13y}=\dfrac{6}{19z}=k\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{7k}\\y=\dfrac{8}{13k}\\z=\dfrac{6}{19k}\end{matrix}\right.\)

Thay vào 2x -y-z=-6, ta được:

\(2\cdot\dfrac{3}{7k}-\dfrac{8}{13k}-\dfrac{6}{19k}=-6\Leftrightarrow\left(\dfrac{6}{7}-\dfrac{8}{13}-\dfrac{6}{19}\right)\cdot\dfrac{1}{k}=-6\Leftrightarrow\dfrac{1}{k}=\dfrac{5187}{64}\Leftrightarrow k=\dfrac{64}{5187}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{7k}=\dfrac{2223}{64}\\y=\dfrac{8}{13k}=\dfrac{399}{8}\\z=\dfrac{6}{19k}=\dfrac{819}{32}\end{matrix}\right.\)

Vậy.............

{số vẫn không đẹp mấy nhỉ T_T!!!}

\(\dfrac{3}{7}.x=\dfrac{8}{13}y=\dfrac{6}{19}z\)

\(\Rightarrow\)\(\dfrac{x}{\dfrac{7}{3}}=\dfrac{y}{\dfrac{13}{8}}=\dfrac{z}{\dfrac{19}{6}}\Rightarrow.\dfrac{2x}{\dfrac{14}{3}}=\dfrac{y}{\dfrac{13}{8}}=\dfrac{z}{\dfrac{19}{6}}\)

AD tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{2x}{\dfrac{14}{3}}=\dfrac{y}{\dfrac{13}{8}}=\dfrac{z}{\dfrac{19}{6}}=\dfrac{2x-y-z}{\dfrac{14}{3}-\dfrac{13}{8}-\dfrac{19}{6}}=\dfrac{-6}{\dfrac{-3}{24}}=48\)

\(\Rightarrow\)x=112;y=78;z=152

a) không hiểu, khoảng trắng đó là ám chỉ cái gì???: x,y,z ở tử hay mẫu?

b) mấy bài rối rối thế này cứ đặt k cho an toàn

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

thay vào \(x^2-3y^2+2z^2=325\), được: ​

\(\left(4k\right)^2-3\left(\dfrac{5k}{2}\right)^2+2\left(2k\right)^2=325\Leftrightarrow16k^2-\dfrac{75}{4}k^2+8k^2=325\Leftrightarrow\dfrac{21}{4}k^2=325\Leftrightarrow k=\pm\dfrac{10\sqrt{273}}{21}\)

{sao số xấu vậy nhỉ?}

\(\Rightarrow\left\{{}\begin{matrix}x=4k=\pm\dfrac{40\sqrt{273}}{21}\\y=\dfrac{5k}{2}=\pm\dfrac{25\sqrt{273}}{21}\\z=2k=\dfrac{20\sqrt{273}}{21}\end{matrix}\right.\)

(bạn xem tớ có làm đúng đề của bạn không??)

THANK YOU!!

Ta có: x : y : z : t = 15 : 7 : 3 : 1

⇒ \(\dfrac{x}{15}\) = \(\dfrac{y}{7}\) = \(\dfrac{z}{3}\) = \(\dfrac{t}{1}\)

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{x}{15}\) = \(\dfrac{y}{7}\) = \(\dfrac{z}{3}\) = \(\dfrac{t}{1}\) = \(\dfrac{x-y+z-t}{15-7+3-1}\) = \(\dfrac{10}{10}\) = 1

⇒ \(\dfrac{x}{15}\) = 1 ⇒ x = 15

\(\dfrac{y}{7}\) = 1 ⇒ y = 7

\(\dfrac{z}{3}\) = 1 ⇒ z = 3

\(\dfrac{t}{1}\) = 1 ⇒ t = 1

Vậy x = 15 ; y = 7 ; z = 3 ; t = 1

Chúc bạn An Lê Khánh học tốt!