cho y=f(x)=5-x+26
Tìm x khi y=26
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15};\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\\ \Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x+y-z}{10+15-12}=\dfrac{-26}{13}=-2\\ \Rightarrow\left\{{}\begin{matrix}x=-20\\y=-30\\z=-24\end{matrix}\right.\)
a, thay x=-2;x=6;x=-4 vào ta được:
f(-2)=-2*2=-4
f(6)=2*6=12
f(-4)=-4*2=-8
b,khi y=6 thì x=6/2=3
khi y=8 thì x=8/2=4
c,khi x=2 thì y=2*2=4
khĩ=5 thì y=2*5=10
\(a,f\left(1\right)=3\cdot1^2+1+1=5\\ f\left(-\dfrac{1}{3}\right)=3\cdot\left(-\dfrac{1}{3}\right)^2-\dfrac{1}{3}+1=\dfrac{1}{3}-\dfrac{1}{3}+1=1\\ f\left(\dfrac{2}{3}\right)=3\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{2}{3}+1=\dfrac{4}{3}-\dfrac{2}{3}+1=\dfrac{5}{3}\\ f\left(-2\right)=3\cdot\left(-2\right)^2-2+1=11\\ f\left(-\dfrac{4}{3}\right)=3\cdot\left(-\dfrac{4}{3}\right)^2-\dfrac{4}{3}+1=\dfrac{16}{3}-\dfrac{4}{3}+1=5\)
\(b,f\left(\dfrac{2}{3}\right)=\left|2\cdot\dfrac{2}{3}-9\right|-3=\dfrac{23}{3}-3=\dfrac{14}{3}\\ f\left(-\dfrac{5}{4}\right)=\left|2\cdot\left(-\dfrac{5}{4}\right)-9\right|-3=\dfrac{23}{2}-3=\dfrac{17}{2}\\ f\left(-5\right)=\left|2\left(-5\right)-9\right|-3=19-3=16\\ f\left(4\right)=\left|2\cdot4-9\right|-3=1-3=-2\\ f\left(-\dfrac{3}{8}\right)=\left|2\cdot\left(-\dfrac{3}{8}\right)-9\right|-3=\dfrac{39}{4}-3=\dfrac{27}{4}\)
\(c,x=0\Rightarrow y=2\cdot0^2-7=-7\\ x=-3\Rightarrow y=2\cdot\left(-3\right)^2-7=11\\ x=-\dfrac{1}{2}\Rightarrow y=2\cdot\left(-\dfrac{1}{2}\right)^2-7=\dfrac{-13}{2}\\ x=\dfrac{2}{3}\Rightarrow y=2\cdot\left(\dfrac{2}{3}\right)^2-7=-\dfrac{55}{9}\)
\(5-x+26=y\)
\(\Leftrightarrow5-x+26=26\)
\(\Leftrightarrow5-x=26-26\)
\(\Leftrightarrow5-x=0\)
\(\Leftrightarrow x=5-0\)
\(\Leftrightarrow x=5\)
Vậy x = 5
Ps : Ko chắc