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16 tháng 8
Ta có: \(\frac{A}{10^{10}}=\frac{10^{20}-6}{10^{20}-6\cdot10^{10}}=\frac{10^{20}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}\)
\(\frac{B}{10^{10}}=\frac{10^{21}-6}{10^{21}-6\cdot10^{10}}=\frac{10^{21}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)
Ta có: \(10^{20}<10^{21}\)
=>\(10^{20}-6\cdot10^{10}<10^{21}-6\cdot10^{10}\)
=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)
=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}+1>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}+1\)
=>\(\frac{A}{10^{10}}>\frac{B}{10^{10}}\)
=>A>B
23 tháng 8
Bài 8:
a: \(5^3=125;3^5=243\)
mà 125<243
nên \(5^3<3^5\)
b: \(7\cdot2^{13}<8\cdot2^{13}=2^3\cdot2^{13}=2^{16}\)
c: \(27^5=\left(3^3\right)^5=3^{3\cdot5}=3^{15}\)
\(243^3=\left(3^5\right)^3=3^{5\cdot3}=3^{15}\)
Do đó: \(27^5=243^5\)
d: \(625^5=\left(5^4\right)^5=5^{4\cdot5}=5^{20}\)
\(125^7=\left(5^3\right)^7=5^{3\cdot7}=5^{21}\)
mà 20<21
nên \(625^5<125^7\)
Bài 9:
a: \(3^{x}\cdot5=135\)
=>\(3^{x}=\frac{135}{5}=27=3^3\)
=>x=3(nhận)
b: \(\left(x-3\right)^3=\left(x-3\right)^2\)
=>\(\left(x-3\right)^3-\left(x-3\right)^2=0\)
=>\(\left(x-3\right)^2\cdot\left\lbrack\left(x-3\right)-1\right\rbrack=0\)
=>\(\left(x-3\right)^2\cdot\left(x-4\right)=0\)
=>\(\left[\begin{array}{l}x-3=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\left(nhận\right)\\ x=4\left(nhận\right)\end{array}\right.\)
c: \(\left(2x-1\right)^4=81\)
=>\(\left[\begin{array}{l}2x-1=3\\ 2x-1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=4\\ 2x=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-1\left(loại\right)\end{array}\right.\)
d: \(\left(5x+1\right)^2=3^2\cdot5+76\)
=>\(\left(5x+1\right)^2=9\cdot5+76=45+76=121\)
=>\(\left[\begin{array}{l}5x+1=11\\ 5x+1=-11\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=10\\ 5x=-12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-\frac{12}{5}\left(loại\right)\end{array}\right.\)
e: \(5+2^{x-3}=29-\left\lbrack4^2-\left(3^2-1\right)\right\rbrack\)
=>\(2^{x-3}+5=29-\left\lbrack16-9+1\right\rbrack\)
=>\(2^{x-3}+5=29-8=21\)
=>\(2^{x-3}=16=2^4\)
=>x-3=4
=>x=4+3=7(nhận)
f: \(3+2^{x-1}=24-\left\lbrack4^2-\left(2^2-1\right)\right\rbrack\)
=>\(2^{x-1}+3=24-\left\lbrack16-4+1\right\rbrack=24-13=11\)
=>\(2^{x-1}=11-3=8=2^3\)
=>x-1=3
=>x=4(nhận)
Bài 6:
a: \(5\cdot5\cdot5\cdot5\cdot5\cdot5=5^6\)
b: \(27\cdot14\cdot7\cdot2=27\cdot14\cdot14=3^3\cdot14^2\)
c: \(x\cdot x\cdot x\cdot y=x^3\cdot y\)
d: \(5^3\cdot5^4=5^{3+4}=5^7\)
e: \(7^8:7^2=7^{8-2}=7^6\)
f: \(42^7:6^7\cdot49=7^7\cdot49=7^7\cdot7^2=7^{7+2}=7^9\)
16 tháng 8
Bài 5:
a: \(37\cdot146+46\cdot2-46\cdot37\)
\(=37\left(146-46\right)+46\cdot2\)
\(=37\cdot100+92=3700+92=3792\)
b: \(2\cdot5\cdot71+5\cdot18\cdot2+10\cdot11\)
\(=10\cdot71+10\cdot18+10\cdot11\)
\(=10\left(71+18+11\right)=10\cdot100=1000\)
c: \(135+360+65+40\)
=135+65+360+40
=200+400
=600
d: \(27\cdot75+25\cdot27-450\)
\(=27\left(75+25\right)-450\)
=2700-450
=2250
Bài 4:
a: \(32\cdot163+32\cdot837\)
\(=32\cdot\left(163+837\right)\)
\(=32\cdot1000=32000\)
b: \(2\cdot3\cdot4\cdot5\cdot25=2\cdot5\cdot4\cdot25\cdot3=3\cdot10\cdot100=3000\)
c: \(25\cdot27\cdot4=27\cdot100=2700\)
Bài 3:
a: \(128\cdot19+128\cdot41+128\cdot40\)
\(=128\cdot\left(19+41+40\right)=128\cdot100=12800\)
b: \(375+693+625+307\)
=375+625+693+307
=1000+1000
=2000
c: \(37+42-37+22\)
=37-37+42+22
=0+64
=64
d: \(21\cdot32+21\cdot68\)
\(=21\cdot\left(32+68\right)=21\cdot100=2100\)
Bài 2:
a: \(17\cdot85+15\cdot17-120\)
\(=17\left(85+15\right)-120\)
=1700-120
=1580
b: \(189+73+211+127\)
=189+211+73+127
=400+200
=600
c: \(38\cdot73+27\cdot38\)
\(=38\left(73+27\right)=38\cdot100=3800\)
Bài 1:
a: \(28\cdot76+23\cdot28-28\cdot13\)
\(=28\left(76+23-13\right)=28\cdot86=2408\)
b: \(39\cdot50+25\cdot39+75\cdot61\)
\(=39\left(50+25\right)+75\cdot61\)
\(=39\cdot75+75\cdot61=75\left(39+61\right)=75\cdot100=7500\)
c: \(32\cdot163+837\cdot32\)
\(=32\left(163+837\right)=32\cdot1000=32000\)
d: \(63+118+37+82\)
=63+37+118+82
=100+200
=300
20 tháng 8
c: \(\left(x-1\right)^3=\left(-9\right)^3\)
=>x-1=-9
=>x=-9+1=-8
f: \(3x-2^3=7+\left(-9\right)\)
=>3x-8=7-9=-2
=>3x=-2+8=6
=>x=2
23 tháng 8
bài 3:
a: \(C=5+5^2+5^3+\cdots+5^{20}\)
\(=5\left(1+5+5^2+\cdots+5^{19}\right)\) ⋮5
b: \(C=5+5^2+5^3+\cdots+5^{20}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{19}+5^{20}\right)\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdots+5^{19}\left(1+5\right)\)
\(=6\left(5+5^3+\cdots+5^{19}\right)\) ⋮6
c: \(C=5+5^2+5^3+\cdots+5^{20}\)
\(=\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+\cdots+\left(5^{17}+5^{18}+5^{19}+5^{20}\right)\)
\(=5\left(1+5+5^2+5^3\right)+5^5\left(1+5+5^2+5^3\right)+\cdots+5^{17}\left(1+5+5^2+5^3\right)\)
\(=\left(1+5+5^2+5^3\right)\left(5+5^5+\cdots+5^{17}\right)=156\cdot\left(5+5^5+\cdots+5^{17}\right)\)
\(=13\cdot12\cdot\left(5+5^5+\cdots+5^{17}\right)\) ⋮13
Bài 2:
a: \(B=3+3^2+3^3+\cdots+3^{120}\)
\(=3\left(1+3+3^2+3^3+\cdots+3^{119}\right)\) ⋮3
b: \(B=3+3^2+3^3+\cdots+3^{120}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\cdots+\left(3^{119}+3^{120}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+\cdots+3^{119}\left(1+3\right)\)
\(=4\left(3+3^3+\cdots+3^{119}\right)\) ⋮4
c: \(B=3+3^2+3^3+\cdots+3^{120}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{118}+3^{119}+3^{120}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+\cdots+3^{118}\left(1+3+3^2\right)\)
\(=13\left(3+3^4+\cdots+3^{118}\right)\) ⋮13
Bài 1:
a: \(A=2+2^2+2^3+\ldots+2^{20}\)
\(=2\left(1+2+2^2+\cdots+2^{19}\right)\) ⋮2
b: \(A=2+2^2+2^3+\ldots+2^{20}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{19}+2^{20}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{19}\left(1+2\right)\)
\(=3\left(2+2^3+\cdots+2^{19}\right)\) ⋮3
c: \(A=2+2^2+2^3+\ldots+2^{20}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\cdots+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{17}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+\ldots+2^{17}\right)=5\cdot3\cdot\left(2+2^5+\cdots+2^{17}\right)\) ⋮5
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 8
Bài 1:
a; A = 2 + \(2^2\) + 2\(^3\) + ... + 2\(^{20}\)
A = 2 x (1+ 2+ 2\(^2\) + ... + 2\(^{19}\))
A ⋮ 2(đpcm)
b; A = 2 + \(2^2\) + 2\(^3\) + ... + 2\(^{20}\)
Xét dãy số: 1; 2;...; 20 đây là dãy số cách đều với khoảng cách là:
2 - 1 = 1
Số số hạng của dãy số trên là:
(20 - 1) : 1+ 1 = 20(số)
Vì 20 : 2 = 10
Vậy nhóm hai số hạng liên tiếp của A vào nhau khi đó ta có:
A = (2+ 2\(^2\)) + (2\(^3\) + 2\(^4\)) + ... + (2\(^{19}+\) 2\(^{20}\))
A = 2.(1 + 2) + 2\(^3\).(1+ 2) + ... + 2\(^{19}\) .(1 + 2)
A = 2.3 + 2\(^3\).3 + ... + 2\(^{19}\).3
A = 3.(2+ 2\(^3\) + ... + 2\(^{19}\))
A ⋮ 3 (đpcm)
c; A = 2 + \(2^2\) + 2\(^3\) + ... + 2\(^{20}\)
Xét dãy số: 1; 2; 3;...; 20
Dãy số trên có 20 số hạng:
Vì 20 : 4 = 5
Vậy nhóm 4 hạng tử của A thành một nhóm khi đó:
A = (2+ 2\(^2\) + 2\(^3\) + 2\(^4\)) + ... + (2\(^{17}+2^{18}+2^{19}+2^{20}\))
A = 2.(1 + 2 + 2\(^2\) + 2\(^3\)) + ... + 2\(^{17}\).(1 + 2 + 2\(^2\) + 2\(^3\))
A = (1+ 2 +2\(^2\) + 2\(^3\)).(2+ ...+ 2\(^{17}\))
A = (1 + 2 + 4 + 8).(2+ ...+ 2\(^{17}\))
A = (3+ 4 + 8).(2+ ...+ 2\(^{17}\))
A = (7 + 8)(2+ ...+ 2\(^{17}\))
A = 15.(2+ ...+ 2\(^{17}\))
A ⋮ 5(đpcm)
Ta có: \(10A=\frac{10^{21}-60}{10^{21}-6}=\frac{10^{21}-6-54}{10^{21}-6}=1-\frac{54}{10^{21}-6}\)
\(10B=\frac{10^{22}-60}{10^{22}-6}=\frac{10^{22}-6-54}{10^{22}-6}=1-\frac{54}{10^{22}-6}\)
Ta có: \(10^{21}-6<10^{22}-6\)
=>\(\frac{54}{10^{21}-6}>\frac{54}{10^{22}-6}\)
=>\(-\frac{54}{10^{21}-6}<-\frac{54}{10^{22}-6}\)
=>\(-\frac{54}{10^{21}-6}+1<-\frac{54}{10^{22}-6}+1\)
=>10A<10B
=>A<B