Ta có: \(\hept{\begin{cases}\left(\sqrt{u^2+2}+u\right)\left(\sqrt{u^2+2}-u\right)=2\\\left(\sqrt{v^2-2v+3}+v-1\right)\left(\sqrt{v-2v+3}-v+1\right)=2\end{cases}}\)

Theo đề bài thì ta có:

\(\left(u+\sqrt{u^2+2}\right)\left(v-1+\sqrt{v^2-2v+3}\right)=2\)

Từ đây ta có hệ:

\(\hept{\begin{cases}\sqrt{u^2+2}-u=\sqrt{v^2-2v+3}+v-1\left(1\right)\\\sqrt{u^2+2}+u=\sqrt{v^2-2v+3}-v+1\left(2\right)\end{cases}}\)

Lấy (1) - (2) ta được: \(u+v=1\)

Ta có: \(u^3+v^3+3uv=1\)

\(\Leftrightarrow3uv+u^2-uv+v^2=1\)

\(\Leftrightarrow\left(u+v\right)^2=1\)(đúng)

\(\Rightarrow\)ĐPCM

\(Z_L=\omega L=10\Omega\)

\(Z_C=\frac{1}{\omega C}=20\Omega\)

Ta có giản đồ véc tơ

Ta có: \(U_L=U_R=\frac{U_C}{2}\)

Từ giản đồ véc tơ ta có:

\(U_0=U_{0L}\sqrt{2}=20\sqrt{2}\sqrt{2}=40V\)

u trễ pha \(\frac{3\pi}{4}\) với uL

\(\Rightarrow u=40\cos\left(100\pi t+\frac{\pi}{2}-\frac{3\pi}{4}\right)\)

\(\Rightarrow u=40\cos\left(100\pi t-\frac{\pi}{4}\right)\)(V)

Chọn B.

Vì cái này có hai chiều lên ta phải CM hai lần 

(+) nếu 2 (u^2 - v^2) = 3uv => u = 2v 

TA có 2( u^2 - v^2)  = 3uv => 2u^2 - 2v^2 - 3uv = 0 => 2u^2 - 4uv + uv - 2v^2  = 0 

=> 2u(u - 2v) + v ( u - 2v) = 0 

=> ( 2u + v )(  u - 2v) = 0 

=> 2u + v = 0 hoặc u - 2v = 0 => u = 2v 

(+) CM ngược lại 

ĐK : u, v > 0 , u khác v

\(=\frac{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}{\sqrt{u}+\sqrt{v}}-\frac{\left(\sqrt{u}+\sqrt{v}\right)\left(u-\sqrt{uv}+v\right)}{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}\)

\(=\sqrt{u}-\sqrt{v}-\frac{u-\sqrt{uv}+v}{\sqrt{u}-\sqrt{v}}\)

\(=\frac{u-2\sqrt{uv}+v-u+\sqrt{uv}-v}{\sqrt{u}-\sqrt{v}}=\frac{-\sqrt{uv}}{\sqrt{u}-\sqrt{v}}\)

Bài 2.

a) 101³ = (100+1)³ = 100³+3.100².1+3.100.1²+1³ 

   = 1000000+30000+300+1 = 1030301

b) 299³ = (300-1)³ = 300³-3.300².1+3.300.1²-1³

   = 27000000 - 270000 + 900 -1 = 26730899

c) 99³ = (100-1)³ = 100³-3.100².1+3.100.1²-1

   = 1000000 - 30000 + 300 -1 = 970299

\(1,\\ b,A=\left(u-v\right)^3+3uv\left(u+v\right)\\ A=u^3-3u^2v+3uv^2-v^3+3u^2v+3uv^2=u^3-v^3\\ c,6\left(c-d\right)\left(c+d\right)+2\left(c-d\right)^2-\left(c-d\right)^3\\ =6c^2-6d^2+2c^2-4cd+2d^2-c^3+3c^2d-3cd^2+d^3\\ =8c^2-c^3-4d^2-4cd+3c^2d-3cd^2+d^3\)

\(2,\\ a,101^3=\left(100+1\right)^3\\ =100^3+3\cdot10000\cdot1+3\cdot100\cdot1+1\\ =1000000+30000+300+1=1030301\\ b,299^3=\left(300-1\right)^3\\ =300^3-3\cdot90000\cdot1+3\cdot300\cdot1-1\\ =27000000-270000+900-1\\ =26730899\\ c,99^3=\left(100-1\right)^3\\ =100^3-3\cdot10000\cdot1+3\cdot100\cdot1-1\\ =1000000-30000+300-1=970299\)

Bài 1:
a.

$A=u^3-3u^2v+3uv^2-v^3+3uv^2+3u^2v$

$=u^3+6uv^2-v^3$

c.

$C=(c-d)[6(c+d)+2(c-d)-(c-d)^2]$

$=(c-d)[8c+4d-(c^2-2cd+d^2)]=(c-d)(-c^2+2cd-d^2+8c+4d)$