標題:
secondary MATHS
發問:
1.solve logA(ab)?logB(ab)+logA(a-b)?logB(a-b)=0 the capital letters stand for the base(i.e.A,B),where a,b are not equal to 1 and a>b>0,and leave yr ans. in surd form. 2.logA(b)+logB(a)=5/2....(1) log2(a)+log2(B)=6.......(2) the capital letter A,B and 2 are the bases3 It is given that... 顯示更多 1.solve logA(ab)?logB(ab)+logA(a-b)?logB(a-b)=0 the capital letters stand for the base(i.e.A,B),where a,b are not equal to 1 and a>b>0,and leave yr ans. in surd form. 2.logA(b)+logB(a)=5/2....(1) log2(a)+log2(B)=6.......(2) the capital letter A,B and 2 are the bases 3 It is given that sin^2(x) and cos^2(x) are the roots of the equation x^2+ax+1=0,where a is a real constant. a:Show that a=-1 b:Find the value of 1/sin^2(X)+1/cos^2(x). c:Hence,find the sum to infinity of the series 3+{sin^2(x)+cos^2(x)}+{sin^4(x)+cos^4(x)}+{sin^6(x)+cos^6(x)+......
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最佳解答:
1. general equation: logx(y) = [log y] / [log x] [logA(ab)][logB(ab)] + [logA(a-b)][logB(a-b)] = 0 [log(ab)/logA][log(ab)/logB] + [log(a-b)/logA][log(a-b)/logB] = 0 {[log(ab)]^2 + [log(a-b)]^2} / [logA logB] = 0 [log(ab)]^2 + [log(a-b)]^2 = 0 => ab = 1 and a-b = 1 => (1+b)b = 1 => b^2 + b - 1 = 0 b = [sqrt(5)-1]/2 or -[sqrt(5)+1]/2 a = 2/[sqrt(5)-1] or -2/[sqrt(5)+1] a = [sqrt(5)+1]/2 or -[sqrt(5)-1]/2 2. (what does the question want to find?) 3a. x^2 + ax + 1 = 0 as [sin x]^2 and [cos x]^2 are the roots of equation above sum of roots: [sin x]^2 + [cos x]^2 = -a = 1 => a = -1 3b. product of roots: [sin x]^2 [cos x]^2 = 1 1/[sin x]^2 + 1/[cos x]^2 = {[sin x]^2 + [cos x]^2} / {[sin x]^2 [cos x]^2} = 1 3c. 3+{sin^2(x)+cos^2(x)}+{sin^4(x)+cos^4(x)}+{sin^6(x)+cos^6(x)+...... rearrange 1 + [1 + (sinx)^2 + (sinx)^4 + (sinx)^6 +...] + [1 + (cosx)^2 + (cosx)^4 + (cosx)^6 +...] [1 + (sinx)^2 + (sinx)^4 + (sinx)^6 +...] = 1/[1-(sinx)^2] = 1/(cosx)^2 [1 + (cosx)^2 + (cosx)^4 + (cosx)^6 +...] = 1/[1-(cosx)^2] = 1/(sinx)^2 3+{sin^2(x)+cos^2(x)}+{sin^4(x)+cos^4(x)}+{sin^6(x)+cos^6(x)+...... = 1 + [1/(cosx)^2 + 1/(sinx)^2] = 2
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