Sửa đề: \(\frac12\cdot2^{n}+4\cdot2^{n}=9\cdot2^5\)

=>\(2^{n}\left(4+\frac12\right)=9\cdot2^5\)

=>\(2^{n}\cdot\frac92=9\cdot2^5\)

=>\(2^{n}=9\cdot2^5:\frac92=9\cdot2^5\cdot\frac29=2^6\)

=>n=6

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 \(243^3< 125^5\)

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

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

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

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

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

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)|\(k\in N\)}

1a

Many school nowadays require their students to spend an amount of time doing community service

2c

Volunteering gives children a sense of responsibility....

Volunteer work can also help children learn improtant lessons about themselves and about life

3a

For example, community work can make them relise what they are good at and what they enjoy doing most => their likes and their ability

A volunteer work can even help children decide what they want to do when they grow up => their ability

Doing voluntary work can reduce stress and improve the children's mental health => their health

4d

They might feel happier when they can bring happiness to other people

5a

cả dẫn chứng nhé

ΔABC=ΔDEF

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

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

ΔABC=ΔDEF

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

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

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

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

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