Tìm x biết :
( 2x -1 )\(^{10}\) = 49\(^5\)
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`(5x+1)=36/49`
`<=> 5x = 36/49-1`
`<=> 5x = -13/49`.
`<=> x = -13/245.`
Vậy `x = -13/245`.
`b, x-2/9 = 2/3`.
`<=> x = 2/3 + 2/9`
`<=> x = 8/9`.
Vậy `x = 8/9`.
c: (8x-1)^(2x+1)=5^(2x+1)
=>8x-1=5
=>8x=6
=>x=3/4
d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0
=>x-3,5=0 và y-0,1=0
=>x=3,5 và y=0,1
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/ \(\left(9-2x\right)^3=27\)
\(\left(9-2x\right)^3=3^3\)
\(9-2x=3\)
\(2x=9-3\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
Ta có
(2x-1)^10 = 49^5
=> (2x-1)^10 = (7^2)^5
=> (2x-1)^10 = 7^10
=> 2x-1 = 7
=> 2x= 8
=> x=4
a) \(\Rightarrow\left(2x-3\right)^2=49\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c) \(\Rightarrow x\left(x-5\right)+2\left(x-5\right)=0\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a, ⇒ (2x - 3)2 = 49
⇒ (2x - 3)2 = \(\left(\pm7\right)^2\)
⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0
⇒ (x - 5).(2x + 7) = 0
⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c, ⇒ x2 - 5x + 2x - 10 = 0
⇒ (x2 - 5x) + (2x - 10) = 0
⇒ x.(x - 5) +2.(x - 5) = 0
⇒ (x - 5).(x + 2)=0
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a)=2
b)=0.3333333333
c)=9
d)= chua tinh
e)=33
g)=5
tk minh nha ban nha
a) (2x-5)+17=16
2x-5=16-17
2x-5=-1
2x=-1+5
2x=4
x=4:2
x=2
b) 10-2.(4-3x)=-4
2.(4-3x)=10--4
2.(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
\(\left(2x-3\right)^2=7^2\)
\(2x-3=7\)
\(2x=10\)
\(x=5\)
Vậy x=5
a: \(\left(2x-3\right)^2-49=0\)
\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
( 2x - 1) 10 = 495
=> (2x)10 - (110) = 282475249
2x10 - 1 = 282475249
2x10 = 282475250
=> x10 = 141237625
=> x = ........... tự làm nhé
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
=>(2x-1)10=(72)5
=>(2x-1)10=710
=> 2x-1=7
=>2x=8
=>x=4
tick nha
(2x-1)10=495
(2x-1)2.5=495
((2x-1)2)5=49
(2x-1)2=72
(2x-1)=7
2x=7+1
2x=8
x=8:2
x=4