A=1+3+3^2+3^3+...+3^100
B=1 phần 2+(1 phần 2)^2+(1 phần 2)^3+...+(1 phần 2)^99
C=5^100-5^99+5^98-5^97+...+5^2-5+1
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VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
26 tháng 8 2019
\(A=1+3+3^2+3^3+...+3^{99}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A-A=2A=\left(3+3^2+3^3+...+3^{100}\right)-\left(\text{}\text{}\text{}1+3^2+3^3+...+3^{99}\right)\)
\(\Rightarrow2A=3^{100}-1\Rightarrow A=\frac{3^{100}-1}{2}\)
N
26 tháng 8 2019
\(DK:x\ge2\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{3x-5}=x-2\)
\(\Leftrightarrow4x-6+2\sqrt{\left(x-1\right)\left(3x-5\right)}=x^2-4x+4\)
\(\Leftrightarrow2\sqrt{3x^2-8x+5}=x^2-8x+10\)
\(\Leftrightarrow4\left(3x^2-8x+5\right)=x^4+64x^2+100-16x^3-160x+20x^2\)
\(\Leftrightarrow12x^2-32x+20=x^4-16x^3+84x^2-160x+100\)
\(\Leftrightarrow x^4-16x^3+72x^2-128x+80=0\)
\(\Leftrightarrow\left(x^4-10x^3\right)-\left(6x^3-60x^2\right)+\left(12x^2-120x\right)-\left(8x-80\right)=0\)
\(\Leftrightarrow x^3\left(x-10\right)-6x^2\left(x-10\right)+12x\left(x-10\right)-8\left(x-10\right)=0\)
\(\Leftrightarrow\left(x-10\right)\left(x^3-6x^2+12x-8\right)=0\)
\(\Leftrightarrow\left(x-10\right)\left(x-2\right)^3=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=10\left(n\right)\\x=2\left(n\right)\end{cases}}\)
Vay PT co 2 nghiem \(x=10,x=2\)
NT
0
NP
1
26 tháng 8 2019
Bạn tham khảo link này nha !
https://olm.vn/hoi-dap/detail/228561471798.html
a
\(A=1+3+3^2+3^3+....+3^{100}\)
\(3A=3+3^2+3^3+3^4+.....+3^{101}\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(B=1-\frac{1}{2^{99}}\)
c
\(C=5^{100}-5^{99}+5^{98}-5^{97}+....+5^2-5+1\)
\(5C=5^{101}-5^{100}+5^{99}-5^{98}+....+5^3-5^2+5\)
\(6C=5^{101}+1\)
\(C=\frac{5^{101}+1}{6}\)
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow\frac{1}{2}B=\)\(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow B-\frac{1}{2}B=\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\right]-\left[\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(\Rightarrow\frac{1}{2}B=\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\Rightarrow B=\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\right].2\)