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#)Giải :
1)Áp dụng tính chất dãy tỉ số bằng nhau :
\(\frac{x}{2}=\frac{y}{1}=\frac{z}{3}=\frac{3x-2y+4z}{6-2+12}=\frac{16}{16}=1\Leftrightarrow\hept{\begin{cases}\frac{x}{2}=1\\\frac{y}{1}=1\\\frac{z}{3}=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\\z=3\end{cases}}}\)
Vậy x = 2; y = 1; z = 3
2)Áp dụng tính chất dãy tỉ số bằng nhau :
\(\frac{x}{1}=\frac{y}{6}=\frac{z}{3}=\frac{2x-3y+4z}{2-18+12}=\frac{-24}{-4}=6\Leftrightarrow\hept{\begin{cases}\frac{x}{1}=6\\\frac{y}{6}=6\\\frac{z}{3}=6\end{cases}\Leftrightarrow\hept{\begin{cases}x=6\\y=36\\z=18\end{cases}}}\)
Vậy x = 6; y = 36; z = 18
3)Áp dụng tính chất dãy tỉ số bằng nhau :
\(\frac{x}{0,5}=\frac{y}{0,3}=\frac{x-y}{0,5-0,3}=\frac{1}{0,2}=5\Leftrightarrow\hept{\begin{cases}\frac{x}{0,5}=5\\\frac{y}{0,3}=5\\\frac{z}{0,2}=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=2,5\\y=1,5\\z=1\end{cases}}}\)
Vậy x = 2,5; y = 1,5; z = 1
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\(a,\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\) và 2x + 3y - z = 124
Ta có : \(\frac{x}{3}=\frac{y}{4}\)=> \(\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{7}\)=> \(\frac{y}{20}=\frac{z}{28}\)
=> \(\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\)
=> \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
=> \(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{124}{62}=2\)
Đến đây là tìm x,y,z rồi
b. Ta có : \(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=x+y+z(1)\)
Áp dụng tính chất bằng nhau của tỉ lệ thức ta được :
\(\frac{x+y+z}{2\left[x+y+z\right]}=x+y+z(2)\)
Nếu x + y + z = 0 thì từ \((1)\)suy ra x = 0 , y = 0 , z = 0
Nếu x + y + z \(\ne\)0 thì từ \((2)\)ta suy ra : \(\frac{1}{2}=x+y+z\), khi đó \((1)\)trở thành :
\(\frac{x}{\frac{1}{2}-x+1}=\frac{y}{\frac{1}{2}-y+1}=\frac{z}{\frac{1}{2}-z-2}=\frac{1}{2}\)
Do đó : 2x = \(\frac{3}{2}-x\)=> \(x=\frac{1}{2}\); 2y = \(\frac{3}{2}-y\)=> \(y=\frac{1}{2}\); 2z = \(-\frac{3}{2}-z\)=> \(z=-\frac{1}{2}\)
Vậy có hai đáp số \((0,0,0)\)và \((\frac{1}{2};\frac{1}{2};-\frac{1}{2})\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo t/c dãy tỉ số bằng nhau :
\(\frac{x}{5}=\frac{y}{3}=\frac{z}{2}=\frac{2x+3y+4z}{10+9+8}=\frac{54}{27}=2\)
\(\Rightarrow x=2.5=10\)
\(y=2.3=6\)
\(z=2.2=4\)
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\(\dfrac{x}{-3}=\dfrac{y}{5}\)⇒\(\dfrac{x}{-6}=\dfrac{y}{10}\)
\(\dfrac{y}{2}=\dfrac{z}{7}\)⇒\(\dfrac{y}{10}=\dfrac{z}{35}\)
⇒\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)
⇒\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\)
⇒\(\left\{{}\begin{matrix}x=-6.-6=36\\y=-6.10=-60\\z=-6.35=-210\end{matrix}\right.\)
\(a,\dfrac{x}{-3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{-6}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{7}\Rightarrow\dfrac{y}{10}=\dfrac{z}{35}\\ \Rightarrow\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}=\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\\ \Rightarrow\left\{{}\begin{matrix}x=36\\y=-60\\z=-210\end{matrix}\right.\)
\(b,6x=4y=z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y+z}{4-9+12}=\dfrac{42}{7}=6\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=18\\z=72\end{matrix}\right.\)
\(c,x=-2y\Rightarrow\dfrac{x}{-2}=y\Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}\\ 7y=2z\Rightarrow\dfrac{y}{2}=\dfrac{z}{7}\\ \Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}=\dfrac{2x}{-8}=\dfrac{3y}{6}=\dfrac{2x-3y+z}{-8+6+7}=\dfrac{42}{5}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{168}{5}\\y=\dfrac{84}{5}\\z=\dfrac{294}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
+ Nếu x = 6
\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)
+ Nếu x = 4
\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)
Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)
b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{4}{3}\)
Thay vào ta được:
\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)
\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)
\(\Leftrightarrow14y=-4\)
\(\Rightarrow y=-\frac{2}{7}\)
Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn bảo bạn làm câu a r nên mik thôi còn câu b là:
ta có
x-1/2 = y-2/3 = z-3/4 = 2x-2/4 = z-3/a
áp dụng t/c của dãy tỉ số = nhau, ta có:
2x-2+3y-6-z+3 / 4+9-4 = 2x+3y-z-5 / 9 = 50-5 / 9 =45 / 5 = 5
=>
x-1 / 2 = 5=>x-1=10 => x=11
y-2 / 3 = 5 => y-2 = 15 => y = 17
z-3 / 4 = 5=> z-3 = 20 =>z =23
tick nha bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)
2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)
3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)
Áp dụng t/c dtsbn:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)