Tìm x biết:
a) ( x - 1)3 = 343
b)(x - 2)4 = 4069
c) ( x - 4)2 = ( x - 4)4
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2.Tìm x, biết
a,(x-1)3=3 (x-1)3=343
=> (x-1)3=73
=> x -1 = 7
=> x = 8
B, (X-2)4=4096
(X-2)4= 84
=> x - 2 = 8
=> x = 10
C,(2x2-13)4=(-5)4
=> 2x2-13 = -5
=> 2x2 = 8
=> x2 = 4
=> x = 2
Study well
a. (x-1)3 = 343
=> (x-1)3 = 73
=> x - 1 = 7
=> x = 7 + 1
=> x = 8
a ) ( x - 1 )3 = 343
73 = 343
x + 1 = 7
x = 7 - 1
x = 6
b ) ( x - 2 )4 = 4096
84 = 4096
x - 2 = 8
x = 8 + 2
x = 10
\(\left(x-1\right)^3=343=7^3\) => x -1 =7 => x =8
\(\left(x-2\right)^4=4096=8^4=\left(-8\right)^4\)\(\Rightarrow\orbr{\begin{cases}x-2=8\\x-2=-8\end{cases}\Rightarrow\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)
a) \(\frac{3}{7}-\frac{1}{7}x=\frac{2}{3}\)
=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)
=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)
b) \(3x^2-2=72\)=> 3x2 = 74 => x2 = 74/3 => x không thỏa mãn
c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)
=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)
=> \(\left(19x+50\right):14=9\)
=> \(19x+50=126\)
=> \(19x=76\)
=> x = 4
d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)
=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)
=> \(x\left(2+4+8+16+32\right)=343\)
=> x . 62 = 343
=> x = 343/62
Bài 1 :
a) 72x-1 = 343
=> 72x-1 = 73
=> 2x - 1 = 3 => 2x = 4 => x = 2
b) (7x - 11)3 = 25.32 + 200
=> (7x - 11)3 = 32.9 + 200
=> (7x - 11)3 = 488
xem kĩ lại đề này :vvv
c) 174 - (2x - 1)2 = 53
=> (2x - 1)2 = 174 - 53
=> (2x - 1)2 = 174 - 125 = 49
=> (2x - 1)2 = (\(\pm\)7)2
=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
Mà x \(\in\)N nên x = 4( thỏa mãn điều kiện)
Bài 2 :
a) x5 = 32 => x5 = 25 => x = 2
b) (x + 2)3 = 27
=> (x + 2)3 = 33
=> x + 2 = 3 => x = 3 - 2 = 1
c) (x - 1)4 = 16
=> (x - 1)4 = 24
=> x - 1 = 2 => x = 3 ( vì đề bài cho x thuộc N nên thỏa mãn)
d) (x - 1)8 = (x - 1)6
=> (x - 1)8 - (x - 1)6 = 0
=> (x - 1)6 [(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=\left(\pm1\right)^2\end{cases}}\)
+) x - 1 = 1 => x = 2 ( tm)
+) x - 1 = -1 => x = 0 ( tm)
Vậy x = 1,x = 2,x = 0
a) \(\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\Rightarrow\frac{28}{27}=x:\frac{3}{7}\Rightarrow x=\frac{4}{9}\)
b) \(\left(x-\frac{4}{7}\right)^3=343\Rightarrow\left(x-\frac{4}{7}\right)^3=7^3\Rightarrow x-\frac{4}{7}=7\Rightarrow x=\frac{53}{7}\)
c) \(x^5=x^3\Leftrightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}\)
e) \(\left(x-1\right)^4=16\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^4=2^4\\\left(x-1\right)^4=\left(-2\right)^4\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
a: =>9x=5(1-4x)
=>9x=5-20x
=>29x=5
hay x=5/29
b: =>x/5=5/x
=>x2=25
=>x=5 hoặc x=-5
c: =>(x+2)3=1/343
=>x+2=1/7
hay x=-13/7
a) $(x-1)^3$=343
$=>(x-1)^3=7^3$
$=>x-1=7$
$=>x=8$
b) (số sai rồi nha bạn, ra lẻ lắm)
c) $(x-4)^2=(x-4)^4$
$=>(x-4)^4-(x-4)^2=0$
$=>(x-4)^2[(x-4)^2-1]=0$
\(=>\left[{}\begin{matrix}\left(x-4\right)^2=0\\\left(x-4\right)^2-1=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}\left(x-4\right)^2=0\\\left(x-4\right)^2=1\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x-4=0\\x-4=1\\x-4=-1\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=4\\x=5\\x=3\end{matrix}\right.\)
a,\(\left(x-1\right)^3=343\)
\(\Rightarrow x-1=7\Rightarrow x=8\)
b, \(\left(x-2\right)^4=4096\)
\(\Rightarrow x-2=8\Rightarrow x=10\)
c, \(\left(x-4\right)^2=\left(x-4\right)^4\)
\(\Rightarrow\left(x-4\right)^2=\left[\left(x-4\right)^2\right]^2\)
\(\Rightarrow x-4=\left(x-4\right)^2\)
\(\Rightarrow\left(x-4\right)-\left(x-4\right)^2=0\)
\(\Rightarrow\left(x-4\right)\left(1-x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\5-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Chúc bạn học tốt!!!