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a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
Mình chỉ ghj đáp za thôj nên thông cảm nha
b)1953368
c)225
d)32
\(a,=4^{10}.4^{10}.4^{45}\)
\(=4^{65}\)
\(b,=5^9+3^5\)
\(=1953125+243\)
\(=1953368\)
\(c,=1+8+27+64+125\)
\(=225\)
\(d,=32^5:32^4\)
\(=32\)
A, 89-(73-x)=20
73-x =89-20
73-x =69
x = 73-69
x = 4
B, 62-(2x-3):3 = 23
(2x-3):3 = 62-23
(2x-3):3 = 39
2x-3 = 39.3
2x-3 = 117
2x = 117+3
2x = 120
x = 120:2
x = 60
C, 4(x-3) = 72 - 110
4(x-3) = 49 - 1
4(x-3) = 48
x-3 = 48:4
x-3 = 12
x = 12+3
x = 15
D, 32 (x+4)-52= 5.22
9(x+4)-25 = 5.4
9(x+4)-25 = 20
9(x+4) = 20+25
9(x+4) = 45
x+4 = 45:9
x+4 = 5
x = 5-4
x = 1
(Làm có hơi dài dòng nhưng đầy đủ lắm)
a, 89-(73-x)=20 b. 62-( 2x -3): 3=23 c, 4(x-3) = 72-11o d, 32 (x+4)-52 = 5.22
(73-x)=89-20 (2x - 3 ):3 = 62-23 4(x-3)=49-1 32(x+4) - 52= 5.4=20
73-x = 69 (2x - 3): 3= 39 4(x-3) =48 32(x+4)-25 =20
x = 73-69 (2x - 3)= 39: 3 x-3 = 48 :8 32(x+4)=20+25=45
x = 4 2x -3 =13 x-3 = 6 9(x+4)=45
2x= 13 + 3 x=6+3 x+4=45:9
2x= 16 x= 9 x+4=5
x=16:2 x=5-4
x= 8 x=1
chỗ nào ko hiểu cứ hỏi mik nhé
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
1.A=2^2+2^4+...+2^2010
=> 2^2 A= 2^4+2^6+..+2^2012
=> 2^2 A - A=( 2^4+2^6+..+2^2012 ) -(2^2+2^4+...+2^2010 )
=> 3A= 2^2012 -2^2
=> A= (2^2012-2^2)/3
B=3-3^2+3^3-...-3^2010
=>3B= 3^2 -3^3+3^4-...-3^2011
=> 3B + B = (3^2 -3^3+3^4-...-3^2011) +(3-3^2+3^3-...-3^2010)
=> 4B =3-3^2011
=> B= (3-3^2011)/4
2.
A=3+3^2+..+3^100
=> 3A =3^2+3^3+...+3^101
=> 3A- A = (3^2+3^3+...+3^101)-(3+3^2+..+3^100)
=> 2A=3^101 -3
=> 2A+3 =3^101 mà 2A+3 =3^n
=> n=101
\(1,S=3+3^2+3^3+...+3^{20}\)(1)
\(\Rightarrow3S=3^2+3^3+3^4+...+3^{21}\)(2)
Lấy (2) -(1) ta có :
\(\Rightarrow2S=3^{21}-3\)
\(\Rightarrow S=\frac{3^{21}-3}{2}\)
\(3,A=1.2.3+2.3.4+3.4.5+...+\left(n-1\right)n\left(n+1\right)\)
\(\Rightarrow4A=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+\left(n-1\right)n\left(n+1\right)\left[\left(n+2\right)-\left(n-2\right)\right]\)
\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+...+\left(n-1\right)n\left(n+1\right)\left(n+2\right)-\left(n-2\right)\left(n-1\right)n\left(n+1\right)\)
\(\Rightarrow4A=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)