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Đề bài là j vậy ak ??

Đề bài ???????

\(\frac{x-y}{x+y}=\frac{z-x}{z+x}\Rightarrow\left(x-y\right)\left(z+x\right)=\left(z-x\right)\left(x+y\right).\)

\(\Rightarrow xz+x^2-yz-xy=zx+zy-x^2-xy\)

\(\Rightarrow x^2-yz=zy-x^2\)

\(\Rightarrow2x^2=2yz\Rightarrow x^2=yz\)

Ta có : \(\frac{x-y}{x+y}=\frac{z-x}{z+x}=k\)

=> \(\hept{\begin{cases}x-y=k(x+y)\\z-x=k(z+x)\end{cases}}\)

=> \(\hept{\begin{cases}x(1-k)=-y(1+k)\\z(1-k)=-x(1+k)\end{cases}}\)

=> \(\frac{x(1-k)}{z(1-k)}=\frac{-y(1+k)}{-x(1+k)}\)

=> \(\frac{x}{z}=\frac{y}{x}\Rightarrow x^2=yz\)

\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+...+\frac{1}{50.56}\)

\(A=\frac{1}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+...+\frac{6}{50.56}\right)\)

\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+...+\frac{1}{50}-\frac{1}{56}\right)\)

\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{56}\right)\)

\(A=\frac{1}{6}.\frac{3}{28}\)

\(A=\frac{1}{56}\)

\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)

\(B=5.\left(\frac{9}{12.21}+\frac{9}{21.30}\right)-4.\left(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64}\right)\)

\(B=5.\left(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64}\right)\)

\(B=5.\left(\frac{1}{12}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{64}\right)\)

\(B=5.\frac{1}{20}-4.\frac{5}{192}\)

\(B=\frac{1}{4}-\frac{5}{48}\)

\(B=\frac{7}{48}\)

Ta có \(\frac{A}{B}=\frac{1}{56}\div\frac{7}{48}=\frac{1}{56}\times\frac{48}{7}=\frac{6}{49}\)

Lấy \(\frac{6}{49}-\frac{1}{8}=-\frac{1}{392}< 0\)

\(\Rightarrow\frac{6}{49}< \frac{1}{8}\) hay \(\frac{A}{B}< \frac{1}{8}\)

\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+....+\frac{1}{50.56}\)

\(=\frac{1}{6}.(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+....+\frac{6}{50.56})\)

\(=\frac{1}{6}.(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+....+\frac{1}{50}-\frac{1}{56})\)

\(=\frac{1}{6}.(\frac{1}{8}-\frac{1}{56})\)

\(=\frac{1}{6}.(\frac{7}{56}-\frac{1}{56})\)

\(=\frac{1}{6}.\frac{6}{56}\)

\(=\frac{1}{56}\)

\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)

\(=5(\frac{9}{12.21}+\frac{9}{21.30})-4(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64})\)

\(=5(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30})-4(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64})\)

\(=5(\frac{1}{12}-\frac{1}{30})-4(\frac{1}{24}-\frac{1}{64})\)

\(=5(\frac{5}{60}-\frac{2}{60})-(\frac{4}{24}-\frac{4}{64})\)

\(=5.\frac{1}{20}-(\frac{1}{6}-\frac{1}{16})\)

\(=\frac{1}{4}-(\frac{8}{48}-\frac{3}{48})\)

\(=\frac{1}{4}-\frac{5}{48}\)

\(=\frac{12}{48}-\frac{5}{48}=\frac{7}{48}\)

\(\frac{A}{B}=\frac{1}{56}\div\frac{7}{48}\)

\(=\frac{1}{56}.\frac{48}{7}\)

\(=\frac{6}{49}=\frac{48}{392}\)bé hơn \(\frac{49}{392}=\frac{1}{8}\)

Vậy \(\frac{A}{B}\)bé hơn \(\frac{1}{8}\)

Chúc bạn học tốt

\(\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right).....\left(\frac{1}{100}-1\right)\)

\(=\left(\frac{1}{2}-\frac{2}{2}\right)\cdot\left(\frac{1}{3}-\frac{3}{3}\right)\cdot.....\cdot\left(\frac{1}{100}-\frac{100}{100}\right)\)

\(=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot.....\cdot\left(-\frac{99}{100}\right)\)

\(=-\left(\frac{1}{2}\cdot\frac{2}{3}\cdot.....\cdot\frac{99}{100}\right)\)

\(=-\left(\frac{1}{100}\right)=-\frac{1}{100}\)

\(x^2-5x+6=0\)

\(\left(x-3\right)\left(x-2\right)=0\)

\(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)

\(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

\(x^2-5x+6=0\)

\(x^2-2x-3x+6=0\)

\(\left(x-2\right)\left(x-3\right)=0\)

=> x=2 hoặc x=3