1/1×2 + 1/2×3 + 1/3×4 +......+ 1/99×100
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ĐK: x \(\ne\) -1
Đặt t = x+1 (t \(\ne\) 0)
=> x = t-1
PT tương đương
(t-1)2 + \(\frac{\left(t-1\right)^2}{t^2}\)= 3
<=> t2 - 2t + 1 + 1 - \(\frac{2}{t}\)+ \(\frac{1}{t^2}\) = 3
<=> t2 + \(\frac{1}{t^2}\)- 2(t + \(\frac{1}{t}\)) = 1
Đặt z = t + \(\frac{1}{t}\) (|z| >= 2)
z2 = t2 + \(\frac{1}{t^2}\)+ 2
PT tương đương
z2 - 2 - 2z = 1
<=> z2 - 2z -3 = 0
<=> z = -1 hoặc z = 3
z = -1 (Ko có t nào thỏa mãn t + \(\frac{1}{t}\)= -1)
z = 3
=> t + \(\frac{1}{t}\) = 3
<=> t2 - 3t + 1 = 0 (vì t \(\ne\) 0)
<=> t = \(\frac{3-\sqrt{5}}{2}\) hoặc t = \(\frac{3+\sqrt{5}}{2}\)
x = t - 1
=> x = \(\frac{1-\sqrt{5}}{2}\) hoặc x = \(\frac{1+\sqrt{5}}{2}\)
\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2\left(z-y+y-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2\left(z-y\right)+y^2\left(y-x\right)+z^2\left(x-y\right)\)
\(=\left[x^2\left(y-z\right)-y^2\left(y-z\right)\right]+\left[y^2\left(y-x\right)-z^2\left(y-x\right)\right]\)
\(=\left(x^2-y^2\right)\left(y-z\right)+\left(y^2-z^2\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(y-z\right)+\left(y-z\right)\left(y+z\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left[\left(x+y\right)-\left(y+z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
\(\frac{1}{1+2a+3ab+4abc}+\frac{2}{2+3b+4bc+bcd}+\frac{3}{3+4c+cd+2acd}+\frac{4}{4+d+2ad+3abd}\)
= \(\frac{1}{1+2a+3ab+4abc}+\frac{2a}{2a+3ab+4abc+abcd}+\frac{3ab}{3ab+4abc+abcd+2abacd}\)
\(+\frac{4abc}{4abc+abcd+2aabcd+3abcabd}\)
= \(\frac{1}{1+2a+3ab+4abc}+\frac{2a}{2a+3ab+4abc+1}+\frac{3ab}{3ab+4abc+1+2a}+\frac{4abc}{4abc+1+2a+3ab}\)
= \(\frac{1+2a+3ab+4abc}{1+2a+3ab+4abc}=1\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=1.\frac{99}{100}\)
\(=\frac{99}{100}\)
lớp 8 hả mk bn kia sai zùi