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\(A=\left[-a^5.\left(-a^5\right)\right]^2+\left[-a^2.\left(-a^2\right)\right]^5=0\)O
=>\(\left(-a^{10}\right)^2+\left(-a^4\right)^5=a^{20}-a^{20}=0\)
\(B;\left(-1\right)^n.a^{a+k}=\left(-a\right)^n.a^k\)
\(=\left(-1\right)^n.a^n.a^k=\left(-1.a\right)^n.a^k\)
=\(\left(-a^n\right).a^k\)
Sai rồi thê này nè
a/ \(\frac{1}{a\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\)
Ta co: \(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)
b/ \(\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
Ta co: \(\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}=\frac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\frac{2}{a\left(a+1\right)\left(a+2\right)}\)
Ta có: VP = \(a\left(b^2-2bc+c^2\right)+b\left(c^2-2ac+a^2\right)+c\left(a^2-2ab+b^2\right)\)
= \(ab^2+ac^2+bc^2+ba^2+ca^2+cb^2-6abc\)(1)
\(VT=\left(ab+b^2+ac+bc\right)\left(c+a\right)-8abc\)
\(=abc+b^2c+ac^2+bc^2+a^2b+b^2a+a^2c+abc-8abc\)
= \(ab^2+ac^2+bc^2+ba^2+ca^2+cb^2-6abc\)(2)
Từ (1) ; (2) => VT = VP
Vậy đẳng thức luôn đúng.
Giải:
a) Biến đổi VP, ta có:
\(\dfrac{1}{a}-\dfrac{1}{a+1}\)
\(=\dfrac{1.\left(a+1\right)}{a.\left(a+1\right)}-\dfrac{a.1}{a.\left(a+1\right)}\)
\(=\dfrac{a+1}{a.\left(a+1\right)}-\dfrac{a}{a.\left(a+1\right)}\)
\(=\dfrac{a+1-a}{a.\left(a+1\right)}\)
\(=\dfrac{1}{a.\left(a+1\right)}\) (đpcm)
b) Biến đổi VP, ta được:
\(\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{1\left(a+2\right)}{a\left(a+1\right)\left(a+2\right)}-\dfrac{1.a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{a+2}{a\left(a+1\right)\left(a+2\right)}-\dfrac{a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{a+2-a}{a\left(a+1\right)\left(a+2\right)}\)
\(=\dfrac{2}{a\left(a+1\right)\left(a+2\right)}\) (đpcm)
Chúc bạn học tốt!!!
Ta có: \(\left(-1\right)^n\cdot a^{n+k}\)
\(=\left(-1\right)^n\cdot a^n\cdot a^k\)
\(=\left(-1\cdot a\right)^n\cdot a^k\)
\(=\left(-a\right)^n\cdot a^k\)(đpcm)