Question

a)Tìm x,biết 2l3x-1l+1=5
b) Tìm số tự nhiên y biết 3^y+3^y+2=810
c)Tính giá trị của biểu thức  M=3x²-5y+1 tại ,x=-3,y=4

Answer 1

a/\(2\left|3x-1\right|+1=5\)
\(\Rightarrow2\left|3x-1\right|=4\)
\(\Rightarrow\left|3x-1\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x=1\)
Vậy x = 1
b/\(3^y+3^{y+2}=810\)
\(\Rightarrow3^y+3^y\cdot3^2=810\)
\(\Rightarrow3^y\left(1+3^2\right)=810\)
\(\Rightarrow3^y\cdot10=810\)
\(\Rightarrow3^y=81\)
\(\Rightarrow y=4\)
c/Thay x = -3, y = 4 vào M, ta có:
\(M=3\cdot\left(-3\right)^2-5\cdot4+1\)
\(=3\cdot9-20+1\)
\(=27-20+1\)
\(=8\)

Answer 2

a)Ta có:

\(2\left|3x-1\right|+1=5\)

\(\Rightarrow2\left|3x-1\right|=4\)

\(\Rightarrow\left|3x-1\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) Ta có:

\(3^y+3^{y+2}=810\)

\(\Rightarrow3^y\left(1+3^2\right)=810\)

\(\Rightarrow3^y.10=810\)

\(\Rightarrow3^y=81\)

\(\Rightarrow y=4\)

c) Thay \(x=-3;y=4\) ta được:

\(M=3\left(-3\right)^2-5.4+1=3.9-20+1=27-20+1=8\)