Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Sai thì thôi nhé!
a) \(f\left(-3\right)=\frac{2}{3}\times-3-\frac{1}{2}=-2-\frac{1}{2}=\frac{-4}{2}-\frac{1}{2}=\frac{-5}{2}\)
\(f\left(\frac{3}{4}\right)=\frac{2}{3}\times\frac{3}{4}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0\)
b) \(f\left(x\right)=\frac{1}{2}\Leftrightarrow\frac{2}{3}\times x-\frac{1}{2}=\frac{1}{2}\Leftrightarrow\frac{2}{3}\times x=1\Leftrightarrow x=1:\frac{2}{3}\Leftrightarrow x=1\times\frac{3}{2}\Leftrightarrow x=\frac{3}{2}\)
c)\(y=f\left(x\right)=\frac{2}{3}\times x-\frac{1}{2}\left(1\right)\)
\(A\left(\frac{3}{4};-\frac{1}{2}\right)\)
\(A\left(\frac{3}{4};\frac{-1}{2}\right)\Rightarrow\hept{\begin{cases}x_A=\frac{3}{4}\\y_A=\frac{-1}{2}\end{cases}}\)
Thay \(x_A=\frac{3}{4}\)vào (1) ta có:
\(y=f\left(x\right)=\frac{2}{3}\times\frac{3}{4}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0\ne y_A\)
Vậy điểm A không thuộc đồ thì hàm số \(y=f\left(x\right)=\frac{2}{3}\times x-\frac{1}{2}\)
\(B\left(0,5;-2\right)\)
\(B\left(0,5;-2\right)\Rightarrow\hept{\begin{cases}x_B=0,5\\y_B=-2\end{cases}}\)
Thay \(x_B=0,5\)vào (1) ta có:
\(y=f\left(x\right)=\frac{2}{3}\times0,5-\frac{1}{2}=\frac{1}{3}-\frac{1}{2}=\frac{2}{6}-\frac{3}{6}=\frac{-1}{6}\ne y_B\)
Vậy điểm B không thuộc đồ thị hàm số \(y=f\left(x\right)=\frac{2}{3}\times x-\frac{1}{2}\)
a, 5x = 2y
\(\Rightarrow\frac{x}{2}=\frac{y}{5}=k\)
\(\Rightarrow\hept{\begin{cases}x=2k\\y=5k\end{cases}\Rightarrow}\hept{\begin{cases}x^3=\left(2k\right)^3\\y^2=\left(5k\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}x^3=8k^3\\y^2=25k^2\end{cases}}\)
=> 8k3 . 25k2 = 200
=>200k5 = 200
=> k5 = 1
=> k = 1
\(\Rightarrow\hept{\begin{cases}x=2k=2.1=2\\y=5k=5.1=5\end{cases}}\)
b, Đặt \(\frac{x}{3}=\frac{y}{4}=k\)
\(\Rightarrow\hept{\begin{cases}x=3k\\y=4k\end{cases}}\Rightarrow\hept{\begin{cases}x^2=\left(3k\right)^2\\y^2=\left(4k\right)^2\end{cases}\Rightarrow\hept{\begin{cases}x^2=9k^2\\y^2=16k^2\end{cases}}}\)
=> 9k2 + 16k2 = 100
=> 25k2 = 100
=> k2 = 4
=> k = ±2
=> +) x = 3k = 3 . 2 = 6
+) x = 3k = 3 . (-2) = -6
=> +) y = 4k = 4 . 2 = 8
+) y = 4k = 4 . (-2) = -8
c, Đặt \(\frac{x}{5}=\frac{y}{2}=\frac{z}{-3}=k\)
\(\Rightarrow\hept{\begin{cases}x=5k\\y=2k\\z=-3k\end{cases}}\)
=> 5k . 2k . (-3)k = 240
=> -30k3 = 240
=> k3 = -8
=> k = -2
\(\Rightarrow\hept{\begin{cases}x=5k=5.\left(-2\right)=-10\\y=2k=2.\left(-2\right)=-4\\z=-3k=-3.\left(-2\right)=6\end{cases}}\)
a) \(\left(x-\frac{2}{5}\right).\left(x+\frac{3}{7}\right)<0\)
\(\Rightarrow x-\frac{2}{5}<0\) hoặc \(x-\frac{2}{5}>0\)
\(x+\frac{3}{7}>0\) \(x+\frac{3}{7}<0\)
\(\Rightarrow x<\frac{2}{5}\) hoặc \(x>\frac{2}{5}\)
\(x>-\frac{3}{7}\) \(x<-\frac{3}{7}\)
\(\Rightarrow-\frac{3}{7} hoặc \(x\in rỗng\)
vậy \(-\frac{3}{7}
b) \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)
\(\frac{-1}{12}\le x\le\frac{1}{4}\)
\(\frac{-1}{12}\le x\le\frac{3}{12}\)
\(\Rightarrow x=\frac{-1}{12};0;\frac{1}{12};\frac{2}{12};\frac{3}{12}\)
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
2) Vì \(\frac{x}{y}=\frac{5}{7}\Rightarrow\frac{x}{5}=\frac{y}{7}\)
Áp dụng tc của dãy tỉ số bằng nhau ta có:
\(\frac{x}{5}=\frac{y}{7}=\frac{x-y}{5-7}=\frac{7}{-2}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-7}{2}.5=\frac{-35}{2}\\y=\frac{-7}{2}.7=\frac{-1}{2}\end{cases}}\)
Vậy ..
1) ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x^3}{8}=\frac{y^3}{27}=\frac{z^3}{64}.\)
ADTCDTSBN
\(\frac{x^3}{8}=\frac{y^3}{27}=\frac{z^3}{64}=\frac{x^3+y^3-z^3}{8+27-64}=\frac{-29}{-29}=1\)
=>....
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)Và x3+y3-z3=-29
Vì \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
=> \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x^3}{8}=\frac{y^3}{27}=\frac{z^3}{64}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{x^3}{8}=\frac{y^3}{17}=\frac{z^3}{65}=\frac{x^3+y^3-z^3}{8+17-64}=\frac{14}{39}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{14}{39}\Rightarrow x=\frac{28}{39}\\\frac{y}{3}=\frac{14}{39}\Rightarrow y=\frac{14}{13}\\\frac{x}{4}=\frac{14}{39}\Rightarrow z=\frac{56}{39}\end{cases}}\)
Vậy x =\(\frac{28}{39}\)
y = \(\frac{14}{13}\)
z = \(\frac{56}{39}\)