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Cho mk ké chút nha
13/10 nhân 4/9 cộng 4/15 nhân 7/18 trừ 4/9 nhân 2 nhân 1/3
\(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(=\frac{2^2}{2}.\frac{3^2}{8}.....\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(=\frac{2.2.3.3.....\left(n+1\right)\left(n+1\right)}{1.3.2.4.....n\left(n+1\right)}\)
\(=\frac{2.3....\left(n+1\right)}{1.2.3....n}.\frac{2.3...\left(n+1\right)}{3.4.5....\left(n+2\right)}\)
\(=\left(n+1\right)\frac{2}{n+2}\)
\(=\frac{2n+2}{n+2}\)
Câu hỏi của Nghĩa Nguyễn - Toán lớp 9 - Học toán với OnlineMath
Đặt \(A=1+2+2^2+2^3+......+2^{31}+2^{32}\)
Ta có:
\(A=1+2+2^2+2^3+......+2^{31}+2^{32}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{32}+2^{33}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{32}+2^{33}\right)-\left(1+2+2^2+2^3+......+2^{31}+2^{32}\right)\)
\(\Leftrightarrow A=2^{33}-1\)
Mặt khác \(A=2^n-1\)
\(\Rightarrow2^{33}-1=2^n-1\)
\(\Rightarrow2^{33}-1-2^n+1=0\)
\(\Rightarrow2^{33}-2^n=0\)
\(\Rightarrow2^{33}=2^n\)
\(\Rightarrow n=33\)
Vậy n=33
Bài 2:
a) \(\left(x+5\right)^2=x^2+10x+25\)
b) \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)
c) \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)
d) \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)
e) \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)
f) \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)
Bài 1:
$M=(2a+b)^2-(b-2a)^2=[(2a+b)-(b-2a)][(2a+b)+(b-2a)]$
$=4a.2b=8ab$
$N=(3a+1)^2+2a(1-2b)+(2b-1)^2$
$=(9a^2+6a+1)+2a-4ab+(4b^2-4b+1)$
$=9a^2+8a+4b^2-4b-4ab+2$
$A=(m-n)^2+4mn=m^2-2mn+n^2+4mn$
$=m^2+2mn+n^2=(m+n)^2$
\(d=\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)....\left(1+\dfrac{1}{n^2+2n}\right)\)
\(d=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...........\dfrac{n^2+2n+1}{n^2+2n}\)
\(d=\dfrac{2^2}{3}.\dfrac{3^2}{8}.\dfrac{4^2}{15}......\dfrac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(d=\dfrac{2^2.3^2.4^2......\left(n+1\right)^2}{3.8.15.....n\left(n+2\right)}\)
\(d=\dfrac{2.2.3.3.4.4......\left(n+1\right)\left(n+1\right)}{1.3.2.4.3.5......n\left(n+2\right)}\)
\(d=\dfrac{2.3.4......\left(n+1\right)}{1.2.3......n}.\dfrac{2.3.4.....\left(n+1\right)}{3.4.5.....\left(n+2\right)}\)
\(d=\left(n+1\right)\dfrac{2}{n+2}\)
\(d=\dfrac{2n+2}{n+2}\)
mí bạn giải nhanh hộ mk nhoa