Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y-z}{6+5-3}=\dfrac{54}{8}=\dfrac{27}{4}\)
+\(\dfrac{x}{6}=\dfrac{27}{4}\Rightarrow x=\dfrac{27.6}{4}=\dfrac{81}{2}\)
+\(\dfrac{y}{5}=\dfrac{27}{4}\Rightarrow y=\dfrac{27.5}{4}=\dfrac{135}{4}\)
+\(\dfrac{z}{3}=\dfrac{27}{4}\Rightarrow z=\dfrac{27.3}{4}=\dfrac{81}{4}\)
Vậy \(x=\dfrac{81}{2};y=\dfrac{135}{4};z=\dfrac{81}{4}\)
2,Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}=\dfrac{x+2y-3c}{2+2.3+3.4}=\dfrac{-20}{20}=-1\)
+\(\dfrac{x}{2}=-1\Rightarrow x=-1.2=-2\)
+\(\dfrac{y}{3}=-1\Rightarrow y=-1.3=-3\)
+\(\dfrac{c}{4}=-1\Rightarrow c=-1.4=-4\)
Vậy \(x=-2;y=-3;c=-4\)
a) Ta có: \(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\)
\(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{20}=\dfrac{z}{32}\)
\(\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)
\(\Rightarrow\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}\)
Áp dụng tc dãy tỉ số bằng nhau:
\(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}=\dfrac{2x+5y-2z}{14+100-64}=2\)
Do \(\left\{{}\begin{matrix}\dfrac{2x}{14}=2\\\dfrac{5y}{100}=2\\\dfrac{2z}{64}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=14\\y=40\\z=64\end{matrix}\right.\).
b) \(5x=8y=20z\Rightarrow\dfrac{5x}{40}=\dfrac{8y}{40}=\dfrac{20z}{40}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}\)
Áp dụng...
\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x-y-z}{8-5-2}=3\)
....
c) \(\dfrac{6}{11}x=\dfrac{9}{2}y=\dfrac{18}{5}z\Rightarrow\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}\)
...
b: 2x^3-1=15
=>2x^3=16
=>x=2
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
=>y-25=32; z+9=50
=>y=57; z=41
d: 3/5x=2/3y
=>9x=10y
=>x/10=y/9=k
=>x=10k; y=9k
x^2-y^2=38
=>100k^2-81k^2=38
=>19k^2=38
=>k^2=2
TH1: k=căn 2
=>\(x=10\sqrt{2};y=9\sqrt{2}\)
TH2: k=-căn 2
=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)
\(a,\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{x}{7}\) và \(x+y+z=138\)
\(\dfrac{x}{5}=\dfrac{y}{6}\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}\) \(\left(1\right)\)
\(\dfrac{y}{8}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{24}=\dfrac{z}{21}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y+z}{20+24+21}=\dfrac{138}{65}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{138}{65}\\\dfrac{y}{24}=\dfrac{138}{65}\\\dfrac{z}{21}=\dfrac{138}{65}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{553}{13}\\y=\dfrac{3312}{65}\\z=\dfrac{2898}{65}\end{matrix}\right.\)
Vậy.......
Câu 1: Mình chỉnh sửa lại đầu bài của bạn nha. Không biết có đúng không. Nếu để đầu bài như bạn thì mình không làm ra được. Mog góp ý !!!!
Áp dụng t/c DTSBN ta có:
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\)
\(=\dfrac{x+y+x}{y+z+1+x+z+1+x+y-2}=\dfrac{x+y+x}{2x+2y+2z}=\dfrac{1}{2}\)
=>\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\left(1\right)\)
=>\(\dfrac{y}{x+z+1}=\dfrac{1}{2}\left(2\right)\)
=>\(\dfrac{z}{x+y-2}=\dfrac{1}{2}\left(3\right)\)
=> x+y+z = 1/2 (4)
Ta có : Từ (1) => 2x = y+z+1 kết hợp (4)
=> 2x = 1/2-x+1
=> 3x = 3/2 => x=1/2
Ta có: Từ (2) => 2y = x+z+1
=> 2y + y = x+y+z+1
=> 3y = 1/2+1 (theo 4) => 3y=3/2
=> y=1/2
Ta có : Từ (4) => x+y+z=1/2
=>1/2 + 1/2 +z = 1/2
=> z=-1/2
Vậy ( x;y;z)=(1/2;1/2;-1/2)
a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)
\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)
\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)
Đến đây tự làm tiếp nhé
b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)
=> x = 75, y = 50, z = 30
c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)
\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)
\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)
\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)
=> x=... , y=... , z=...
d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)
Ta có: xy = 90 => 2k.5k = 90 => 10k2 = 90 => k2 = 9 => k = 3 hoặc -3
Với k = 3 => x = 6, y = 15
Với k = -3 => x = -6, y = -15
Vậy...
e, Tương tự câu d
b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)
=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)
\(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)
\(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)
\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
a)\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{12}\Leftrightarrow\dfrac{-x}{-8}=\dfrac{y}{5}=\dfrac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{-x}{-8}=\dfrac{y}{5}=\dfrac{z}{12}=\dfrac{-x+y+z}{-8+5+12}=\dfrac{60}{9}=\dfrac{20}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}.8=\dfrac{160}{3}\\y=\dfrac{20}{3}.5=\dfrac{100}{3}\\z=\dfrac{20}{3}.12=80\end{matrix}\right.\)
b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Leftrightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-20}{-4}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.2=10\\y=5.3=15\\z=5.4=20\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}4x=3y\\7y=5z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{y}{20}\\\dfrac{y}{20}=\dfrac{z}{28}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{x-y+z}{15-20+28}=\dfrac{-46}{23}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2.15=-30\\y=-2.20=-40\\z=-2.28=-56\end{matrix}\right.\)
2: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{4x+y-z}{4\cdot3+12-16}=\dfrac{8}{8}=1\)
Do đó: x=3; y=12; z=16