1+1=2 nha bạn!

ủa bạn học lớp 8 là có phép tính hay là sao?

2 nha
~HT~

24. 

Ta có: \(3k^2+3k+1=k^3+3k^2+3k+1-k^3=\left(k+1\right)^3-k^3\)

Do đó \(a_k=\frac{\left(k+1\right)^3-k^3}{\left(k^2+k\right)^3}=\frac{\left(k+1\right)^3-k^3}{k^3.\left(k+1\right)^3}=\frac{1}{k^3}-\frac{1}{\left(k+1\right)^3}\)

Áp dụng ta được: 

\(P=a_1+a_2+...+a_9\)

\(=\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{2^3}-\frac{1}{3^3}+...+\frac{1}{9^3}-\frac{1}{10^3}\)

\(=1-\frac{1}{10^3}=\frac{999}{1000}\)

23. Ta có: 

\(B=\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{\left(6^2-1\right)}.....\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)

\(=\frac{1.1.3.3.5.5.....\left(2n+1\right)\left(2n+1\right)}{\left(1.3\right).\left(3.5\right).\left(5.7\right).....\left[\left(2n+1\right)\left(2n+3\right)\right]}\)

\(=\frac{\left[1.3.5.....\left(2n+1\right)\right].\left[1.3.5.....\left(2n+1\right)\right]}{\left[1.3.5.....\left(2n+1\right)\right].\left[3.5.7.....\left(2n+3\right)\right]}\)

\(=\frac{1}{2n+3}\)

a,Do AE////DC nên FEFDFEFD=AFFCAFFC(1)

Do AD////CG nên FDFGFDFG=AFFGAFFG(2)

Từ (1) và 2 ) =>

FDFGFDFG=FEFDFEFD

=>FD²=EF .FG

b,Ta có E là trung điểm của AB nên AE=EB=18
Ta có: AE^2+AD^2=DE^2
nên 1872=DE^2 nên DE=30
lại có t/g AED=T/G BEG(G.C.G) nên EG=DE nên DG=2DE=60

\(abc=a+b+c\Leftrightarrow\frac{abc}{abc}=\frac{a+b+c}{abc}\)

\(\Leftrightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=Q\)

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\Rightarrow P=3^2-2Q=9-2=7\)

a) \(18x^2y-12x^3\)

\(=6x^2.3y-6x^2.2x\)

\(=6x^2\left(3y-2x\right)\)

b) \(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(5x+3\right)\left(x-y\right)\)

ta có :

\(P=a^3+b^3+ab=\left(a+b\right)^3-3ab\left(a+b\right)+ab\)

\(=1-2ab\ge1-\frac{\left(a+b\right)^2}{2}=\frac{1}{2}\)

dấu bằng xảy ra khi \(a=b=\frac{1}{2}\)

\(P=a^3+b^3+ab\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+ab\)

\(=1-3ab+ab=1-2ab\)

\(\ge1-\frac{\left(a+b\right)^2}{2}=\frac{1}{2}\)

a, mx^2+ my - nx^2-ny

=>(mx^2+my)-(nx^2+ny)

=>m(x^2+y)-n(x^2+y)

=>(x^2+y)(m-n)

TL

Biến đổi được mx² + tny + nx² - ny = ( x² + y ) ( m -n)

HT