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a ) \(\left(x-\frac{1}{4}\right)^4=\frac{1}{256}\)
\(\left(x-\frac{1}{4}\right)=\sqrt[4]{\frac{1}{256}}\)
\(\left(x-\frac{1}{4}\right)=\frac{1}{4}\)
\(x=\frac{1}{4}+\frac{1}{4}\)
\(x=\frac{1}{2}\)
b ) \(\left(3x-2\right)^5=-234\)
\(\left(3x-2\right)=-\sqrt[5]{234}\)
\(\left(3x-2\right)=-2,977441049\)
\(3x=-0,9774410485\)
\(x=-0,3258136828\)
a.
(x-1/4)^4=1/256
(x-1/4)^4=(1/4)^4
x-1/4=1/4
x=1/4+1/4
x=2/4
\(\text{a, 3(x+1)+4x=10}\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow7x+3=10\)
\(\Rightarrow7x=10-3=7\)
\(\Rightarrow x=1\)
c, x+1/10+x+2/9=x+3/8+x+4/7
=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)
=> x+11/10 + x+11/9 = x+11/8 + x+11/7
...............
a) \(3\left(x+1\right)+4x=10\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow3x+4x=10-3\)
\(\Rightarrow7x=7\)
\(\Rightarrow x=7\)
ai k mình k lại [ chỉ 3 người đầu tiên mà trên 10 điểm hỏi đáp ]
3(1-4x)(x-1) + 4(3x-2)(x+3) = -27
<=> (3-12x)(x-1) + (12x-8)(x+3) = -27
<=> 3x-3-12x2+12x+12x2+36x-8x-24+27 = 0
<=> (-12x2+12x2)+(3x+12x+36x)+(27-3+36) = 0
<=> 51x+60 =0
<=> 51x = -60
<=> x= -60 : 51
<=> x= -20/17
\(\left(x-3\right)^2=16\)
\(\Rightarrow\left(x-3\right)^2=4^2\)
\(\Rightarrow x-3=4\)
\(\Rightarrow x=4+3\)
\(\Rightarrow x=7\)
a) 3(1-4x) (x-1) +4(3x-2) (x+3)=-27
b) 5(2x+3) (x+2)-2(5x-4) (x-1)=75
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
+) \(\left(x-3\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^2=4^2\\\left(x-3\right)^2=\left(-4\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x-3=4\\x-3=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)
Vậy x = 7 hoặc x = -1
+) \(\left(1-3x\right)^3=-64\)
\(\Rightarrow\left(1-3x\right)^3=\left(-4\right)^3\)
\(\Rightarrow1-3x=-4\)
\(\Rightarrow3x=1+4\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=5:3\)
\(\Rightarrow x=\frac{5}{3}\)
Vậy \(x=\frac{5}{3}\)
+) \(x^{13}=27.x^{10}\)
\(\Rightarrow x^{13}:x^{10}=27\)
\(\Rightarrow x^3=27\)
\(\Rightarrow x^3=3^3\)
\(\Rightarrow x=3\)
Vậy x = 3
+) \(\left(4x-1\right)^2=\left(1-4x\right)^4\)
\(\Rightarrow\left(4x-1\right)^2=\left(4x-1\right)^4\)
\(\Rightarrow\left(4x-1\right)^2-\left(4x-1\right)^4=0\)
\(\Rightarrow\left(4x-1\right)^2\left[1-\left(4x-1\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(4x-1\right)^2=0\\1-\left(4x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(4x-1\right)^2=0\\\left(4x-1\right)^2=1\end{cases}}\)
TH 1 : \(\left(4x-1\right)^2=0\Rightarrow4x-1=0\Rightarrow4x=1\Rightarrow x=\frac{1}{4}\)
TH 2 : \(\left(4x-1\right)^2=1\Rightarrow\orbr{\begin{cases}4x-1=1\\4x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}4x=2\\4x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=0\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{4};\frac{1}{2};0\right\}\)
_Chúc bạn học tốt_
a, (x-3)^2 = 16
=> (x-3)^2=4^2
=> x-3=4
=> x= 4+3
=> x = 7 .Vậy x =7
b,(1-3x)^3 = 64
=> ( 1-3x)^3 = 4^3
=> 1-3x = 4
=> 3x = 1-4
=> 3x = -3
=> x = -1 . Vậy x = -1
c, x^13 = 27.x^10
=> x^13 : x^10 = 27
=> x^3 = 3^3
=> x = 3 . Vậy x = 3