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b)ta có: \(\frac{x}{5}=\frac{y}{4}=\frac{z}{-6}\Rightarrow\frac{x^3}{125}=\frac{y^3}{64}=\frac{z^3}{-216}=\frac{x^3}{125}=\frac{y^3}{64}=\frac{3z^3}{-648}\)
ADTCDTSBN
có: \(\frac{x^3}{125}=\frac{3z^3}{-648}=\frac{x^3+3z^3}{125+\left(-648\right)}=\frac{-14121}{-523}=27\)
=> x3/125 = 27 => x3 = 3 375 => x = 15
y3/64 = 27 => y3 = 1 728 => y = 12
z3/-216 =27 => z3 = -5 832 => z3 = -18
KL:...
câu c thì mk ko bk! sr bn nha!
a) ta có: \(\frac{x}{y}=\frac{7}{20}\Rightarrow x20=y7\Rightarrow\frac{x}{7}=\frac{y}{20}\Rightarrow\frac{x}{49}=\frac{y}{140}\)
\(\frac{y}{z}=\frac{7}{3}\Rightarrow y3=z7\Rightarrow\frac{y}{7}=\frac{z}{3}\Rightarrow\frac{y}{140}=\frac{z}{60}\)
\(\Rightarrow\frac{x}{49}=\frac{y}{140}=\frac{z}{60}\)
ADTCDTSBN
có: \(\frac{x}{49}=\frac{y}{140}=\frac{z}{60}=\frac{x-y+z}{49-140+60}=\frac{-155}{-31}=5\)
=> x/49 = 5 => x = 245
y/140 = 5 => y = 700
z/60 = 5 => z = 300
KL:...
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Ta có: \(\frac{4}{3x-2y}=\frac{3}{2z-4x}=\frac{2}{4y-3z}\)
\(\Rightarrow\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\)
\(=\frac{4.\left(3x-2y\right)}{4.4}=\frac{3.\left(2z-4x\right)}{3.3}=\frac{2.\left(4y-3z\right)}{2.2}\)
\(=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{16+9+4}=\frac{0}{29}=0\)
\(\Rightarrow\begin{cases}12x-8y=0\\6z-12x=0\\8y-6z=0\end{cases}\)\(\Rightarrow\begin{cases}12x=8y\\6z=12x\\8y=6z\end{cases}\)\(\Rightarrow12x=8y=6z\)
= \(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}=\frac{x+y-z}{\frac{1}{12}+\frac{1}{8}-\frac{1}{6}}=\frac{-10}{\frac{1}{24}}=-10.24=-240\)
\(\Rightarrow\begin{cases}x=-240.\frac{1}{12}=-20\\y=-240.\frac{1}{8}=-30\\z=-240.\frac{1}{6}=-40\end{cases}\)
Vậy x = -20; y = -30; z = -40
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a )
Ta có :
\(\hept{\begin{cases}\frac{x}{5}=\frac{y}{6}\\\frac{y}{8}=\frac{z}{7}\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{20}=\frac{y}{24}\\\frac{y}{24}=\frac{z}{21}\end{cases}}}\)
và \(x+y-z=69\)
ADTCDTSBN , ta có :
\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{20}=3\\\frac{y}{24}=3\\\frac{z}{21}=3\end{cases}\Rightarrow\hept{\begin{cases}x=3.20=60\\y=3.24=72\\z=3.21=63\end{cases}}}\)
Vậy ...
b )
Ta có :
\(5y=72\Rightarrow y=\frac{72}{5}=14,4\)
\(\Rightarrow x=14,4.3:2=21,6\)
và \(3x+5y-7z=30\)
Thay vào làm tiếp :
c )
\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\)
\(=\frac{3\left(x-1\right)}{6}=\frac{4\left(y+3\right)}{16}=\frac{5\left(z-5\right)}{30}\)
\(=\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}\)
\(=\frac{5z-25-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\)( ADTCDTSBN )
\(=\frac{5z-25-3x+3-4y-12}{8}=\frac{5z-3x-4y-34}{8}\)
\(=\frac{50-34}{8}=\frac{16}{8}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{x-1}{2}=2\\\frac{y+3}{4}=2\\\frac{z-5}{6}=2\end{cases}\Rightarrow\hept{\begin{cases}x-1=2.2=4\\y+3=2.4=8\\z-5=2.6=12\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\y=5\\z=17\end{cases}}}\)
Vậy ...
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)
suy ra: x/5 = 45 => x = 225
y/7 = 45 => y = 315
z/9 = 45 => z = 405
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Ta có : \(\frac{4}{3x-2y}=\frac{3}{2z-4x}=\frac{2}{4y-3z}\)
=> \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3}{2}\)
=> \(\frac{4\left(3x-2y\right)}{4\cdot4}=\frac{3\left(2z-4x\right)}{3\cdot3}=\frac{2\left(4y-3z\right)}{2\cdot2}\)
=> \(\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{12x-8y+6z-12x+8y-6z}{29}=\frac{0}{29}=0\)
=> \(\hept{\begin{cases}12x-8y=0\\6z-12x=0\\8y-6z=0\end{cases}}\Rightarrow\hept{\begin{cases}12x=8y\\6z=12x\\8y=6z\end{cases}}\Rightarrow12x=8y=6z\)
=> \(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}\)
Đặt \(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}=k\Rightarrow\hept{\begin{cases}x=\frac{1}{12}k\\y=\frac{1}{8}k\\z=\frac{1}{6}k\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{k}{12}\\y=\frac{k}{8}\\z=\frac{k}{6}\end{cases}}\)
=> \(x+y-z=\frac{k}{12}+\frac{k}{8}-\frac{k}{6}\)
=> \(\frac{k}{24}=-10\)
=> \(k=-240\)
Từ đó suy ra : \(x=-\frac{240}{12}=-20\),y = \(-\frac{240}{8}=-30\),z = \(-\frac{240}{6}=-40\)
VÌ \(\frac{4}{3x-2y}=\frac{3}{2z-4x}=\frac{2}{4y-3z}\)
\(\Rightarrow\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\)
\(=\frac{4\left(3x-2y\right)}{4.4}=\frac{3\left(2z-4x\right)}{3.3}=\frac{2\left(4y-3z\right)}{2.2}\)
\(=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)
\(=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{29}=0\)
\(\Rightarrow3x-2y=0\) (1)
\(2z-4x=0\)
\(4y-3z=0\) (2)
TỪ (1) VÀ (2) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y-z}{2+3-4}=\frac{-10}{1}=-10\)
\(\Rightarrow\frac{x}{2}=-10\Rightarrow x=-20\)
\(\frac{y}{3}=-10\Rightarrow y=-30\)
\(\frac{z}{4}=-10\Rightarrow z=-40\)
a) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{-4}=\frac{x-y-z}{2-3+4}=\frac{27}{3}=9\)
=> \(\hept{\begin{cases}\frac{x}{2}=9\\\frac{y}{4}=9\\\frac{z}{-4}=9\end{cases}}\) => \(\hept{\begin{cases}x=9.2=18\\y=9.3=27\\z=9.\left(-4\right)=-36\end{cases}}\)
Vậy ...
a, ÁP DỤNG DÃY TỈ SỐ BĂNG NHAU TA CÓ
\(\frac{x}{2}=\frac{y}{3}=\frac{x}{-4}=\frac{x-y-z}{2-3+4}=\frac{27}{3}=9\)
\(\Rightarrow\hept{\begin{cases}x=9.2=18\\y=9.3=27\\z=9.\left(-4\right)=-36\end{cases}}\)