Ta có \(\left(x+1\right)^2\ge0\)

\(\Rightarrow\sqrt{\left(x+1\right)^2+7}\ge\sqrt{7}\)

Dấu "=" xảy ra khi x=-1

Vậy........................................

\(\sqrt{\left(x+1\right)^2+7}\)

\(\Rightarrow2\sqrt{2}\)

hok tốt

b)\(\frac{\sqrt{27}}{\sqrt{12}}+\frac{1}{2}\)

\(=\frac{\sqrt{3}.\sqrt{9}}{\sqrt{3}.\sqrt{4}}+\frac{1}{2}\)

\(=\frac{\sqrt{9}}{\sqrt{4}}+\frac{1}{2}\)

\(=\frac{3}{2}+\frac{1}{2}\)

\(\frac{4}{2}=2\)

a) \(\sqrt{45}.\sqrt{15}.\sqrt{27}\)

\(=\left(\sqrt{15}\right)^2.\left(\sqrt{3}\right)^2.\sqrt{9}\)

\(=15.3.3\)

\(=135\)

Ta có AB = BC (gt)

Suy ra: ∆ABC cân.

Nên A1ˆ=C1ˆA1^=C1^ (1)

Lại có \(\widehat{A_1}=\widehat{A_2}\) (2) (vì AC là tia phân giác của ˆAA^)

Từ (1) và (2) suy ra \(\widehat{C_1}=\widehat{A_2}\)

nên BC // AD (do \(\widehat{A_1};\widehat{C_2}\) ở vị trí so le trong)

Vẽ hình :

\(\text{[}\left(x\right)^y\text{]}^{\text{z}}=x^{yz}\) 

hệ số = 1 

\(A=1+2+2^2+...+2^{2018}\)

\(2A=2+2^2+...+2^{2019}\)

\(2A-A=\left[2+2^2+...+2^{2019}\right]-\left[1+2+2^2+...+2^{2018}\right]\)

\(A=2^{2019}-1\)

#)Giải :

\(A=1+2+2^2+2^3+...+2^{2018}\)

\(2A=2+2^2+2^3+2^4+...+2^{2019}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2019}\right)-\left(1+2+2^2+2^3+...+2^{2018}\right)\)

\(A=2^{2019}-1\)

\(B=3+3^2+3^3+...+3^{2017}\)

\(3B=3^2+3^3+3^4+...+3^{2018}\)

\(3B-B=\left(3^2+3^3+3^4+...+3^{2018}\right)-\left(3+3^2+3^3+...+3^{2017}\right)\)

\(2B=3^{2018}-3\)

\(B=\frac{3^{2018}-3}{2}\)

3A = 2.3.3 + 3.4.3 + 4.5.3 + ... + 99.100.3 

3A = 2.3.(4-1) + 3.4.(5 -2) + 4.5.(6 - 3) + ... + 99.100.(101 - 98) 

3A = 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + 4.5.6 - 3.4.5 + ... + 99.100.101 - 98.99.100 

3A = (2.3.4 + 3.4.5 + 4.5.6 + ... + 99.100.101) - (1.2.3 + 2.3.4 + 3.4.5 + ... + 98.99.100) 

3A = 99.100.101 - 1.2.3

3A = 999900 - 6 

3A = 999894 

A = 999894 : 3 

A = 333298