Bài giải

        \(1\frac{1}{3}\text{ x }2\frac{1}{4}\text{ x }\text{ }3\frac{1}{5}\text{ x }4\frac{1}{6}\text{ x }5\frac{1}{7}\text{ x }6\frac{1}{8}\text{ x }7\frac{1}{9}\text{ x }8\frac{1}{10}\)

\(=\frac{4}{3}\text{ x }\frac{9}{4}\text{ x }\frac{16}{5}\text{ x }\frac{25}{6}\text{ x }\frac{36}{7}\text{ x }\frac{49}{8}\text{ x }\frac{64}{9}\text{ x }\frac{81}{10}\)

\(=\frac{4\text{ x }9\text{ x }16\text{ x }25\text{ x }36\text{ x }49\text{ x }64\text{ x }81}{3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9}\)

\(=\frac{4\text{ x }3^2\text{ x }4^2\text{ x }5^2\text{ x }6^2\text{ x }7^2\text{ x }8^2\text{ x }9^2}{3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9}\)                   ( Nếu bạn chưa học lũy thừa thì bạn không cần viết bước này ) 

\(=4\text{ x }3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9\)

\(=725760\)

                                                         Bài giải

        \(1\frac{1}{3}\text{ x }2\frac{1}{4}\text{ x }\text{ }3\frac{1}{5}\text{ x }4\frac{1}{6}\text{ x }5\frac{1}{7}\text{ x }6\frac{1}{8}\text{ x }7\frac{1}{9}\text{ x }8\frac{1}{10}\)

\(=\frac{4}{3}\text{ x }\frac{9}{4}\text{ x }\frac{16}{5}\text{ x }\frac{25}{6}\text{ x }\frac{36}{7}\text{ x }\frac{49}{8}\text{ x }\frac{64}{9}\text{ x }\frac{81}{10}\)

\(=\frac{4\text{ x }9\text{ x }16\text{ x }25\text{ x }36\text{ x }49\text{ x }64\text{ x }81}{3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9}\)

\(=\frac{4\text{ x }3^2\text{ x }4^2\text{ x }5^2\text{ x }6^2\text{ x }7^2\text{ x }8^2\text{ x }9^2}{3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9}\)                   ( Nếu bạn chưa học lũy thừa thì bạn không cần viết bước này ) 

\(=4\text{ x }3\text{ x }4\text{ x }5\text{ x }6\text{ x }7\text{ x }8\text{ x }9\)

\(=725760\)

mình chịu

\(1\frac{1}{3}\times2\frac{1}{4}\times...\times8\frac{1}{10}\)

\(=\frac{4}{3}\times\frac{9}{4}\times...\times\frac{81}{10}\)

\(=\frac{4\times9\times...\times81}{3\times4\times...\times10}\)

\(=\frac{2\times2\times3\times3\times...\times9\times9}{3\times4\times5\times...\times10}\)

\(=\frac{\left(2\times3\times4\times...\times9\right)\times\left(2\times3\times4\times...\times9\right)}{3\times4\times5\times...\times10}\)

\(=\frac{2\times2\times3\times...\times9}{10}\)

\(=2\times3\times4\times6\times...\times9\)

\(=72576\)

Đặt \(\frac{x_1-1}{5}=\frac{x_2-2}{4}=\frac{x_3-3}{3}=\frac{x_4-4}{2}=\frac{x_5-5}{1}=k\)

Áp dụng TC DTSBN ta có :

\(k=\frac{\left(x_1-1\right)+\left(x_2-2\right)+\left(x_3-3\right)+\left(x_4-4\right)+\left(x_5-5\right)}{5+4+3+2+1}\)

\(=\frac{x_1+x_2+x_3+x_4+x_5-15}{15}=\frac{30-15}{15}=1\)

\(\frac{x_1-1}{5}=1\Rightarrow x_1=6;\frac{x_2-2}{4}=1\Rightarrow x_2=6;\frac{x_3-3}{3}=1\Rightarrow x_3=6;\frac{x_4-4}{2}=1\Rightarrow x_4=6;\frac{x^5-5}{2}=1\Rightarrow x_5=6\)

Vậy \(x_1=x_2=x_3=x_4=x_5=6\)

=48 nha

B1 Nhân biểu tthức trên với 1/2 được

1/6+1/12/+1/20+....+1/110=1/2×3+1/3×4+.....+1/10×11

=1/2-1/3+1/3-1/4+.....+1/10-1/11=1/2-1/11=9/22

B2

B.1/1×2+1/2×3+.....+1/99×100=1-1/2 +1/2-1/3+....+1/99-1/100=1-1/100=99/100

Phần a sai đề phải llà1/2×4+1/4×6+.....+1/8×10 mới làm đc nhé

dễ thế

Câu 1:

a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^n}{c^n}=\frac{b^n}{d^n}=\frac{a^n+b^n}{c^n+d^n}=\frac{a^n-b^n}{c^n-d^n}\)

b,Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{b}{d}\cdot\frac{a}{c}\Rightarrow\frac{a^2}{b^2}=\frac{ab}{cd}\)

\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ac}{cd}=\frac{b^2}{d^2}\)

\(\Rightarrow\frac{ac}{bd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)

Ta lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a+b}{c+d}\cdot\frac{a+b}{c+d}\Rightarrow\frac{ab}{cd}=\left(\frac{a+b}{c+d}\right)^2\left(2\right)\)

Từ (1) và (2) => \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)

Câu 2:

\(\frac{a1}{a2}=\frac{a2}{a3}=....=\frac{a2017}{a2018}=\frac{a1+a2+...+a2017}{a2+a3+....+a2018}\)

\(\Rightarrow\frac{a1}{a2}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(1\right)\)

\(\frac{a2}{a3}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(2\right)\)

..............

\(\frac{a2017}{a2018}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(2017\right)\)

Nhân các vế (1),(2)....(2017) ta được:

\(\frac{a1}{a2}\cdot\frac{a2}{a3}\cdot\cdot\cdot\cdot\cdot\frac{a2017}{a2018}=\frac{a1}{a2018}=\left(\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\right)^{2017}\)

Vậy...

Câu 3:

\(x_2^2=x_1x_3\Rightarrow\frac{x1}{x2}=\frac{x2}{x3}\)

\(x_3^2=x_2x_4\Rightarrow\frac{x2}{x3}=\frac{x3}{x4}\)

\(x_4^2=x_3x_5\Rightarrow\frac{x3}{x4}=\frac{x4}{x5}\)

\(x_5^2=x_4x_6\Rightarrow\frac{x4}{x5}=\frac{x5}{x6}\)

Đến đây thfi làm giống câu 2

cho x₁, x₂ , x₃ là 3 số thực khác 0 thỏa mãn x₁ + x₂ + x₃ = a ; x₁x₂ + x₂x₃ + x₁x₃ = 0 ; x₁x₂x₃ = b

CMR: a/b < 0

a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)

\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)

\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=0-1\)

\(\Rightarrow x=-1\)

Vậy \(x=-1.\)

Mình chỉ làm câu a) thôi nhé.

Chúc bạn học tốt!

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