\(\left(2x+1\right)^3=-125\)
\(< =>\left(2x+1\right)^3=\left(-5\right)^3\)
\(< =>2x+1=-5\)
\(< =>2x=-5-1=-6\)
\(< =>x=-\frac{6}{2}=-3\)
Bài làm:
a) \(\left(2x+1\right)^3=-125\)
\(\Leftrightarrow\left(2x+1\right)^3=\left(-5\right)^3\)
\(\Rightarrow2x+1=-5\)
\(\Leftrightarrow2x=-6\)
\(\Rightarrow x=-3\)
b) \(\left(7-x\right)^2-\left(-11\right)=15\)
\(\Leftrightarrow\left(7-x\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}7-x=2\\7-x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=9\end{cases}}\)
c) \(2020^x-2019=-2019\)
\(\Leftrightarrow2020^x=0\)
=> ko tồn tại x thỏa mãn PT
\(\left(7-x\right)^2-\left(-11\right)=15\)
\(< =>\left(7-x\right)^2=15-11=4\)
\(< =>\left(7-x\right)^2=2^2=\left(-2\right)^2\)
\(< =>\orbr{\begin{cases}7-x=2\\7-x=-2\end{cases}}\)
\(< =>\orbr{\begin{cases}x=5\\x=9\end{cases}}\)
a. \(\left(2x+1\right)^3=-125\)
\(\Rightarrow\left(2x+1\right)^3=-5^3\)
\(\Rightarrow2x+1=-5\)
\(\Rightarrow2x=-6\)
\(\Rightarrow x=-3\)
b. \(\left(7-x\right)^2-\left(-11\right)=15\)
\(\Rightarrow\left(7-x\right)^2=4\)
\(\Rightarrow\left(7-x\right)^2=2^2\)
\(\Rightarrow7-x=2\)
\(\Rightarrow x=5\)
c. \(2020^x-2019=-2019\)
\(\Leftrightarrow2020^x=0\left(vo-ly\right)\)
=> x vô nghiệm
(2x+1)^3=-125
(2x+1)^3=(-5)^3
2x+1=5
2x=5-1
2x=4
x4:2
x=2
vậy x=2
\(2020^x-2019=-2019\)
\(< =>2020^x=-2019+2019\)
\(< =>2020^x=0\)vô lí
vậy biểu thức vô nghiệm
(7-x)^2+11=15
(7-x)^2=15-11
(7-x)^2=4
(7-x)^2=2^2
7-x=4
x=7-4
x=3
vậy x=3
2020^x-2019=-2019
2020^x=-2019+2019
2020^x=0
vậy x=0
a, \(\left(2x+1\right)^3=-125\Leftrightarrow\left(2x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow2x+1=-5\Leftrightarrow x=-3\)
b, \(\left(7-x\right)^2+11=15\Leftrightarrow\left(7-x\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}7-x=2\\7-x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=9\end{cases}}}\)
c, \(2020^x-2019=-2019\Leftrightarrow2020^x=0\)( vô nghiệm )
