CMR:
a) 1/a(a+1)=a/1-1/a+1
b)2/a(a+1)(a+2)=1/a(a+1)-1/(a+1)(a+2)
Những câu hỏi liên quan
cho a^1/a^2=a^2/a^3=.......=a^2021/a^2021
cmr:a^1/a^2021=(a^1+a^2+.....+a^2020/a^2+a^3+.....+a^2021)^2020
Ta có \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}=\frac{a+a^2+....+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\frac{a}{a^2}=\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\left(\frac{a}{a^2}\right)^{2020}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)
=> \(\frac{a}{a^2}.\frac{a}{a^2}...\frac{a}{a^2}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(2020 thừa số \(\frac{a}{a^2}\))
=> \(\frac{a}{a^2}.\frac{a^2}{a^3}...\frac{a^{2020}}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(Vì \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}\))
=> \(\frac{a}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(đpcm)
a, Cho A= 1/99 + 2/98 + 3/47 + .......... + 98/2 + 99/1
B= 1/2 + 1/3 + 1/4 + ..........+ 1/99 + 1/100
Tính B/A
b, Cho A= 1/49 + 2/48 + 3/47 +.......+ 48/2 +49/1
B= 1 + 2/3 + 2/4 +......+ 2/49 + 2/50
Tính A/B
a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)
\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B
=>B/A=1/100
b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)
\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)
\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
=>A/B=25
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9 tháng 8 2021 lúc 15:02
cho a,b >0, a+b=1
B= 1/a^2+b^2 + 1/ab + 2ab
C=1/a^2+b^2 + 1/ab + 4ab
D=1/a^2+b^2 + 1/ab + 5ab
CMR:
a,\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{2}\)
b,Cho a+b=1,a>0,b>0 CMR:\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)\(\ge9\)
Chứng minh đẳng thức sau :
a. \(\left[\dfrac{1}{a-1}-\dfrac{2a}{\left(a^2+1\right)\left(a-1\right)}\right]:\dfrac{a^2+a+1}{a^2+1}=\dfrac{a-1}{a^2+a+11}\) VỚI a ≠ 1
b. \(\left(\dfrac{1-x^3}{1-x}-x\right):\dfrac{1+x}{1-x-x^2+x^3}=\left(1-x^2\right)\left(1+x^2\right)\)
Câu a bạn sửa lại đề 11→1
\(a,VT=\dfrac{a^2-2a+1}{\left(a-1\right)\left(a^2+1\right)}\cdot\dfrac{a^2+1}{a^2+a+1}\\ =\dfrac{\left(a-1\right)^2}{\left(a-1\right)\left(a^2+a+1\right)}=\dfrac{a-1}{a^2+a+1}=VP\)
\(b,=\left[\dfrac{\left(1-x\right)\left(x^2+x+1\right)}{1-x}-x\right]\cdot\dfrac{\left(1+x\right)\left(1-x^2\right)}{1+x}\\ =\dfrac{\left(x^2+1\right)\left(1+x\right)\left(1-x^2\right)}{1+x}=\left(x^2+1\right)\left(1-x^2\right)=VP\)
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Cho a,b,c khác 0 thỏa mãn a\(\left(\dfrac{1}{c}+\dfrac{1}{b}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-2\)
a(1b+1c)+b(1c+1a)+c(1a+1b)=−2
và a3+b3+c3=1. CMR \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Cho a,b,c \(\ge1.CMR:a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)+2\left(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{1+c^2}\right)\ge9\)
Chính bài của em:
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chứng minh a^2n+1+b^2n+1=(a+b)(a^2n-1b+a^2n-1b^2)
cho a,b,c>0 tmdk 1/a+1/b+1/c<=3.cmr:a/1+b^2+b/1+c^2+c/1+a^2+1/2(ab+bc+ca)>+3
cho A=1+((2x^2+x-1/1-x)-(2x^3-x+x^2/1-x^3):2x-1/x^2-x) a)Hãy rút gọn A b)tính x để A=5 c)cmr:A>2/3