\(a.5\cdot3^x=5\cdot3^4\\ =>3^x=\dfrac{5\cdot3^4}{5}=3^4\\ =>x=4\\ b.7\cdot4^x=7\cdot4^3\\ =>4^x=\dfrac{7\cdot4^3}{7}=4^3\\ =>x=3\\ c.\dfrac{3}{5}\cdot4^x=7\cdot4^3\\ =>4^x=\dfrac{7\cdot4^3}{\dfrac{3}{5}}=\dfrac{35}{3}\cdot4^3\\ =>\dfrac{4^x}{4^3}=\dfrac{35}{3}\\ =>4^{x-3}=\dfrac{35}{3}\\ =>x-3=log_4\dfrac{35}{3}\\ =>x=log_4\dfrac{35}{3}+3\\ d.\dfrac{3}{2}\cdot5^x=\dfrac{3}{2}\cdot5^{12}\\ =>5^x=\dfrac{5^{12}\cdot\dfrac{3}{2}}{\dfrac{3}{2}}=5^{12}\\ =>x=12\) 

e: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)

=>\(9\cdot5^x=9\cdot5^6\)

=>\(5^x=5^6\)

=>x=6

f: \(5\cdot3^x=7\cdot3^5-2\cdot3^5\)

=>\(5\cdot3^x=5\cdot3^5\)

=>\(3^x=3^5\)

=>x=5

g: \(5\cdot3^{x+6}=2\cdot3^5+3\cdot3^5\)

=>\(5\cdot3^{x+6}=5\cdot3^5\)

=>\(3^{x+6}=3^5\)

=>x+6=5

=>x=-1

\(x\left(2x+\dfrac{-4}{10}\right)\) = 0

\(\left[{}\begin{matrix}x=0\\2x-\dfrac{4}{10}=10\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\2x=\dfrac{4}{10}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=\dfrac{4}{10}:2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {0; \(\dfrac{1}{5}\)}

\(x\left(2x+\dfrac{-4}{10}\right)=0\\ =>x\left(2x+\dfrac{-2}{5}\right)=0\\ =>2x\left(x-\dfrac{1}{5}\right)=0\\ TH1:2x=0\\ =>x=0\\ TH2:x-\dfrac{1}{5}=0\\ =>x=\dfrac{1}{5}\)

\(14-8x^2=\dfrac{3}{2}\\ =>8x^2=14-\dfrac{3}{2}\\ =>8x^2=\dfrac{25}{2}\\ =>x^2=\dfrac{25}{2}:8\\ =>x^2=\dfrac{25}{16}\\ =>x=\pm\dfrac{5}{4}\)

\(-\dfrac{3}{11}.\dfrac{5}{7}+\dfrac{5}{7}.-\dfrac{8}{11}+\dfrac{19}{7}\)

`=` \(\dfrac{5}{7}.\left(\dfrac{-3}{11}+\dfrac{-8}{11}\right)+\dfrac{19}{7}\)

`=` \(\dfrac{5}{7}.\dfrac{-11}{11}+\dfrac{19}{7}\)

`=` \(\dfrac{5}{7}.\left(-1\right)+\dfrac{19}{7}\)

`=` \(-\dfrac{5}{7}+\dfrac{19}{7}\)

`=` \(\dfrac{14}{7}\)

`=  2`

Giúp mình với please

\(a.\left(x-2\right)\left(x^2+x-1\right)-x\left(x^2-1\right)\\ =\left(x^3+x^2-x-2x^2-2x+2\right)-\left(x^3-x\right)\\ =x^3-x^2-3x+2-x^3+x\\ =-x^2-2x+2\\ b.\left(2x-9\right)\left(2x+9\right)-4x^2\\ =\left[\left(2x\right)^2-9^2\right]-4x^2\\ =4x^2-81-4x^2\\ =-81\\ c.2x^2+3\left(x-1\right)\left(x-1\right)\\ =2x^2+3\left(x-1\right)^2\\ =2x^2+3\left(x^2-2x+1\right)\\ =2x^2+3x^2-6x+3\\ =5x^2-6x+3\)

a; (\(x\) - 2)(\(x^2\) + \(x\) - 1) - \(x\)(\(x^2\) - 1)

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

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

= 0 - \(x^2\) - (3\(x\) - \(x\)) + 2

= - \(x^2\) - 2\(x\) + 2

GIÚP MÌNH VÓI MÌNH THẤY ĐỀ BÀI CÓ GÌ ĐÓ SAI MONG CÁC BẠN SỦA GÚP VÀ GIẢ ,VẼ HÌNH NỮA NHÉ

MÌNH CẢM ƠN

a) Xét tam giác ABH và tam giác DBH có:

 AB = BD (g.t)

 BH chung

 HA = HD (g.t)

b) Ta có: Góc BHA = Gó BHD =90*

=> HE là trung trực

=> EA = ED

=> Tam giác AED cân