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`a, 3/4 - 5/4 :(x-1) =1/2`
`=> 5/4:(x-1)= 3/4 -1/2`
`=> 5/4:(x-1)= 3/4 - 2/4`
`=> 5/4:(x-1)= 1/4`
`=> x-1= 5/4 : 1/4`
`=> x-1=5`
`=>x=5+1`
`=>x=6`
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`(1/2-x)^2 -2^2 =12`
`=> (1/2-x)^2 = 12+4`
`=> (1/2-x)^2= 16`
`=> (1/2-x)^2 =4^2`
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=4\\\dfrac{1}{2}-x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
__
`(1/2)^(2x-1) =1/16`
`=> (1/2)^(2x-1) = (1/2)^4`
`=> 2x-1=4`
`=> 2x=4+1`
`=>2x=5`
`=>x=5/2`
\(a,\dfrac{3}{4}-\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{2}\)
\(\dfrac{5}{4}:\left(x-1\right)=\dfrac{3}{4}-\dfrac{1}{2}\)
\(\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{4}\)
\(x-1=\dfrac{5}{4}:\dfrac{1}{4}\)
\(x-1=5\)
\(x=6\)
\(\left(\dfrac{1}{2}-x\right)^2-2^2=12\)
\(\left(\dfrac{1}{2}-x\right)^2-4=12\)
\(\left(\dfrac{1}{2}-x\right)^2=16\)
\(\left(\dfrac{1}{2}-x\right)^2=4^2hoặc\left(\dfrac{1}{2}-x\right)^2=\left(-4\right)^2\)
\(\dfrac{1}{2}-x=4hoặc\dfrac{1}{2}-x=-4\)
=>1/2 -x =4 1/2 -x= -4
=> x=1/2-4 x=1/2-(-4)
=>x=-7/2 x=9/2
vậy x∈{-7/2 ; 9/2}
\(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{16}\)
\(=>\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^4\)
\(=>2x-1=4\)
\(=>2x=5\)
\(=>x=\dfrac{5}{2}\)
\(a,\dfrac{3}{2}\cdot x-1=\dfrac{1}{2}x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{1}{2}x=-\dfrac{3}{5}+1\)
\(\Rightarrow\left(\dfrac{3}{2}-\dfrac{1}{2}\right)x=-\dfrac{3}{5}+\dfrac{5}{5}\)
\(\Rightarrow x=\dfrac{2}{5}\)
\(b,\dfrac{1}{2}x+\dfrac{1}{2}\left(x-2\right)=\dfrac{3}{4}-2x\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{1}{2}x+2x-1=\dfrac{3}{4}\)
\(\Rightarrow\left(\dfrac{1}{2}+\dfrac{1}{2}+2\right)x=\dfrac{3}{4}+1\)
\(\Rightarrow3x=\dfrac{7}{4}\)
\(\Rightarrow x=\dfrac{7}{4}:3\)
\(\Rightarrow x=\dfrac{7}{12}\)
\(c,\left(x-\dfrac{1}{2}\right)-\dfrac{1}{4}=0\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{2}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
\(d,4^{x-3}+1=17\)
\(\Rightarrow4^{x-3}=17-1\)
\(\Rightarrow4^{x-3}=16\)
\(\Rightarrow4^{x-3}=4^2\)
\(\Rightarrow x-3=2\)
\(\Rightarrow x=2+3\)
\(\Rightarrow x=5\)
#Toru
`3/2 x -1 =1/2x -3/5`
`=> 3/2x -1/2x = -3/5 +1`
`=> 2/2x= -3/5 + 5/5`
`=> x= 2/5`
__
`1/2x +1/2(x-2) = 3/4 -2x`
`=> 1/2x + 1/2x - 2/2 = 3/4 -2x`
`=> 1/2x +1/2x +2x = 3/4 + 1`
`=> 1/2x +1/2x + 4/2x = 3/4 +4/4`
`=> 6/2x = 7/4`
`=> x= 7/4 : 3`
`=>x=7/12`
__
`(x-1/2) -1/4=0`
`=> x-1/2=1/4`
`=> x=1/4 +1/2`
`=> x= 1/4 +2/4`
`=>x=3/4`
__
`4^(x-3) +1=17`
`=> 4^(x-3) =17-1`
`=> 4^(x-3)=16`
`=> 4^(x-3)=4^2`
`=> x-3=2`
`=>x=2+3`
`=>x=5`
b) ( x - 4 )2 = ( x - 4 )4
( x - 4 )2 = ( x - 42 )2
=> ( x - 4 )2 = ( x - 16 )2
=> x - 4 = x - 16
=> x = 22 . 42 = 22 . ( 22 )2 = 22 . 24 = 26 = 64
=> x = 64
a) \(\left(x-1\right)^3=125\)
\(\Leftrightarrow\)\(\left(x-1\right)^3=5^3\)
\(\Leftrightarrow\)\(x-1=5\)
\(\Leftrightarrow\) \(x=5+1\)
\(\Leftrightarrow\) \(x=6\)
Vậy \(x=6\)
b) chưa ra - hihi ^^
\(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{8}{x}\)
\(\Rightarrow x^2=8\cdot2\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x^2=4^2\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Muốn tạo số chia hết cho 4 thì 2 chữ số tận cùng phải chia hết cho 4
Gọi các số cần tìm có dạng \(\overline{abc}\left(a,b,c\in N;0< a< 10;0\le b,c< 10\right)\)
Mà \(\overline{abc}⋮4\Rightarrow\overline{bc}\in\left\{00;04;12;16;20;24;40;44;60;64\right\}\)
Với mỗi cặp \(\overline{bc}\) ta có \(a\in\left\{1;2;4;6\right\}\left(4\text{ cách chọn}\right)\)
Vậy có thể tạo \(4\cdot10=40\) số thỏa yêu cầu đề
Nhân vô rồi chuyển dấu lên và nhóm nhân -1ra ngoài rồi trg ngoặc là dãy có quy luật giải dãy đó r nhân phá ngoặc
`#hungg`
\(Q\left(x\right)=ax^5+2x^4-2x^5-x^2+6x-3+x^4\\ =\left(ax^5-2x^5\right)+\left(2x^4+x^4\right)-x^2+6x-3\\ =\left(a-2\right)x^5+3z^4-x^2+6x-3\)
Để `Q(x)` có bậc 4 thì \(a-2=0\Rightarrow a=2\)
=>x-2/3=-1/12
hay x=7/12
\(\dfrac{3}{4}-\left(x-\dfrac{2}{3}\right)=\dfrac{5}{6}\\ \Rightarrow x-\dfrac{2}{3}=\dfrac{3}{4}-\dfrac{5}{6}\\ \Rightarrow x-\dfrac{2}{3}=-\dfrac{1}{12}\\ \Rightarrow x=\dfrac{2}{3}-\dfrac{1}{12}\\ \Rightarrow x=\dfrac{7}{12}\)