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$(2x - \dfrac13)^5 = \dfrac{32}{243}$
$(2x - \dfrac13)^5 = \dfrac{2^5}{3^5}$
$(2x - \dfrac13)^5 = \left(\dfrac23\right)^5$
$2x - \dfrac13 = \dfrac23$
$2x = 1$
$x = \dfrac12$.
a, \(\frac{\left(5-2x\right)}{3}=\frac{\left(4x-1\right)}{-5}\)
\(\Leftrightarrow-5(5-2x)=3\left(4x-1\right)\)
\(\Leftrightarrow10x-25=12x-3\)
\(\Leftrightarrow10x-12x=25-3\)
\(\Leftrightarrow-2x=22\)
\(\Leftrightarrow x=-11\)
b, \(\frac{\left(12-3x\right)}{32}=\frac{6}{\left(4-x\right)}\)
\(\Leftrightarrow\frac{3\left(4-x\right)}{32}=\frac{6}{\left(4-x\right)}\)
\(\Leftrightarrow3(4-x)\left(4-x\right)=32.6\)
\(\Leftrightarrow(4-x)\left(4-x\right)=32.2\)
\(\Leftrightarrow(4-x)^2=64\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=8\\4-x=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=12\end{cases}}\)
c, \(\frac{\left(10-2x\right)}{6}=\frac{27}{\left(5-x\right)}\)
\(\Leftrightarrow\frac{2\left(5-x\right)}{6}=\frac{27}{\left(5-x\right)}\)
\(\Leftrightarrow2(5-x)\left(5-x\right)=27.6\)
\(\Leftrightarrow(5-x)\left(5-x\right)=27.3\)
\(\Leftrightarrow(5-x)^2=81\)
\(\Leftrightarrow\orbr{\begin{cases}5-x=9\\5-x=-9\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=14\end{cases}}\)
a, \(\frac{5-2x}{3}=\frac{4x-1}{-5}\Leftrightarrow-25+10x=12x-3\Leftrightarrow-22-2x=0\Leftrightarrow x=-11\)
b, \(\frac{12-3x}{32}=\frac{6}{4-x}\Leftrightarrow\frac{12-3x}{32}=\frac{18}{12-3x}\)
\(\Leftrightarrow\left(12-3x\right)^2=576\Leftrightarrow12-3x=\pm2\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=\frac{14}{3}\end{cases}}\)
c, \(\frac{10-2x}{6}=\frac{27}{5-x}\Leftrightarrow\frac{10-2x}{6}=\frac{54}{10-2x}\)
\(\Leftrightarrow\left(10-2x\right)^2=324\Leftrightarrow10-2x=\pm18\)\(\Leftrightarrow\orbr{\begin{cases}x=14\\x=-4\end{cases}}\)
a, \(x^2\) - 19 = 5.9
\(x^2\) - 19 = 45
\(x^2\) = 45 + 19
\(x^2\) = 64
\(x^2\) = 82
\(x\) = 8
b, (2\(x\) + 1)3 = -0,001
(2\(x\) + 1)3 = (-0,1)3
2\(x\) + 1 = -0,1
2\(x\) = -0,1 - 1
2\(x\) = - 1,1
\(x\) = -1,1: 2
\(x\) = - 0,55
a ) \(\left(2x-1\right)^4=81\)
\(\Leftrightarrow\left(2x-1\right)^4=3^4\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow x=2\)
Vậy \(x=2.\)
b ) \(\left(x-1\right)^5=-32\)
\(\Leftrightarrow\) \(\left(x-1\right)^5=-2^5\)
\(\Leftrightarrow x-1=-2\)
\(\Leftrightarrow x=-1\)
Vậy \(x=-1.\)
c ) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[\left(1-2x+1\right)\left(1+2x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[\left(2-2x\right).2x\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\2-2x=0\\2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x=2\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\\x=0\end{matrix}\right.\)
Vậy ...............
a) (2x-1)4=81
\(\Leftrightarrow\)\(\left[\begin{array}{} (2x-1)^4=(3)^4\\ (2x-1)^4=(-3)^4 \end{array}\right.\)
\(\Rightarrow\)\(\left[\begin{array}{} 2x-1=3\\ 2x-1=-3 \end{array}\right.\)
\(\Rightarrow\)\(\left[\begin{array}{} 2x=3+1\\ 2x=-3+1 \end{array}\right.\)
\(\Rightarrow\)\(\left[\begin{array}{} 2x=4\\ 2x=-2 \end{array}\right.\)
\(\Rightarrow\)\(\left[\begin{array}{} x=4:2\\ x=-2:2 \end{array}\right.\)
\(\Rightarrow\)\(\left[\begin{array}{} x=2\\ x=-1 \end{array}\right.\)
Vậy x=2 hoặc x=-1
b) (x-1)5= -32
\(\Leftrightarrow\)\( (x-1)^5=(-2)^5 \)
\(\Rightarrow\)\( (x-1)=-2 \)
\(\Rightarrow\)\( x=-2+1 \)
\(\Rightarrow\)\( x=-1 \)
Vậy x=-1
c) ( 2x-1)6= ( 2x-1)8
\(\Leftrightarrow\) (2x-1)6=(2x-1)8.
\(\Leftrightarrow\)(2x-1)8-(2x-1)6=0.
\(\Leftrightarrow\)(2x-1)6)[(2x-1)2-1]=0.
\(\Leftrightarrow\)(2x-1)6(2x-1+1)(2x+1+1)=0.
\(\Leftrightarrow\)(2x-1)62x(2x+2)=0.
\(\Leftrightarrow\)(2x-1)6<=>2x(2x-1)=0.\(\Rightarrow x=\dfrac{1}{2}\)
hoặc 2x=0\(\Rightarrow\)x=0
hoặc 2x+2=0\(\Rightarrow\)2x=-2\(\Leftrightarrow\)x=-2:2\(\Leftrightarrow\)x=-1
Vậy x=\(\dfrac{1}{2}\)hoặc x=0 hoặc x=-1
Chúc bạn học tốt !!!
a) (2x-1)^3=27
b) (2x-1)^4=81
c) (x-2)^5=-32
d) (3x-1)^4=(3x-1)^6
đ) 5^x +5^x+2=650
g) 3^x-1 +5.3^x-1=162
a) (2x-1)3 = 27
(2x-1)3 = 93
2x-1 = 9
2x = 9+1
2x = 10
x = 10:5
x = 2
Vậy x = 2
b) (2x-1)4 = 81
(2x-1)4 = (\(\pm\)34)
2x-1 = \(\pm\)3
Trường hợp 1:
2x-1 = 3
2x = 3+1
2x = 4
x = 4:2
x = 2
Trường hợp 2:
2x-1 = -3
2x = -3+1
2x = -2
x = -2:2
x = -1
Vậy x \(\in[_{ }2;-1]\)
Vì không tìm thấy ngoặc nhọn nên mình dùng tạm ngoặc vuông nhé
Vậy \(( - 32{x^5} + 1):( - 2x + 1) = 16{x^4} + 8{x^3} + 4{x^2} + 2x + 1\).
\(1,\frac{x+1}{x-2}=\frac{3}{4}\)
\(\Rightarrow3x-6=4x+4\)
\(\Rightarrow3x-4x=4+6\)
\(\Rightarrow-x=10\Leftrightarrow x=-10\)
\(2,\frac{x-1}{3}=\frac{x+3}{5}\)
\(\Rightarrow5x-5=3x+9\)
\(\Rightarrow5x-3x=9+5\)
\(\Rightarrow2x=14\Leftrightarrow x=7\)
\(3,\frac{2x+3}{24}=\frac{3x-1}{32}\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow72x-64x=24+96\)
\(\Rightarrow8x=120\)
\(\Rightarrow x=15\)
5 - (2x ? 1)3 = 32
Cho mình hỏi chổ (?) là dấu gì vậy bạn, bạn bổ sung nhé.
2x 1 là sao nhỉ?
là 2\(x\) hay 2 x 1