`#3107.101117`

a)

`x \div y \div z = 4 \div 3 \div 9`

`=> x/4 = y/3 = z/9`

`=> x/4 = (3y)/9 = (4z)/36`

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

`x/4 = (3y)/9 = (2z)/8 = (x - 3y + 4z)/(4 - 9 + 36) = 62/31 = 2`

`=> x/4 = y/3 = z/9 = 2`

`=> x = 4*2 = 8` $\\$ `y = 3*2 = 6` $\\$ `z = 9*2 = 18`

Vậy, `x = 8; y = 6; z = 18`

c)

\(x \div y \div z = 1 \div 2 \div 3\)

`=> x/1 = y/2 = z/3`

`=> (4x)/4 = (3y)/6 = (2z)/6`

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

`(4x)/4 = (3y)/6 = (2z)/6 = (4x - 3y + 2z)/(4 - 6 + 6) = 36/4 = 9`

`=> x/1 = y/2 = z/3 = 9`

`=> x = 1*9=9` $\\$ `y = 2*9 = 18` $\\$ `z = 3*9 = 27`

Vậy, `x = 9; y = 18; z = 27`

Các câu còn lại cậu làm tương tự nhé.

thôi sai rồi

người ta cho có x với z mà

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

=>\(\left\{{}\begin{matrix}\dfrac{4x-3y}{5}=\dfrac{5y-4z}{3}\\\dfrac{4x-3y}{5}=\dfrac{3z-5x}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3\left(4x-3y\right)=5\left(5y-4z\right)\\4\left(4x-3y\right)=5\left(3z-5x\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x-9y-25y+20z=0\\16x-12y-15z+25x=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x-34y+20z=0\\41x-12y-15z=0\end{matrix}\right.\)

mà x-y+z=200 nên ta có hệ phương trình:

\(\left\{{}\begin{matrix}12x-34y+20z=0\\41x-12y-15z=0\\x-y+z=200\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}36x-102y+60z=0\\164x-48y-60z=0\\60x-60y+60z=12000\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}200x-150y=0\\-24x-42y=-12000\\x-y+z=200\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-3y=0\\4x+7y=2000\\x-y+z=200\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-10y=-2000\\4x-3y=0\\x-y+z=200\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=200\\4x=3y\\x-y+z=200\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=200\\x=\dfrac{3}{4}y=150\\150-200+z=200\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=200\\x=150\\z=250\end{matrix}\right.\)

\(\dfrac{4x-3y}{5}=\dfrac{5y-4z}{6}=\dfrac{-4x-2y+4z}{7}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4x-3y}{5}=0\Leftrightarrow x=\dfrac{3y}{4}\\\dfrac{5y-4z}{6}=0\Leftrightarrow z=\dfrac{5y}{4}\end{matrix}\right.\)

Ta có \(x+y+z=36\)

\(\Leftrightarrow\dfrac{3y}{4}+y+\dfrac{5y}{4}=36\)

\(\Rightarrow y=12\)

\(\Rightarrow\left\{{}\begin{matrix}x=9\\z=15\end{matrix}\right.\)

Bạn ơi chỗ này có thể giải thích cho mình được hôngtheer

+) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)

=> \(\hept{\begin{cases}\frac{x^2}{9}=4\\\frac{y^2}{16}=4\end{cases}}\) => \(\hept{\begin{cases}x^2=4.9=36\\y^2=4.16=64\end{cases}}\) => \(\hept{\begin{cases}x=\pm6\\y=\pm8\end{cases}}\)

Vậy ...

1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)

⇒ x=4;y=6;z=8

\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)

\(2,\) Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)

\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)

\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)

b) Ta có: \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}.\)

\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{15}=\frac{z}{21}.\)

=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\) và \(x+y+z=92.\)

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

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2.\)

\(\left\{{}\begin{matrix}\frac{x}{10}=2\Rightarrow x=2.10=20\\\frac{y}{15}=2\Rightarrow y=2.15=30\\\frac{z}{21}=2\Rightarrow z=2.21=42\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(20;30;42\right).\)

c) Ta có: \(2x=3y=5z.\)

=> \(\frac{x}{3}=\frac{y}{5}=\frac{z}{2}\) và \(x+y-z=95.\)

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

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{2}=\frac{x+y-z}{3+5-2}=\frac{95}{6}.\)

\(\left\{{}\begin{matrix}\frac{x}{3}=\frac{95}{6}\Rightarrow x=\frac{95}{6}.3=\frac{95}{2}\\\frac{y}{5}=\frac{95}{6}\Rightarrow y=\frac{95}{6}.5=\frac{475}{6}\\\frac{z}{2}=\frac{95}{6}\Rightarrow z=\frac{95}{6}.2=\frac{95}{3}\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(\frac{95}{2};\frac{475}{6};\frac{95}{3}\right).\)

Chúc bạn học tốt!

a) \(\frac{x}{2}=\frac{y}{5}=\frac{z}{4}\)

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

\(\frac{x}{2}=\frac{y}{5}=\frac{z}{4}=\frac{2x}{4}=\frac{3y}{15}=\frac{z}{4}=\frac{2x-3y+z}{4-15+4}=\frac{112}{7}=16\)

\(\frac{x}{2}=16=>x=32\)

\(\frac{y}{5}=16=>x=80\)

\(\frac{z}{4}=16=>z=64\)

Câu b) tương tự chỉ cần thay số vào nha bạn