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Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
a. \(6^2:4.3+2.5^2\)
= \(36:12+2.25\)
= \(3+50\)
=\(53\)
b. \(2.\left(5.4^2-18\right)\)
= \(2.\left(5.16-18\right)\)
= \(2.\left(80-18\right)\)
= \(2.62\)
= \(124\)
c. \(80:\left\{\left[\left(11-2\right).2\right]+2\right\}\)
\(=80:\left\{\left[9.2\right]+2\right\}\)
\(=80:\left\{18+2\right\}\)
\(=80:20\)
\(=4\)
\(2^5.4=2^5.2^2=2^7\);\(5^4.25=5^4.5^2=5^6\); \(16^3.2^3=\left(2^4\right)^3.2^3=2^{12}.2^3=2^{15}\)
\(625^5:25^7=\left(5^4\right)^5:\left(5^2\right)^7=5^{20}:5^{14}=5^6\); \(25^6:125^3=\left(5^2\right)^6:\left(5^3\right)^3=5^{12}:5^9=5^3\)
\(12^4.3^4=\left(2^2.3\right)^4.3^4=2^8.3^4.3^4=2^8.3^8=6^8\); \(9^6:3^2=\left(3^2\right)^6:3^2=3^{12}:3^2=3^{10}\)
\(2^3.2^4.2=2^8\)
Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
a, \(\left|2x+1\right|=3\)
=> 2x + 1 = 3 hoặc 2x + 1 = -3
=> 2x = 3 - 1 hoặc 2x = -3 - 1
=> 2x = 2 hoặc 2x = -4
=> x = 1 hoặc x = -2
b, \(2\frac{1}{2}x+1\frac{1}{2}=2\frac{2}{3}\)
=> \(\frac{5}{2}x+\frac{3}{2}=\frac{8}{3}\)
=> \(\frac{5}{2}x=\frac{8}{3}-\frac{3}{2}\)
=> \(\frac{5}{2}x=\frac{16-9}{6}\)
=> \(\frac{5}{2}x=\frac{7}{6}\)
=> \(x=\frac{7}{6}:\frac{5}{2}=\frac{7}{6}\cdot\frac{2}{5}=\frac{7}{3}\cdot\frac{1}{5}=\frac{7}{15}\)
c, \(3\cdot5^{x-3}+1=16\)
=> 3 . 5x-3 = 16 - 1
=> 3 . 5x-3 = 15
=> 5x-3 = 15 : 3
=> 5x-3 = 5
=> x - 3 = 5 : 5
=> x - 3 = 1
=> x = 1 + 3 = 4
d, \((x-1)^2=25\)
=> \((x-1)^2=5^2\)
=> x - 1 = 5 hoặc x - 1 = -5
=> x = 6 hoặc x = -4
e, \((-2)^2+\left|3x+1\right|=(-28)\cdot7\)
=> 4 + |3x + 1| = -196
=> |3x + 1| = -196 - 4 = -200
=> |3x + 1| = -200
Không thỏa mãn điều kiện
125(28+72)-25(3^2.4+64)
=125.100-25(9.4+64)
=125.100-25.(36+64)
=125.100-25.100
=12500-2500
=10000
a) \(\left(x-6\right)^3=\left(x-6\right)^2\Leftrightarrow\orbr{\begin{cases}x-6=1\Leftrightarrow x=7\\x-6=0\Leftrightarrow x=6\end{cases}}\)
b) \(\left(7.x-11\right)^3=2^5.5^2+200\)
\(\Leftrightarrow\left(7.x-11\right)^3=800+200\)
\(\Leftrightarrow\left(7.x-11\right)^3=1000\)
\(\Leftrightarrow\left(7.x-11\right)^3=10^3\)
\(\Leftrightarrow7x-11=10\Leftrightarrow7x=21\Leftrightarrow x=3\)
c) \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)
\(\Leftrightarrow3+2^{x-1}=24-\left[4^2-3\right]\)
\(\Leftrightarrow3+2^{x-1}=24-13\)
\(\Leftrightarrow3+2^{x-1}=11\)
\(\Leftrightarrow2^{x-1}=8\Leftrightarrow2^{x-1}=2^3\Leftrightarrow x-1=3\Leftrightarrow x=4\)
a)\(3^x.3=243\Leftrightarrow3^x=81\Leftrightarrow3^x=3^4\Leftrightarrow x=4\)
b) \(2^x.16^2=1024\Leftrightarrow2^x.256=1024\Leftrightarrow2^x=4\Leftrightarrow2^x=2^2\Leftrightarrow x=2\)
c) \(64:4^x=16^8\Leftrightarrow4^x=67108864\Leftrightarrow4^x=4^{13}\Leftrightarrow x=13\)
d) \(2^x=16\Leftrightarrow2^x=2^4\Leftrightarrow x=4\)
1) (x+2)2=36
(x+2)2=62
=>x+2=6 hoặc x+2=-6
x=6-2 hoặc x=-6-2
x=4 hoặc x=-8
2)\(\left(\frac{3}{4}\right)^x=\frac{2^8}{3^4}\)
\(\left(\frac{3}{4}\right)^x:\frac{2^8}{3^4}=0\)
\(\left(\frac{3}{4}\right)^x.\frac{3^4}{2^8}=0\)
không có giá trị x nào thỏa mãn vì \(\left(\frac{3}{4}\right)^x>0;\frac{3^4}{2^8}>0\)
5(x-2)(x+3)=1
5(x-5)(x+3)=50
=>(x-2)(x+3)=0
=>x-2=0 hoặc x+3=0
x=2 hoặc x=-3
(x-2)8=(x-2)6
vì 8 là mũ chẵn nên (x-2)8=(x-2)6 khi:
x-2=0 hoặc x-2=1 hoặc x-2=-1
x=2 hoặc x=1/2 hoặc x=1
a, 13.(x-9)=169
x-9 =169:13
x-9 =13
b, 230+[16+(x,-5)]=315.2\(^3\)
230+[16+(x,-5)]=2520
16+(x,-5)=2520-230
16+(x-5)=2290
x-5 =2290-16
x-5 =2274
x =2274+5
x =2279
c, 13x-3\(^2\)x=2003\(^1\)+1\(^{2016}\)
13x-9x=2004
(13-9)x=2004
4 . x=2004
x=2004:4
x=501
Bài 2:
a. $=58(75+50-25)=58.100=5800$
b. $=27.8+4.9-5=216+36-5=247$
Bài 3:
a.
$(x-38):16=12$
$x-38=16\times 12=192$
$x=192+38 = 230$
b.
$10+2x=4^5:4^3=4^2=16$
$2x=16-10=6$
$x=6:2=3$
c.
$440+2(125-x)=546$
$2(125-x)=546-440=106$
$125-x=106:2=53$
$x=125-53=72$
d.
$(x-15):5+20=22$
$(x-15):5=22-20=2$
$x-15=2\times 5=10$
$x=10+15=25$
e.
$2x-138=2^3.3^2=8.9=72$
$2x=72+138 = 210$
$x=210:2=105$