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

\(\dfrac{11}{5}-\left(0,35+x\right)=1\dfrac{1}{2}\\ \dfrac{11}{5}-\left(\dfrac{7}{20}+x\right)=\dfrac{3}{2}\\ \dfrac{11}{5}-\dfrac{7}{20}-x=\dfrac{3}{2}\\ \dfrac{44}{20}-\dfrac{7}{20}-x=\dfrac{3}{2}\\ \dfrac{37}{20}-x=\dfrac{3}{2}\\ x=\dfrac{37}{20}-\dfrac{3}{2}\\ x=\dfrac{7}{20}\)

\(\dfrac{15}{34}+\dfrac{15}{17}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}\)

\(=\dfrac{15}{34}+\dfrac{19}{34}+\dfrac{15}{17}-1-\dfrac{15}{17}+\dfrac{2}{3}\)

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

a: Ta có: \(\widehat{bMB}=\widehat{NMC}\)(hai góc đối đỉnh)

mà \(\widehat{bMB}=50^0\)

nên \(\widehat{NMC}=50^0\)

Ta có: \(\widehat{MNC}+\widehat{aNC}=180^0\)(hai góc kề bù)

=>\(\widehat{MNC}+110^0=180^0\)

=>\(\widehat{MNC}=70^0\)

Xét ΔMNC có \(\widehat{NMC}+\widehat{MNC}+\widehat{C}=180^0\)

=>\(\widehat{C}+50^0+70^0=180^0\)

=>\(\widehat{C}=60^0\)

b: Ta có: \(\widehat{NMB}+\widehat{NMC}=180^0\)(hai góc kề bù)

=>\(\widehat{NMB}+50^0=180^0\)

=>\(\widehat{NMB}=130^0\)

Ta có: MN//AB

=>\(\widehat{CMN}=\widehat{CBA}\)(hai góc đồng vị)

=>\(\widehat{CBA}=50^0\)

BN là phân giác của góc CBA

=>\(\widehat{NBM}=\dfrac{\widehat{ABC}}{2}=25^0\)

Xét ΔNMB có \(\widehat{NMB}+\widehat{BNM}+\widehat{NBM}=180^0\)

=>\(\widehat{MNB}=180^0-130^0-25^0=25^0\)

c: BN là phân giác của góc CBA

=>\(\widehat{ABN}=\dfrac{\widehat{ABC}}{2}=25^0\)

Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(\widehat{BAN}+60^0+50^0=180^0\)

=>\(\widehat{BAN}=70^0\)

Xét ΔBAN có \(\widehat{BAN}+\widehat{ABN}+\widehat{ANB}=180^0\)

=>\(\widehat{ANB}=180^0-75^0-25^0=85^0\)

Bài 4: \(8^{10}\cdot125^{10}< =2^n\cdot5^n< =20^{16}\cdot5^{16}\)

=>\(1000^{10}< =10^n< =100^{16}\)

=>\(10^{30}< =10^n< =10^{32}\)

=>30<=n<=32

mà n là số tự nhiên

nên \(n\in\left\{30;31;32\right\}\)

Bài 1:

1: \(3^{-2}\cdot3^4\cdot3^n=3^7\)

=>\(3^n\cdot3^2=3^7\)

=>n+2=7

=>n=7-2=5

2: \(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)

=>\(2^n\left(\dfrac{1}{2}+4\right)=2^5\cdot9\)

=>\(2^n=9\cdot2^5:\dfrac{9}{2}=2^6\)

=>n=6

Bài 2:

1: \(243>=3^n>=9\)

=>\(3^2< =3^n< =3^5\)

=>2<=n<=5

mà n là số tự nhiên

nên \(n\in\left\{2;3;4;5\right\}\)

2: \(2^{n+3}\cdot2^n=144\)

=>\(2^{2n+3}=144\)

=>\(2n+3=log_2144\)

=>\(2n=log_2144-3\)

=>\(n=\dfrac{log_2144-3}{2}\left(loại\right)\)

Bài 3:

\(10^x:5^y=20^y\)

=>\(10^x=20^y\cdot5^y=100^y=10^{2y}\)

=>x=2y

vậy: \(\left(x;y\right)\in\){(2k;k)|\(k\in N\)}

ΔABC=ΔDEF

=>\(\widehat{A}=\widehat{D}\)

=>\(\widehat{D}=55^0\)

ΔABC=ΔDEF

=>\(\widehat{B}=\widehat{E}\)

=>\(\widehat{B}=75^0\)

Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{C}=180^0-55^0-75^0=50^0\)

=>\(\widehat{F}=\widehat{C}=50^0\)

\(\dfrac{11}{3}+\left|x\right|=\dfrac{9}{4}\)

=>\(\left|x\right|=\dfrac{9}{4}-\dfrac{11}{3}=\dfrac{27}{12}-\dfrac{44}{12}=-\dfrac{17}{12}\)

mà \(\left|x\right|>=0\forall x\)

nên \(x\in\varnothing\)

\(\left(\dfrac{1}{2}\right)^{x+2}=16^{4-2x}\)

=>\(2^{-x-2}=2^{4\left(4-2x\right)}\)

=>-x-2=4*(4-2x)

=>-x-2=16-8x

=>-x+8x=16+2

=>7x=18

=>\(x=\dfrac{18}{7}\)

Bài 2:

a) \(\dfrac{-7}{-13}=\dfrac{7}{13}\) là số hưu tỉ dương

b) \(\dfrac{2}{-17}=-\dfrac{2}{17}\) là số hưu tỉ âm

c) \(-\dfrac{-6}{5}=\dfrac{6}{5}\) là số hưu tỉ dương

Bài 3:

a) \(-2\dfrac{1}{4}=-\left(2+\dfrac{1}{4}\right)=-\dfrac{9}{4}\)

b) \(6\dfrac{2}{3}=6+\dfrac{2}{3}=\dfrac{20}{3}\)

c) \(-3\dfrac{1}{4}=-\left(3+\dfrac{1}{4}\right)=-\dfrac{13}{4}\)

Bài 1:

\(3^{39}< 3^{40}=\left(3^4\right)^{10}=81^{10}\)

\(11^{21}>11^{20}=121^{10}\)

mà 121>81

nên \(11^{21}>3^{39}\)

Bài 2:

\(5^{27}=\left(5^3\right)^9=125^9;2^{63}=\left(2^7\right)^9=128^9\)

mà 125<128

nên \(5^{27}< 2^{63}\)

\(2^{63}=\left(2^9\right)^7=512^7;5^{28}=\left(5^4\right)^7=625^7\)

mà 512<625

nên \(2^{63}< 5^{28}\)

Do đó: \(5^{27}< 2^{63}< 5^{28}\)