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a, x : (-1/2)^3 = -1/2
=> x : (-1/8) = -1/2
=> x = 4
vậy_
b, (3/4)^5.x = (3/4)^7
=> x = (3/4)^7 : (3/4)^5
=> x = (3/4)^2
=> x = 9/16
vậy-
c, (3/5)^8 : x = (-3/5)^6
=> (3/5)^8 : x = (3/5)^6
=> x = (3/5)^8 : (3/5)^6
=> x = (3/5)^2
=> x= 9 /25
1) Tìm số nguyên x, biết :
a) 3x = 94/ 273
3x = 1/3
3x = 3-1
=> x = -1
b) 3x = 98 / 273 . 812
3x = 37.38
3x = 315
=> x = 15
c) 2x - 3 / 410 = 83
2x - 3 = 83.410
2x - 3 = 226
=> x - 3 = 26
=> x = 29
d) 22x - 3 / 410 = 83 . 165
22x - 3 / 410 = 269
22x - 3 = 269 . 410
22x - 3 = 289
=> 2x - 3 = 89
2x = 91
x = 91/2
e) 35 / 3x = 310
3x = 35 : 310
3x = 3-5
=> x = -5
a)\(A=x^2-1\)
\(Nx:\)\(x^2\ge0\)
\(\Rightarrow A_{Min}=0-1=-1\Leftrightarrow x=0\)
b) \(B=x^2-2x+3\)
\(=x\left(x-2\right)+3\)
\(Nx:x\left(x-2\right)\ge0\)
\(\Rightarrow B_{Min}=3\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow x=0\)
c) \(C=\left|2x+1\right|-5\)
\(Nx:\left|2x+1\right|\ge0\Rightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)
\(\Rightarrow C_{Min}=-5\Leftrightarrow x=\frac{-1}{2}\)
d) \(D=3x^2+6x-7\)
\(=3\left(x^2+2x\right)-7\)
\(Nx:Min_{x^2+2x}=-1\Leftrightarrow x=-1\)
\(D_{Min}=-8\Leftrightarrow x=-1\)
\(P\left(x\right)=3x^5+x^4-2x^2+2x-1\)
\(Q\left(x\right)=-3x^5+2x^2-2x+3\)
\(P\left(x\right)+Q\left(x\right)=3x^5+x^4-2x^2+2x-1-3x^5+2x^2-2x+3\)
\(=x^4+2\)
\(P\left(x\right)-Q\left(x\right)=3x^5+x^4-2x^2+2x-1+3x^5-2x^2+2x-3\)
\(=6x^5+x^4-4x^2+4x-4\)
Thu gọn + sắp xếp luôn
P(x) = 3x5 + x4 - 2x2 + 2x - 1
Q(x) = -3x5 + 2x2 - 2x + 3
P(x) + Q(x) = ( 3x5 + x4 - 2x2 + 2x - 1 ) + ( -3x5 + 2x2 - 2x + 3 )
= ( 3x5 - 3x5 ) + x4 + ( 2x2 -- 2x2 ) + ( 2x - 2x ) + ( 3 - 1 )
= x4 + 2
P(x) - Q(x) = ( 3x5 + x4 - 2x2 + 2x - 1 ) - ( -3x5 + 2x2 - 2x + 3 )
= 3x5 + x4 - 2x2 + 2x - 1 + 3x5 - 2x2 + 2x - 3
= ( 3x5 + 3x5 ) + x4 + ( -2x2 - 2x2 ) + ( 2x + 2x ) + ( -1 - 3 )
= 6x5 + x4 - 4x2 + 4x - 4
a) 7x - 2x = 617 : 615 + 44
=> 5x = 36 + 44
=> 5x = 80
=> x = 80 : 5 = 16
b) 9x - 1 = 18 + 1/9 - 1/9 - 9
=> 9x - 1 = 9
=> x - 1 = 1
=> x = 1 + 1 = 2
c) [(6x - 39) : 7] . 4 = 12
=> (6x - 39) : 7 = 12 : 4
=> (6x - 39) : 7 = 3
=> 6x - 39 = 3.7
=> 6x - 39 = 21
=> 6x = 21 + 39
=> 6x = 60
=> x = 60 : 6
=> x = 10
d) 2 - (x - 1) - 3x = 20
=> 2 - x + 1 - 3x = 20
=> 3 - 4x = 20
=> 4x = 3 - 20
=> 4x = -17
=> x = -17 : 4 = -17/4
e) 2|x - 3| + 7 = 56 : 52
=> 2|x - 3| + 7 = 625
=> 2|x - 3| = 625 - 7
=> 2|x - 3| = 618
=> |x - 3| = 618 : 2
=> |x - 3| = 309
=> \(\orbr{\begin{cases}x-3=309\\x-3=-309\end{cases}}\)
=> \(\orbr{\begin{cases}x=312\\x=-306\end{cases}}\)
a ) \(\frac{3^5}{27}=\frac{3^5}{3^3}=\frac{3^3.3^2}{3^3}=3^2=9\)
b ) \(\frac{4^7}{64}=\frac{4^7}{4^3}=\frac{4^3.4^4}{4^3}=4^4=256\)
c ) \(\frac{x^{13}}{x^5}=\frac{x^5.x^8}{x^5}=x^8\)
d ) \(\frac{x^{19}}{x^{18}}=\frac{x^{18}.x}{x^{18}}=x\)
e ) \(\frac{2.x^{10}}{x^7}=\frac{2.\left(x^7.x^3\right)}{x^7}=2.x^3\)
Bài 1:
a)
\(\dfrac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\\ =\dfrac{\left(2^2\right)^2\cdot\left(5^2\right)^2+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^{2\cdot2}\cdot5^{2\cdot2}+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^4\cdot5^4}{2^3\cdot5^2}+\dfrac{2^5\cdot5^3}{2^3\cdot5^2}\\ =2\cdot5^2+2^2\cdot5\\ =2\cdot25+4\cdot5\\ =50+20\\ =70\)
c)
\(\dfrac{\left(1-\dfrac{4}{9}-2\right)\cdot16}{\left(2-3\right)^{-2}}+12\\ =\dfrac{\left(\dfrac{9}{9}-\dfrac{4}{9}-\dfrac{18}{9}\right)\cdot16}{\left(-1\right)^{-2}}+12\\ =\dfrac{\dfrac{-13}{9}\cdot16}{\dfrac{1}{\left(-1\right)^2}}+12\\ =\dfrac{\dfrac{-208}{9}}{1}+12\\ =\dfrac{-208}{9}+12\\ =\dfrac{-208}{9}+\dfrac{108}{9}\\ =\dfrac{100}{9}\)
Bài 2:
a)
\(\left(x+2\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
b)
\(\left(1,78^{2x-2}-1,78^x\right):1,78^x=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-\dfrac{1,78^x}{1,78^x}=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-1=0\\ \Leftrightarrow \dfrac{1,78^{2x-2}}{1,78^x}=1\\ \Leftrightarrow1,78^{2x-2}=1,78^x\\ \Leftrightarrow2x-2=x\\ \Leftrightarrow2x-x=2\\ \Leftrightarrow x=2\)
d) \(5^{\left(x-2\right)\left(x+3\right)}=1\)
\(\Rightarrow5^{\left(x-2\right)\left(x+3\right)}=5^0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(x_1=-3;x_2=2\)
a) (x-1)5=-32
(x-1)5=-25
=>x-1=2
x=2+1
x=3
c) (2x-1)6=(2x-1)8
=> 2x-1 \(\in\left\{1;0\right\}\)
=>2x \(\in\left\{2;1\right\}\)
=> x= 1
d) 5x+5x+2=650
2.5x+2=650
2.5x=650-2
2.5x=648
2.5x= 23.34
=> x \(\in\phi\)