\(4x^2-25+\left(2x+7\right).\left(5-2x\right)\)

\(=\left(2x+5\right).\left(2x-5\right)-\left(2x+7\right).\left(2x-5\right)\)

\(=\left(2x+5-2x-7\right).\left(2x-5\right)\)

\(=-2.\left(2x-5\right)\)

\(a^2x^2-a^2x^2-b^2x^2+b^2y^2\)

\(=a^2.\left(x^2-y^2\right)-b^2.\left(x^2-y^2\right)\)

\(=\left(a^2-b^2\right).\left(x^2-y^2\right)\)

\(=\left(a-b\right).\left(a+b\right).\left(x-y\right).\left(x+y\right)\)

\(x^2-y^2+12y-36\)

\(=x^2-\left(y^2-12y+36\right)\)

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

\(=\left(x-y+6\right).\left(x+y-6\right)\)

\(\left(x+2\right)^2-x^2+2x-1\)

\(=\left(x+2\right)^2-\left(x^2-2x+1\right)\)

\(=\left(x+2\right)^2-\left(x-1\right)^2\)

\(=[x+2-\left(x-1\right)].[x+2+\left(x-1\right)]\)

\(=\left(x+2-x+1\right).\left(x+2+x-1\right)\)

\(=3.\left(2x+1\right)\)

\(16x^2-y^2=\left(4x\right)^2-y^2=\left(4x-y\right).\left(4x+y\right)\)

\(1+27x^3=1^3+\left(3x\right)^3=\left(1+3x\right).\left(1-3x+9x^2\right)\)

chỗ AB < AC kia các bn chuyển thành chữ " và " nhé. Mk viết nhầm

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

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

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