1) (-1/2).(-1/2)^4
2) 6.3^2 - 24:2^3
3) (1/4)^2 . (1/4)^3
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1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
\(2^x:4=32\\ \Rightarrow2^x=128\\ \Rightarrow2^x=2^7\\ \Rightarrow x=7\)
\(3^{x-2}:3=243\\ \Rightarrow3^{x-2}=729\\ \Rightarrow3^{x-2}=3^6\\ \Rightarrow x-2=6\\ \Rightarrow x=8\)
\(256:4^{x+1}=4^2\\ \Rightarrow4^{x+1}=4^2\\ \Rightarrow x+1=2\\ \Rightarrow x=1\)
\(4^{2x-1}:4=4^4\\ \Rightarrow4^{2x-1}=4^5\\ \Rightarrow2x-1=5\\ \Rightarrow x=3\)
\(5^{x-1}:5=5^3\\ \Rightarrow5^{x-1}=5^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
\(3^{2x+1}:3=3^4\\ \Rightarrow3^{2x+1}=3^5\\ \Rightarrow2x+1=5\\ \Rightarrow x=3\)
1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)
=>\(5^{x-2}=16+9=25\)
=>x-2=2
=>x=4
2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)
=>3^x=0
=>x=0
3: \(\Leftrightarrow2^x+2^x\cdot16=272\)
=>2^x*17=272
=>2^x=16
=>x=4
4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)
=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)
=>2^x-1=8
=>x-1=3
=>x=4
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
1, 54 : x - 1 = 5
54 : x = 5+1 = 6
x = 54 : 6 = 9
2, 42 : x + 0 = 8
x = 42 : 8 = 21/4
3, 24 : x - 8 = 0
24 : x = 0 + 8 = 8
x = 24 : 8 = 3
Tk mk nha
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
A= \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{35}+\frac{1}{99}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.6}+...+\frac{2}{9.11}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\)
\(2A=1-\frac{1}{11}=\frac{10}{11}\)
\(A=\frac{10}{11}:2=\frac{5}{11}\)
\(D=\frac{3^2}{1.4}+\frac{3^2}{4.7}+...+\frac{3^2}{13.16}\)
\(D=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{13.16}\right)\)
\(D=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}\right)\)
\(D=3.\left(1-\frac{1}{16}\right)=3.\frac{15}{16}=2\frac{13}{16}\)
1. \(\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^4=\left(-\frac{1}{2}\right)^5=-\frac{1}{32}\)
2. \(6.3^2-24:2^3=6.9-24:8=54-3=51\)
3. \(\left(\frac{1}{4}\right)^2.\left(\frac{1}{4}\right)^3=\left(\frac{1}{4}\right)^5=\frac{1}{1024}\)
1) (-1/2).(-1/2)^4 = ( -1/2)^ 5 = -1/32
2) 6.3^2 - 24:2^3 = 6,9 - 24 : 8 = 54 - 3 = 51
3) (1/4)^2 . (1/4)^3 = ( 1/4)^5 = 1/1024