Sửa lại đề: \(\frac{a+b}{a-b}=\frac{c+a}{c-a}\)

Từ \(\frac{a+b}{a-b}=\frac{c+a}{c-a}\)\(\Rightarrow\frac{a+b}{c+a}=\frac{a-b}{c-a}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a+b}{c+a}=\frac{a-b}{c-a}=\frac{a+b+a-b}{c+a+c-a}=\frac{2a}{2c}=\frac{a}{c}\)

\(\Rightarrow c.\left(a+b\right)=a.\left(c+a\right)\)

\(\Leftrightarrow ca+bc=ac+a^2\)

\(\Leftrightarrow a^2=bc\)( đpcm )

\(x-1=\frac{7}{10}-\frac{7}{10}\cdot x\)

=> \(x-1=\frac{7}{10}-\frac{7x}{10}\)

=> \(x-1=\frac{7-7x}{10}\)

=> 10(x - 1) = 7 - 7x

=> 10x - 10 - 7 + 7x = 0

=> 10x + 7x + (-10 - 7) = 0

=> 17x - 17 = 0

=> 17x = 17 

=> x = 1

Bài làm :

\(x-1=\frac{7}{10}-\frac{7}{10}.x\)

\(\Leftrightarrow x+\frac{7}{10}x=\frac{7}{10}+1\)

\(\Leftrightarrow\frac{17}{10}x=\frac{17}{10}\)

\(\Leftrightarrow x=\frac{17}{10}:\frac{17}{10}\)

\(\Leftrightarrow x=1\)

Vậy x = 1 .

Học tốt

\(\frac{x+2014}{2}+\frac{2x+4028}{7}=\frac{x+2014}{5}+\frac{x+2014}{6}\)

<=> \(\frac{x+2014}{2}+\frac{2\left(x+2014\right)}{7}=\frac{x+2014}{5}+\frac{x+2014}{6}\)

<=> \(\frac{x+2014}{2}+\frac{x+2014}{\frac{7}{2}}=\frac{x+2014}{5}+\frac{x+2014}{6}\)

<=> \(\frac{x+2014}{2}+\frac{x+2014}{\frac{7}{2}}-\frac{x+2014}{5}-\frac{x+2014}{6}=0\)

<=> \(\left(x+2014\right)\left(\frac{1}{2}+\frac{1}{\frac{7}{2}}-\frac{1}{5}-\frac{1}{6}\right)=0\)

Vì \(\frac{1}{2}+\frac{1}{\frac{7}{2}}-\frac{1}{5}-\frac{1}{6}\ne0\)

=> x + 2014 = 0 <=> x = -2014

Bài làm :

\(\frac{x+2014}{2}+\frac{2x+4028}{7}=\frac{x+2014}{5}+\frac{x+2014}{6}\)

\(\Rightarrow\frac{x+2014}{2}+\frac{2x+4028}{7}-\frac{x+2014}{5}-\frac{x+2014}{6}=0\)

\(\Rightarrow\frac{x+2014}{2}+\frac{2.\left(x+2014\right)}{7}-\frac{x+2014}{5}-\frac{x+2014}{6}=0\)

\(\Rightarrow\left(x+2014\right).\left(\frac{1}{2}+\frac{2}{7}-\frac{1}{5}-\frac{1}{6}\right)=0\)

\(\Rightarrow x+2014=0:\left(\frac{1}{2}+\frac{2}{7}-\frac{1}{5}-\frac{1}{6}\right)\)

\(\Rightarrow x+2014=0\)

\(\Rightarrow x=-2014\)

Vậy x = - 2014 .

Học tốt nhé

Đặt \(\frac{x}{5}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=5k\\y=4k\end{cases}}\)

Ta có x² - y² = 1

<=> (5k)² - (4k)² = 1

=> 25k² - 16k² = 1

=> 9k² = 1

=> k² = 1/9

=> k = \(\pm\frac{1}{3}\)

Nếu k = 1/3 => x = 5/3 ; y = 4/3

Nếu k = -1/3 => x = -5/3 ; y = -4/3

Vậy các cặp (x;y) thỏa mãn là (5/3 ; 4/3) ; (-5/3 ; -4/3)

áp dụng tính chất dãy tỉ số bằng nhau ta được

\(\frac{x}{5}=\frac{y}{4}=\frac{x^2}{5^2}=\frac{y^2}{4^2}=\frac{x}{25}=\frac{y}{16}=\frac{x-y}{25-16}=\frac{1}{9}\)

\(x=5.\frac{1}{9}=\frac{5}{9}\)

\(y=4.\frac{1}{9}=\frac{4}{9}\)

vậy \(x=\frac{5}{9};y=\frac{4}{9}\)

\(a.\left(x-1\right)^2=\frac{9}{25}\)

\(\left(x-1\right)^2=\left(\frac{3}{5}\right)^2\)

\(x-1=\frac{3}{5}\)

\(x=\frac{3}{5}+1\)

\(x=\frac{8}{5}\)

\(b.\frac{1}{2}-\frac{1}{3}.x=\frac{4}{5}\)

\(\frac{1}{3}.x=\frac{1}{2}-\frac{4}{5}\)

\(\frac{1}{3}.x=\frac{-3}{10}\)

\(x=\frac{-3}{10}:\frac{1}{3}\)

\(x=\frac{-3}{10}.\frac{3}{1}\)

\(x=\frac{-9}{10}\)

a) \(\left(x-1\right)^2=\frac{9}{25}\)

\(\left(x-1\right)^2=\left(\frac{3}{5}\right)^2\)

\(TH1:x-1=\frac{3}{5}\Rightarrow x=\frac{8}{5}\)

\(TH2:x-1=-\frac{3}{5}\Rightarrow-\frac{3}{5}+1=\frac{2}{5}\)

\(\Rightarrow x\in\left\{\frac{8}{5};\frac{2}{5}\right\}\)

b) \(\frac{1}{2}-\frac{1}{3}x=\frac{4}{5}\)

\(\frac{1}{3}x=\frac{1}{2}-\frac{4}{5}=-\frac{3}{10}\)

\(x=-\frac{3}{10}:\frac{1}{3}\)

\(x=-\frac{9}{10}\)

\(\frac{4.6^{14}}{9^7.8^5}=\frac{2^2.\left(2.3\right)^{14}}{\left(3^2\right)^7.\left(2^3\right)^5}=\frac{2^{16}.3^{14}}{3^{14}.2^{15}}=2\)

\(\frac{4.6^{14}}{9^7.8^5}=\frac{2^2.2^{14}.3^{14}}{\left(3^2\right)^7.\left(2^3\right)^5}=\frac{2^2.2^{14}.3^{14}}{3^{14}.2^{15}}=2\)

học tốt

\(-\frac{3}{8}.\frac{2}{7}+-\frac{2}{7}.\frac{5}{8}+\frac{1}{7}=-\frac{2}{7}.\left(\frac{3}{8}+\frac{5}{8}\right)+\frac{1}{7}=-\frac{2}{7}.1+\frac{1}{7}=-\frac{2}{7}+\frac{1}{7}=-\frac{1}{7}\)

học tốt

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

\(=\frac{1}{36}.\frac{6}{5}+\frac{3}{5}\Rightarrow\frac{1}{30}+\frac{18}{30}=\frac{19}{30}\)

                                                   Xét tam giác ABH có:

                                                         AB^2 = BC.BH

                                                   hay 6^2 = 10.BH

                                                   suy ra: BH=3,6 (cm)

                                Tương tự: xét tam giác AHC có:

                                                         AC^2 = BC.HC

                                                  hay 8^2 = 10.HC

                                                  suy ra: HC=6,4 (cm)

                                               Trong tam giác AHC có:

                                                   AC^2 = AH^2 + HC^2

                                           hay 10^2 = AH^2 + 6,4^2

                                           suy ra: AH= 7,7 (cm)

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