\(\dfrac{2}{3}+\dfrac{7}{4}:x=\dfrac{5}{6}\\ \Rightarrow\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{2}{3}\\\Rightarrow\dfrac{7}{4}:x=\dfrac{1}{6} \\ \Rightarrow x=\dfrac{7}{4}:\dfrac{1}{6}\\ \Rightarrow x=\dfrac{21}{2}\)

Vậy \(x=\dfrac{21}{2}\)

\(\dfrac{2}{3}+\dfrac{7}{4}:x=\dfrac{5}{6}\)

=> \(\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{2}{3}\)

=> \(\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{4}{6}\)

=> \(\dfrac{7}{4}:x=\dfrac{1}{6}\)

=> \(x=\dfrac{7}{4}:\dfrac{1}{6}\)

=> x = \(\dfrac{7}{4}.6\)

=> \(x=\dfrac{21}{2}\)

Vậy ...

\(\dfrac{5^4.18^4}{125.9^5.16}\\ =\dfrac{5^4.\left(2.3^2\right)^4}{5^3.\left(3^2\right)^5.2^4}\\ =\dfrac{5^4.2^4.3^8}{5^3.2^4.3^{10}}\\ =\dfrac{5}{3^2}\\ =\dfrac{5}{9}\)

26. 

\(a.\left(\dfrac{3}{7}\right)^5\cdot x=\left(\dfrac{3}{7}\right)^7\\ =>x=\left(\dfrac{3}{7}\right)^7:\left(\dfrac{3}{7}\right)^5\\ =>x=\left(\dfrac{3}{7}\right)^{7-5}\\ =>x=\left(\dfrac{3}{7}\right)^2\\ =>x=\dfrac{9}{49}\\ b.\left(0,09\right)^3x=-\left(0,09\right)^2\\ =>\left[\left(0,3\right)^2\right]^3\cdot x=-\left[\left(0,3\right)^2\right]^2\\ =>x=-\left(0,3\right)^4:\left(0,3\right)^6\\ =>x=-\left(0,3\right)^{-2}\\ =>x=-\left(\dfrac{10}{3}\right)^2\\ =>x=-\dfrac{100}{9}\)

27:

a: Vì \(0< \dfrac{1}{2}< 1\)

và 40<50

nên \(\left(\dfrac{1}{2}\right)^{40}>\left(\dfrac{1}{2}\right)^{50}\)

b: \(243^3=\left(3^5\right)^3=3^{15};125^5=\left(5^3\right)^5=5^{15}\)

mà 3<5

nên \(243^3< 125^5\)

\(-\dfrac{2}{5}+\dfrac{3}{4}-\dfrac{-1}{6}+\dfrac{-2}{5}\\ =\left(\dfrac{-2}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{3}{4}+\dfrac{1}{6}\right)\\ =\dfrac{-4}{5}+\left(\dfrac{9}{12}+\dfrac{2}{12}\right)\\ =\dfrac{-4}{5}+\dfrac{11}{12}\\ =\dfrac{-48}{60}+\dfrac{55}{60}\\ =\dfrac{55-48}{60}\\ =\dfrac{7}{60}\)

Đặt: \(\dfrac{x}{3}=\dfrac{y}{5}=k=>\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)

Mà:

\(2x^2+y^2=43\\ =>2\cdot\left(3k\right)^2+\left(5k\right)^2=43\\ =>18k^2+25k^2=43\\ =>43k^2=43\\ =>k^2=1\\ =>k=\pm1\\ TH1:k=1=>\left\{{}\begin{matrix}x=3\cdot1=3\\y=5\cdot1=5\end{matrix}\right.\\ TH2:k=-1=>\left\{{}\begin{matrix}x=3\cdot\left(-1\right)=-3\\y=5\cdot\left(-1\right)=-5\end{matrix}\right.\)

a: Xét ΔAHK vuông tại H và ΔDHB vuông tại H có

HA=HD

HK=HB

Do đó: ΔAHK=ΔDHB

b: ΔAHK=ΔDHB

=>\(\widehat{HAK}=\widehat{HDB}\)

=>AK//DB

c: Xét ΔBAD có

BH là đường cao

BH là đường trung tuyến

Do đó: ΔBAD cân tại B

=>BA=BD

d: Xét ΔHAB vuông tại H và ΔHDK vuông tại H có

HA=HD

HB=HK

Do đó: ΔHAB=ΔHDK

=>\(\widehat{HAB}=\widehat{HDK}\)

=>AB//DK

ta có: IK\(\perp\)AC

AB\(\perp\)AC

Do đó: IK//AB

mà DK//AB

và IK,DK có điểm chung là K

nên I,K,D thẳng hàng

Ta có:

\(\dfrac{1}{2}< \dfrac{x}{10}< \dfrac{4}{5}\\ \Rightarrow\dfrac{5}{10}< \dfrac{x}{10}< \dfrac{8}{10}\\ \Rightarrow5< x< 8\)

Vì \(x\) nguyên nên:

\(x\in\left\{6,7\right\}\)

Vậy \(x\in\left\{6,7\right\}\)

\(A=\sqrt{16+9}=\sqrt{25}=5\)

\(125^7:25^{16}\\ =\left(5^3\right)^7:\left(5^2\right)^{16}\\ =5^{3\cdot7}:5^{2\cdot16}\\ =5^{21}:5^{32}\\ =5^{21-32}\\ =5^{-11}\)