sao tia phân giác của góc bac cắt bc tại a

$\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\iff \frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}$

$\Rightarrow \frac{x-30}{10}=\frac{y-15}{5}=\frac{z-21}{7}$

$\Rightarrow \frac{x}{10}-\frac{30}{10}=\frac{y}{5}-\frac{15}{5}=\frac{z}{7}-\frac{21}{7}$

$\Rightarrow \frac{x}{10}-3=\frac{y}{5}-3=\frac{z}{7}-3$

$\Rightarrow \frac{x}{10}=\frac{y}{5}=\frac{z}{7}$

Đặt: $\frac{x}{10}=\frac{y}{5}=\frac{z}{7}=t\Rightarrow \{\begin{array}{c}x=10t\\ y=5t\\ z=7t\end{array}$

$xyz=22400\iff 350t^3=22400\iff t^3=64\Rightarrow t=4$

$\Rightarrow \{\begin{array}{c}x=40\\ y=20\\ z=28\end{array}$




 

Ớ bài này mình lm sai ạ ! 

Ta có \(\frac{x+3}{x+4}>1\)

=> \(\frac{x+3}{x+4}-1>0\)

=> \(\frac{-1}{x+4}>0\)

=> x + 4 < 0

=> x < -4

Vậy khi x < -4 thì \(\frac{x+3}{x+4}>1\)

b) Nếu x < -3/2

=> |3x - 5| = -3x + 5

|2x + 3| = -2x - 3

Khi đó |3x - 5| + |2x + 3| = 7 (1)

<=> -3x + 5 - 2x - 3 = 7

=> -5x = 5

=> x = -1 (loại)

Nếu \(-\frac{3}{2}\le x\le\frac{5}{3}\)

=> |2x + 3| = 2x + 3

|3x - 5| = -3x + 5

Khi đó (1) <=> -3x + 5 + 2x + 3 = 7

<=> -x = -1

=> x = 1 (tm)

Nếu x > 5/3

=> |3x - 5| = 3x - 5

|2x + 3| = 2x + 3

Khi đó (1) <=> 3x - 5 + 2x + 3 = 7

<=> 5x = 9

=> x = 9/5 (tm)

Vậy \(x\in\left\{1;\frac{9}{5}\right\}\)là giá trị cần tìm

a)\(2x-3=\frac{1}{2}\Leftrightarrow2x=\frac{1}{2}+3\Leftrightarrow2x=\frac{7}{2}\Leftrightarrow x=\frac{7}{2}\div2\Leftrightarrow x=\frac{7}{4}\)

b)\(\frac{5}{12}x=\frac{-4}{15}\Leftrightarrow x=\frac{-4}{15}\div\frac{5}{12}\Leftrightarrow x=\frac{-6}{25}\)

c)\(2,7:2x=\frac{11}{5}:\frac{11}{3}\Leftrightarrow2,7:2x=\frac{3}{5}\Leftrightarrow2x=2,7:\frac{3}{5}\Leftrightarrow2x=4,5\Leftrightarrow x=4,5:2\Leftrightarrow x=2.25\)

a) \(\frac{-3}{7}+x=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{3}+\frac{3}{7}\)

\(\Leftrightarrow x=\frac{16}{21}\)

b) \(\frac{-5}{8}+x=\frac{-2^2}{3}\)

\(\Leftrightarrow x=\frac{-4}{3}+\frac{5}{8}\)

\(\Leftrightarrow x=-\frac{17}{24}\)

c) \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)

\(\Leftrightarrow\frac{8}{3}:\frac{x}{4}=20\)

\(\Leftrightarrow\frac{32}{3x}=20\)

\(\Leftrightarrow3x=\frac{8}{5}\)

\(\Leftrightarrow x=\frac{8}{15}\)

a) \(\frac{1}{2}+\frac{5}{6}.3\frac{2}{5}\)

\(=\frac{1}{2}+\frac{5}{6}.\frac{17}{5}\)

\(=\frac{1}{2}+\frac{17}{6}\)

\(=\frac{10}{3}\)

b) \(\frac{1}{5}-\left(\frac{2}{3}+\frac{1}{10}\right)\)

\(=\frac{1}{5}-\frac{23}{30}=\frac{-17}{30}\)

c)

c) \(\frac{2^{15}.9^4}{6^3.8^3}\)= \(\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^3.3^3.2^9}\)

\(=\frac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5=8.243=1944\)

a)\(\frac{3}{7}\left(-\frac{5}{2}\right)+\left(-\frac{3}{7}\right)\)

\(=\left(-\frac{15}{17}\right)+\left(-\frac{3}{7}\right)\)

\(=-\frac{156}{119}\)

b) \(\frac{2}{3}+\frac{3}{4}\left(-\frac{4}{9}\right)\)

\(=\frac{2}{3}+-\frac{1}{3}=\frac{1}{3}\)

A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)

= \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)

= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)

= \(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1482}\right)\)

= \(\frac{1}{2}.\frac{390}{781}=\frac{195}{781}\)

=195/781

Sửa đề 3^x + 3^{x + 2} = 9¹⁷ + 27¹²

=> 3^x(1 + 3² ) = (3²)¹⁷ + (3³)¹²

=> 3²(1 + 3²) = 3³⁴ + 3³⁶

=> 3²(1 +  3²) = 3³⁴(1 + 3²)

=> 3^x = 3³⁴

=> x = 34

Vậy x = 34