Question

Answer 1

a) (x-1)³=3³

=> x-1=3

x=4

b) x²+x=0

x(x+1)=0

=>\(\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

c)(2x+1)²=25=5²=(-5)²

\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

d)(2x-3)²=36=6²=(-6)²

\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

e)5^x+2=625=5⁴

=> x+2=4

x=2

Answer 2

a: Ta có: \(\left(x-1\right)^3=27\)

\(\Leftrightarrow x-1=3\)

hay x=4

b: Ta có: \(x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c: Ta có: \(\left(2x+1\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Answer 3

a. \(\left(x-1\right)^3=27\Leftrightarrow x-1=3\Leftrightarrow x=4\)

b. \(x^2+x=0\Leftrightarrow x\left(x+1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c. \(\left(2x+1\right)^2=25\) \(\Leftrightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

d. \(\left(2x-3\right)^2=36\) \(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

e. \(5^{x+2}=625\) \(\Leftrightarrow5^{x+2}=5^4\) \(\Rightarrow x+2=4\Rightarrow x=2\)

Answer 4

a, (x-1)³= 27

=> (x-1)³=3³

=> x-1=3

=>. x= 3+1

=> x= 4

Vậy x=4

b, x²+x =0

=> x.(x+1)=0

=> \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy x \(\in\left\{0,-1\right\}\)

c, (2x+1)²=25

=> (2x+1)²=5²

=> 2x+1 =5

=> 2x =4

=> x=2

vậy x=2

d, (2x-3)² = 36

=> (2x-3)²=6²

=> 2x-3 =6

=>2x =9

=> x=\(\dfrac{9}{2}\)

Vậy x= \(\dfrac{9}{2}\)

e, 5^x+2= 625

=> 5^x+2= 5⁴

=> x+2 =4

=> x= 2

Vậy x=2

Hoctot

 

Answer 5

d: Ta có: \(\left(2x-3\right)^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-9=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

e: Ta có: \(5^{x+2}=625\)

\(\Leftrightarrow x+2=4\)

hay x=2