\(\dfrac{x}{3}=\dfrac{y}{4},\dfrac{y}{5}\dfrac{z}{7}va2x+3y-z=124\left(timx,y,z\right)\)
GIUP MK NHA!!!
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}vax+y+z=49\left(timx,y,z\right)\)
Ta có:
\(\dfrac{2x}{3}=\dfrac{3y}{4}\Rightarrow y=\dfrac{4}{3}.\dfrac{2x}{3}=\dfrac{8x}{9}\)
\(\dfrac{2x}{3}=\dfrac{4z}{5}\Rightarrow z=\dfrac{5}{4}.\dfrac{2x}{3}=\dfrac{10x}{12}=\dfrac{5x}{6}\)
\(\Rightarrow x+y+z=x+\dfrac{8x}{9}+\dfrac{5x}{6}=49\)
Hay \(\left(18+16+15\right).\dfrac{x}{18}=49\).
tức là $x = 18 $
\(\Rightarrow y=16\)
và \(z=15\)
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{7}\) và 2x+3y-z =124
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}\)
nên \(\dfrac{x}{15}=\dfrac{y}{20}\)(1)
Ta có: \(\dfrac{y}{5}=\dfrac{z}{7}\)
nên \(\dfrac{y}{20}=\dfrac{z}{28}\)(2)
Từ (1) và (2) suy ra \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)
mà 2x+3y-z=124
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{124}{62}=2\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{15}=2\\\dfrac{y}{20}=2\\\dfrac{z}{28}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=30\\y=40\\z=56\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{x}{4}\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}\\ \dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-z}{2.15+3.20-28}=\dfrac{125}{62}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=15.2=30\\y=20.2=40\\z=28.2=56\end{matrix}\right.\)
a,\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{+6x}\)
b, \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\)
c,\(\dfrac{x}{z+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\left(x,y,zkhac0\right)\)
d, \(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}va2x^2+2y^2-z^2=1\)
a, 1+2y / 18 = 1+4y / 24 = 1+6y / 6x
Ta có : 1+2y / 18 = 1+6y / 6x = 1+2y + 1+6y / 18 + 6y
= 2+ 8y / 18+6y = 2 (1+4y) / 2( 9 +3y) = 1+4y/9+3y
Ta lại có : 1 + 4y/24 = 1+4y / 9+3y
=> 24=9+3y => 15=3y => y=5
Vậy y=5
Nhớ like
b, 1+3y/12 = 1+5y/5x = 1+7y/4x
Ta có : 1+3y/12 = 1+7y/4x = 1+3y+1+7y / 12 +4x
= 2 + 10y / 12 +4x = 2 (1+5y) / 2 (6+2x) = 1+5y / 6+2x
Ta lại có: 1+5y / 5x = 1+5y / 6+2x
=> 5x = 6+2x => 3x = 6 => x=2
Vậy x =2
Mình sửa lại câu a
1+2y/18 = 1+6y / 6x = 1+2y+1+6y / 18 + 6x = 2 +8y /18+6x
= 2 (1+4y) / 2 (9 +3x) = 1+4y / 9 +3x
Ta lại có: 1+4y/24 = 1+4y/ 9 +3x
=> 24 = 9 +3x => 15= 3x => x =5
Tìm x,y,z, bt
a, \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}\&x-3y+4z=62\)
b, \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\&2x+3y-5z=-21\)
c,\(\dfrac{x}{y}=\dfrac{3}{4},\dfrac{y}{z}=\dfrac{5}{7}\&2x+3y-z=186\)
d, \(2x=3y=5z\&\left|x+y-z\right|=95\)
a. Có \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}\) => \(\dfrac{x}{4}=\dfrac{3x}{9}=\dfrac{4z}{36}\) và x-3y+4z=62
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{4}=\dfrac{3y}{9}=\dfrac{4z}{36}\)= \(\dfrac{x-3y+4z}{4-9+36}=\dfrac{62}{31}=2\)
=> x=8
3y=18=>y=6
4z=72=>z=18
Vậy x=8 ; y=6 ; z=18
b, Ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\\ =\dfrac{2x+3y-5z}{4+9-20}=\dfrac{-21}{-7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=3\cdot2=6\\y=3\cdot3=9\\z=3\cdot4=12\end{matrix}\right.\\ vậy...\)
Câu c bạn làm tương tự nhé!
d, Ta có : \(\left|x+y-z\right|=95\Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\)
\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(2x=3y=5z=\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}=\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\\ =\dfrac{x+y-z}{15+10-6}=\dfrac{x+y-z}{19}\\ \Rightarrow\left[{}\begin{matrix}x+y-z=95\\x+y-z=-95\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=15\cdot5=75\\y=10\cdot5=50\\z=6\cdot5=30\end{matrix}\right.\\\left\{{}\begin{matrix}x=-5\cdot15=-75\\y=-5\cdot10=-50\\z=-5\cdot6=-30\end{matrix}\right.\end{matrix}\right.\)
Vậy...
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}va2x+3y-z=50\)
\(\dfrac{x-1}{2}\) hay \(x\) - \(\dfrac{1}{2}\) vậy em?
\(\dfrac{x-1}{2}\) = \(\dfrac{y-2}{3}\) = \(\dfrac{z-3}{4}\)
\(\dfrac{2.\left(x-1\right)}{2.2}\) = \(\dfrac{3.\left(y-2\right)}{3.3}\) = \(\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2.\left(x-1\right)}{4}\) = \(\dfrac{2x-2+3y-6-z+3}{4+9-4}\) = \(\dfrac{2x+3y-z-5}{9}\)=\(\dfrac{50-5}{9}\)= 5
⇒ \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)\) = (11;17;23)
cho x,y,z khac 0 va\(\dfrac{x+3y-z}{z}\)= \(\dfrac{y+3z-x}{x}\)=\(\dfrac{z+3x-y}{y}\)
Tính P = \(\left(\dfrac{x}{y}+3\right)\)\(\left(\dfrac{y}{z}+3\right)\)\(\left(\dfrac{z}{x}+3\right)\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{x+3y-z+y+3z-x+z+3x-y}{x+y+z}=\dfrac{3(x+y+z)-(x+y+z)}{x+y+z}=\dfrac{2(x+y+z)}{x+y+z}=2\)
\(\Rightarrow x=y=z=0\)
\(\Rightarrow \) P không xác định. (?)
a,\(\left|3x-4\right|+\left|3y+5\right|=0\) b,\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\) c,\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
d,\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\) e,\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\)
Giups mình giải bài tìm x,y,z này nhé!!! Cảm ơn nhiều ạ!!!
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
Tìm x,y,z biết :
1) \(x:y:z=3:5:\left(-2\right)\) và \(5x-y+3z=-16\)
2) \(\dfrac{x}{2}=\dfrac{y}{-3};\dfrac{z}{3}=\dfrac{y}{4}\) và \(x+y+z=5,2\)
3) \(2x=3y;7z=5y\) và \(3x-7y+5z=30\)
4) \(3x=4y=5z\) và \(x-\left(y+z\right)=-21\)
5) \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) và \(2x+3y-z=50\)
timx,y,z biết:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) va 2x + 3y -z = 50
áp dụng tính chất dảy tỉ số bằng nhau
ta có : \(\dfrac{2\left(x-1\right)+3\left(y-2\right)-\left(z-3\right)}{\left(2.2\right)+\left(3.3\right)-4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}\)
\(=\dfrac{\left(2x+3y-z\right)-5}{9}=\dfrac{50-5}{9}=\dfrac{45}{9}=5\)
suy ra ta có : \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=2.5\\y-2=3.5\\z-3=4.5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10+1\\y=15+2\\z=20+3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\) vậy \(x=11;y=17;z=23\)