\(\dfrac{1}{3^6}=\dfrac{1}{3^4\cdot3^2}=\dfrac{1}{81\cdot9}=\dfrac{1}{729}\)

  \(\dfrac{1}{3^6}\) = \(\dfrac{1}{3^4.3^2}\) = \(\dfrac{1}{81.9}\) = \(\dfrac{1}{729}\) 

Bài 2:

a: \(27^{11}=\left(3^3\right)^{11}=3^{33};81^8=\left(3^4\right)^8=3^{32}\)

mà 33>32

nên \(27^{11}>81^8\)

b: \(625^5=\left(5^4\right)^5=5^{20};125^7=\left(5^3\right)^7=5^{21}\)

mà 20<21

nên \(625^5< 125^7\)

c: \(3^{2n}=\left(3^2\right)^n=9^n;2^{3n}=\left(2^3\right)^n=8^n\)

mà 9>8

nên \(3^{2n}>2^{3n}\)

Bài 3:

a: \(3^{1234}=\left(3^2\right)^{617}=9^{617};2^{1851}=\left(2^3\right)^{617}=8^{617}\)

mà 9>8

nên \(3^{1234}>2^{1851}\)

b: \(6^{30}=\left(6^2\right)^{15}=36^{15}>12^{15}\)

c: \(5^{36}=\left(5^3\right)^{12}=125^{12};11^{24}=\left(11^2\right)^{12}=121^{12}\)

mà 125>121

nên \(5^{36}>11^{24}\)

d: \(6^3=6\cdot6^2< 7\cdot6^2\)

\(\widehat{C}=\widehat{B}+10^0=\widehat{A}+10^0+10^0=\widehat{A}+20^0\)

\(\widehat{D}=\widehat{C}+10^0=\widehat{A}+20^0+10^0=\widehat{A}+30^0\)

Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

=>\(\widehat{A}+\widehat{A}+10^0+\widehat{A}+20^0+\widehat{A}+30^0=360^0\)

=>\(4\cdot\widehat{A}=300^0\)

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

\(\widehat{B}=75^0+10^0=85^0\)

\(\widehat{C}=75^0+20^0=95^0\)

\(\widehat{D}=75^0+30^0=105^0\)

a: Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

=>\(\widehat{C}+\widehat{D}=360^0-110^0-70^0=180^0\)

=>\(\dfrac{1}{3}\cdot\widehat{D}+\widehat{D}=180^0\)

=>\(\dfrac{4}{3}\cdot\widehat{D}=180^0\)

=>\(\widehat{D}=135^0\)

\(\widehat{C}=\dfrac{1}{3}\cdot135^0=45^0\)

b:

Sửa đề: Cho tứ giác ABCD.

Đặt \(\widehat{B}=x;\widehat{C}=y;\widehat{D}=z\)

\(\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{3}=\dfrac{\widehat{D}}{4}\)

=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

=>\(x+y+z=360^0-90^0=270^0\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{270}{9}=30^0\)

=>\(x=2\cdot30^0=60^0;y=3\cdot30^0=90^0;z=4\cdot30^0=120^0\)

Vậy: \(\widehat{B}=x=60^0;\widehat{C}=y=90^0;\widehat{D}=z=120^0\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(\left(-5\right)^5=\left(-5\right)^4\cdot\left(-5\right)=5^4\cdot\left(-5\right)=625\cdot\left(-5\right)=-3125\)

chịu

\(\left(x^2-1\right)\left(x^2+2\right)< 0\)

mà \(x^2+2>0\forall x\)

nên \(x^2-1< 0\)

=>\(x^2< 1\)

=>-1<x<1

Giúp mình 😭

a: ΔABC vuông tại A

=>\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC=\dfrac{1}{2}\cdot6\cdot8=24\left(cm^2\right)\)

b: AE=3CE

mà AE+CE=AC

nên \(CE=\dfrac{1}{4}AC\)

=>\(\dfrac{S_{CBE}}{S_{CAB}}=\dfrac{CE}{CA}=\dfrac{1}{4}=25\%\)