So sánh:
a)4416 và 3219
b) Tính hợp lí: (22410 : 11210 + 8 . 80) : 23
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Bài 1
a: 11/12=1-1/12
23/24=1-1/24
mà -1/12>-1/24
nên 11/12>23/24
b: -3/20=-9/60
-7/12=-35/60
mà -9>-35
nên -3/20>-7/12
a) Ta có: \(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3}\right)^{5\cdot6}=\left(\dfrac{1}{3}\right)^{30}\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{28}>\left(\dfrac{1}{243}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{3^4}\right)^7>\left(\dfrac{1}{243}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\)
mà \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7\)
nên \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)
\(\left(\dfrac{3}{8}\right)^5\&\left(\dfrac{5}{243}\right)^3\)
\(\left(\dfrac{3}{8}\right)^5=\left(\dfrac{90}{240}\right)^5=\dfrac{90^5}{240^5}\)
\(\left(\dfrac{5}{243}\right)^3=\dfrac{5^3}{243^3}\)
\(=>\dfrac{90^5}{240^5}>\dfrac{5^3}{243^3}\)
\(=>\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)
a) 3^2 và 3.2
3^2=9
3.2=6
-> 3^2>3.2
b)2^3 và 3^2
2^3=8
3^2=9
-> 2^3<3^2
c) 3^3 và 3^4
Vì hai số có cùng cơ số nên ta so sánh số mũ
3<4
-> 3^3<3^4
a)ta có 32=9 ; 3.2=6 => 32 > 3.2
b)ta có 23=8 ; 32=9 => 23 < 32
c) ta có 33 và 34
vì 2 số đều cùng 1 cơ số
mà cơ số đầu có số mũ = 3,cơ số còn lại có lũy thừa =4
=> 3<4
=> 33<34
\(A=1+2+2^2+...+2^{2022}\)
\(\Rightarrow2A=2+2^2+...+2^{2023}\)
\(\Rightarrow2A-A=2^{2023}-1\)
\(\Rightarrow A=2^{2023}-1\)
\(\Rightarrow A< 2^{2023}=2^2.2^{2021}=4.2^{2021}< 5^{2021}\)
\(\Rightarrow A< B\)
a: \(log_2\left(M\cdot N\right)=log_2\left(2^5\cdot2^3\right)=log_2\left(2^8\right)=8\)
\(log_2M+log_2N=log_22^5+log_22^3=5+3=8\)
=>\(log_2\left(MN\right)=log_2M+log_2N\)
b: \(log_2\left(\dfrac{M}{N}\right)=log_2\left(\dfrac{2^5}{2^3}\right)=log_2\left(2^2\right)=2\)
\(log_2M-log_2N=log_22^5-log_22^3=5-3=2\)
=>\(log_2\left(\dfrac{M}{N}\right)=log_2M-log_2N\)
1.
a) \(125+80+375+220\)
\(=\left(125+375\right)+\left(80+220\right)\)
\(=500+300\)
\(=800\)
b) \(25.11.8.4.125\)
\(=\left(25.4\right).\left(8.125\right).11\)
\(=100.1000.11\)
\(=1100000\)
c) \(18.74+18.27-18\)
\(=18.\left(74+27-1\right)\)
\(=18.100\)
\(=1800\)
d) \(75.23+75.77-500\)
\(=75.\left(23+77\right)-500\)
\(=75.100-500\)
\(=7500-500\)
\(=7000\)
e) \(150:\left[25.\left(18-4^2\right)\right]\)
\(=150:\left[25.\left(18-16\right)\right]\)
\(=150:\left[25.2\right]\)
\(=150:50\)
\(=3\)
f) \(125.23.2.8.50\)
\(=\left(125.8\right).\left(2.50\right).23\)
\(=1000.100.23\)
\(=2300000\)
g) \(235+88+165+12\)
\(=\left(235+165\right)+\left(88+12\right)\)
\(=400+100\)
\(=500\)
h) \(71.32+71.68-1100\)
\(=71.\left(32+68\right)-1100\)
\(=71.100-1100\)
\(=7100-1100\)
\(=6000\)
i) \(6^7:6^6+4^3.2-24^0\)
\(=6+64.2-1\)
\(=6+128-1\)
\(=133\)
j) \(546-6.\left[158:\left(30+7^2\right)\right]\)
\(=546-6.\left[158:\left(30+49\right)\right]\)
\(=546-6.\left[158:79\right]\)
\(=546-6.2\)
\(=546-12\)
\(=534\)
k) \(268+147+132+253\)
\(=\left(268+132\right)+\left(147+253\right)\)
\(=400+400\)
\(=800\)
2.
a) \(7x=707\)
\(x=707:7\)
\(x=101\)
b) \(6x+5=47\)
\(6x=47-5\)
\(6x=42\)
\(x=42:6\)
\(x=7\)
c) \(35:\left(x+1\right)=7\)
\(x+1=35:7\)
\(x+1=5\)
\(x=5-1\)
\(x=4\)
d) \(12+\left(29-3x\right)=35\)
\(29-3x=35-12\)
\(29-3x=23\)
\(3x=29-23\)
\(3x=6\)
\(x=6:3\)
\(x=2\)
e) \(2.4^x+10^1=138\)
\(2.4^x=138-10\)
\(2.4^x=128\)
\(4^x=128:2\)
\(4^x=64=4^3\)
\(\Rightarrow x=3\)
f) \(7x=140\)
\(x=140:7\)
\(x=20\)
g) \(7\left(15-x\right)+14=84\)
\(7\left(15-x\right)=84-14\)
\(7\left(15-x\right)=70\)
\(15-x=70:7\)
\(15-x=10\)
\(x=15-10\)
\(x=5\)
h) \(3^x.5-15=390\)
\(3^x.5=390+15\)
\(3^x.5=405\)
\(3^x=405:5\)
\(3^x=81=3^4\)
\(\Rightarrow x=4\)
\(=\dfrac{8}{23}+\dfrac{3}{19}-\dfrac{15}{17}-\dfrac{2}{17}+\dfrac{15}{23}=\dfrac{3}{19}\)
\(\left(\dfrac{8}{23}+\dfrac{3}{19}-\dfrac{15}{17}\right)-\left(\dfrac{2}{17}-\dfrac{15}{23}\right)\\ =\dfrac{8}{23}+\dfrac{3}{19}-\dfrac{15}{17}-\dfrac{2}{17}+\dfrac{15}{23}\\ =\left(\dfrac{8}{23}+\dfrac{15}{23}\right)-\left(\dfrac{15}{17}+\dfrac{2}{17}\right)+\dfrac{3}{19}\\ =1-1+\dfrac{3}{19}\\ =\dfrac{3}{19}\)