a) ( 2x - 1)^2= 49
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3.|x+1|-2=4
3.|x+1|=4+2
3.|x+1|=6
|x+1|=6:3
|x+1|=2
Trường hợp 1 x+1=2
x=2-1
x=1
trường hợp 2
x+1=-2
x=(-2)-1
x=-3
==> x thuộc {1; -3}
k mk nha chúc học tốt
a) \(\left(2x-5\right)^2=49\)
\(\left(2x-5\right)^2=\left(\pm7\right)^2\)
\(=>2x-5=7\) hoặc \(2x-5=-7\)
\(\cdot2x-5=7\) \(\cdot2x-5=-7\)
\(2x=5+7\) \(2x=-7+5\)
\(2x=12\) \(2x=-2\)
\(x=12:2\) \(x=-2:2\)
\(x=6\) \(x=-1\)
Vậy x=6 hoặc x=-1
b/ \(\left(2x+5\right)^2-\left(1-2x\right)^2=10\)
\(4x^2+20x+25-\left(1-4x+4x^2\right)=10\)
\(4x^2+20x+25-1+4x-4x^2=10\)
\(24x+24=10\)
\(24x=10-24\)
\(24x=-14\)
\(x=\frac{-14}{24}\)
\(x=\frac{-7}{12}\)
c/ \(\left(9-2x\right)^3=27\)
\(\left(9-2x\right)^3=3^3\)
\(9-2x=3\)
\(2x=9-3\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
a) (x - 2).3⁵ = 3⁷
x - 2 = 3⁷ : 3⁵
x - 2 = 3²
x - 2 = 9
x = 9 + 2
x = 11
b) x² - 2x = 0
x(x - 2) = 0
⇒ x = 0 hoặc x - 2 = 0
*) x - 2 = 0
x = 2
Vậy x = 0; x = 2
c) (2x - 1)² = 49
⇒ 2x - 1 = 7 hoặc 2x - 1 = -7
*) 2x - 1 = 7
2x = 7 + 1
2x = 8
x = 8 : 2
x = 4
*) 2x - 1 = -7
2x = -7 + 1
2x = -6
x = -6 : 2
x = -3
Vậy x = -3; x = 4
\(\left(2x-1\right)^2=49\)
\(\Rightarrow\left(2x-1\right)^2=7^2\)
\(\Rightarrow\left(2x-1\right)^2=7\)
\(\Rightarrow\orbr{\begin{cases}2x-1=-7\\2x-1=7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=-6\\2x=8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
a) \(\left(2x-1\right)^2=49\)
<=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\)
<=> \(\orbr{\begin{cases}x=4\left(n\right)\\x=-3\left(n\right)\end{cases}}\)
b) \(\left(5x-3\right)^2-\left(4x-7\right)^2=0\)
<=> \(\left(5x-3-4x+7\right)\left(5x-3+4x-7\right)=0\)
<=> \(\orbr{\begin{cases}x=-4\left(n\right)\\x=\frac{10}{9}\left(n\right)\end{cases}}\)
(2x - 1)2 = 49
(2x - 1)2 = 72
2x - 1 = 7
2x = 7 + 1
2x = 8
x = 8 : 2
x = 4
\(\left(2x-1\right)^2=49\)
\(\Rightarrow\left(2x-1\right)^2=7^2\)
\(\Rightarrow2x-1=7\)
\(\Rightarrow2x=7+1\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=8:2\)
\(\Rightarrow x=4\)