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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
1) \(2^x=4^3\)
=> 2x = (22)3
=> 2x = 26
=> x = 6
2) \(\left(\frac{1}{7}\right)^x=\left(\frac{1}{343}\right)^3\)
=> \(\left(\frac{1}{7}\right)^x=\left[\left(\frac{1}{7}\right)^3\right]^3\)
=> \(\left(\frac{1}{7}\right)^x=\left(\frac{1}{7}\right)^9\)
=> x = 9
Bài làm:
1) \(2^x=4^3\)
\(\Leftrightarrow2^x=2^6\)
\(\Rightarrow x=6\)
2) \(\left(\frac{1}{7}\right)^x=\left(\frac{1}{343}\right)^3\)
\(\Leftrightarrow\left(\frac{1}{7}\right)^x=\left(\frac{1}{7}\right)^9\)
\(\Rightarrow x=9\)
a, x:(1/2)3=-1/2
x:1/8= -1/2
x= -1/2.1/8
x=-1/16
b,(3/4)5.x=(3/4)7
x=(3/4)7:(3/4)5
x= (3/4)2
c,(2/5)^8:x=(2/5)^6
x=.......
như cái trên nha lm giống thế
A = \(4\left(x-5\right)-x^2\left(x+1\right)-x^3\left(x-3\right)-\left(x-4+x^2\right)\)
A = \(4x-20-x^3-x^2-x^4+3x^3-x+4-x^2\)
A = \(-x^3-3x^3-x^2+x^2-x^4+4x-x-20+4\)
A = \(-4x^3-x^4+4x-16\)
B = \(-3\left(x^2-x+1\right)-2\left(4-x^2\right)-6\left(x+1\right)-x^4-x^3\)
B = \(-3x^2+3x-3-8+2x^2-6x-6-x^4-x^3\)
B = \(-x^4-x^3-3x^2-2x^2+3x-6x-3-8-6\)
B = \(-x^4-x^3-5x^2-3x-17\)
C = \(-\left(x^4+3x^2-2\right)-x^2\left(5-x\right)+3\left(x-1\right)\)
C = \(-x^4-3x^2+2-5x^2+x^3+3x-3\)
C = \(-x^4+x^3-3x^2+5x^2+3x+2-3\)
C = \(-x^4+x^3-2x^2+3x-1\)
#Yiin
\(A=4x-20-x^3-x-x^4+3x^3-x+4-x^2\)
\(=-x^4+2x^3-x^2+2x-16\)
\(B=-3x^2+3x-3-8+2x^2-6x-6-x^4-x^3\)
\(=-x^4-x^3-x^2-3x-17\)
\(C=-x^4-3x^2+2-5x^2+x^3+3x-3\)
\(=-x^4+x^3-8x^2+3x-1\)
Từ đó có:
\(A-B=-x^4+2x^3-x^2+2x-16-\left(-x^4-x^3-x^2-3x-17\right)\)
\(=-x^4+2x^3-x^2+2x-16+x^4+x^3+x^2+3x+17\)\(=3x^3+5x+1\)
\(B-C=-x^4-x^3-x^2-3x-17-\left(-x^4+x^3-8x^2+3x-1\right)\)
\(=-x^4-x^3-x^2-3x-17+x^4-x^3+8x^2-3x+1\)
\(=-2x^3+7x^2-6x-16\)
\(C-A=-x^4+x^3-8x^2+3x-1-\left(-x^4+2x^3-x^2-2x-16\right)\)
\(=-x^4+x^3-8x^2+3x-1+x^4-2x^3+x^2+2x+16\)
\(=-x^3-7x^2+5x+15\)
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!!!