Tìm x biết
a, (2x-1)4 = 16
b, (2x+1)4 = (2x+1)b
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a) \(\left(2x-1\right)^4=16\)
\(\)TH1: \(\left(2x-1\right)^4=2^4\)
\(=>2x-1=2\)
\(2x=2+1\)
\(2x=3\)
\(x=\dfrac{3}{2}\)
TH2: \(\left(2x-1\right)^4=\left(-2\right)^4\)
\(=>2x-1=-2\)
\(2x=-2+1\)
\(2x=-1\)
\(x=\dfrac{-1}{2}\)
Vậy x = \(\dfrac{3}{2}\) hoặc x = \(\dfrac{-1}{2}\)
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b) \(\left(2x+1\right)^3=125\) ( mình nghĩ đề bài đúng là vầy )
\(\left(2x+1\right)^3=5^3\)
\(=>2x+1=5\)
\(2x=5-1\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
Vậy x = \(2\)
a) Rút gọn được VT = 9x + 7. Từ đó tìm được x = 1.
b) Rút gọn được VT = 2x + 8. Từ đó tìm được x = 7 2 .
a) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Rightarrow72-20x-36x-84=30x-240-6x+84\)
\(\Rightarrow\left(72-84\right)-\left(20x+36x\right)=\left(30x-6x\right)-240+84\)
\(\Rightarrow-12-56=24x-56x\)
\(\Rightarrow-12+156=24x+56x\)
\(\Rightarrow144=80x\)
\(\Rightarrow x=144:80\)
\(\Rightarrow x=\frac{9}{5}\)
b) \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)
\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow15x+25-8x+12-5x-6x-36-1=0\)
\(\Rightarrow-4x=0\)
\(\Rightarrow-4.0\)
\(\Rightarrow x=0\)
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a) \(\left|2x+1\right|=\left|x+4\right|\Rightarrow\left[{}\begin{matrix}2x+1=x+4\\2x+1=-x-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
b) \(\left|2x-1\right|=x+4\Rightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
a) (x - 2)^2 = 16
(x - 2)^2 = 4^2 = (-4)^2
=> x - 2 = 4 hoặc x - 2 = -4
=> x = 4 + 2 = 6
=> x = (-4) + 2 = -2
Vậy x = 6 hoặc x = -2
b) (2x - 1)^3 = 8
(2x - 1)^3 = 2^3
=>2x - 1 = 2
2x = 2 + 1 = 3
\(x=\frac{3}{2}\)
c) (2x - 1)^4 = 81
(2x - 1)^4 = 3^4=(-3)^4
=> 2x - 1 = 3 hoặc 2x - 1 = -3
=>2x = 3 + 1 = 4
=>x = 4 : 2 = 2
2x = (-3) + 1 = -2
x = -1
Vậy x = 2 hoặc x = -1
a) (2x-1)4 = 16
=> (2x-1)4 = 24 hoặc (-2)4
=>\(\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2+1\\2x=-2+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
b) (2x+1)4 = (2x+1)6
=> (2x+1)4 = (2x+1)4+2
=> (2x+1)4 = (2x+1)4 . (2x+1)2
=> (2x+1)4 - (2x+1)4 . (2x+1)2 = 0
=> (2x+1)4 . [1 - (2x+1)2] = 0
\(\left[{}\begin{matrix}\left(2x+1\right)^4=0\\1-\left(2x+1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x+1=0\\2x+1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=0\end{matrix}\right.\)