a) Ta có:

\(VT=\left(a-b\right)^2\)

\(=a^2-2\cdot a\cdot b+b^2\)

\(=a^2-2ab+b^2\)

\(=a^2-4ab+2ab+b^2\)

\(=\left(a^2+2ab+b^2\right)-4ab\)

\(=\left(a+b\right)^2-4ab=VP\)

⇒ Đpcm

b) Ta có:

\(VT=\left(x+y\right)^2+\left(x-y\right)^2\)

\(=x^2+2\cdot x\cdot y+y^2+x^2-2\cdot x\cdot y+y^2\)

\(=x^2+2xy+y^2+x^2-2xy+y^2\)

\(=\left(x^2+x^2\right)+\left(2xy-2xy\right)+\left(y^2+y^2\right)\)

\(=2x^2+0+2y^2\)

\(=2x^2+2y^2\)

\(=2\left(x^2+y^2\right)=VP\)

⇒ Đpcm

a: (a-b)^2

=a^2-2ab+b^2

=a^2+2ab+b^2-4ab

=(a+b)^2-4ab

b: (x+y)^2+(x-y)^2

=x^2+2xy+y^2+x^2-2xy+y^2

=2x^2+2y^2

=2(x^2+y^2)

Bài 1: 

c: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)

Câu b bạn sửa lại đề

\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)

a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)

a: (sina+cosa)^2

=sin^2a+cos^2a+2*sina*cosa

=1+sin2a

b: \(cos^4a-sin^4a=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)\)

\(=cos^2a-sin^2a=cos2a\)

a: Ta có: \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4\sqrt{6}}{2}\right)\cdot\dfrac{1}{\sqrt{6}}\)

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

a)    Ta có:

\(\begin{array}{l}{\sin ^4}\alpha  - {\cos ^4}\alpha  = 1 - 2{\cos ^2}\alpha \\ \Leftrightarrow \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)\left( {{{\sin }^2}\alpha  - {{\cos }^2}\alpha } \right) = 1 - 2{\cos ^2}\alpha \\ \Leftrightarrow {\sin ^2}\alpha  - {\cos ^2}\alpha  - 1 + 2{\cos ^2}\alpha  = 0\\ \Leftrightarrow {\sin ^2}\alpha  + {\cos ^2}\alpha  - 1 = 0\\ \Leftrightarrow 1 - 1 = 0\\ \Leftrightarrow 0 = 0\end{array}\)

Đẳng thức luôn đúng

b)    Ta có:

\(\begin{array}{l}\tan \alpha  + \cot \alpha  = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{\cos \alpha .\sin \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{1}{{\sin \alpha .\cos \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\end{array}\)

Đẳng thức luôn đúng

a) Ta có: \({\left( {\sin \alpha  + \cos \alpha } \right)^2} = {\sin ^2}\alpha  + 2\sin \alpha \cos \alpha  + {\cos ^2}\alpha  = 1 + \sin 2\alpha \;\)

b) \({\cos ^4}\alpha  - {\sin ^4}\alpha  = \left( {{{\cos }^2}\alpha  - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2}\alpha  + {{\sin }^2}\alpha } \right) = \cos 2\alpha \;\)

BĐT này sai nha bạn.

Nó chỉ đúng khi \(x>0\)

Với \(x>0\) thì bất đẳng thức tương đương với \(x^2+1\ge2x\)

\(\Leftrightarrow x^2-2x+1\ge0\) \(\Leftrightarrow\left(x-1\right)^2\ge0\) (luôn đúng)

\(\Rightarrow\) Điều cần chứng minh là đúng

a, \(\dfrac{a^2+2ab+b^2}{4}\ge ab\)

\(\Leftrightarrow\)a^2+2ab+b^2>=4ab

\(\Leftrightarrow\)a^2-2ab+b^2>=0

\(\Leftrightarrow\)(a-b)^2>=0 (luôn đúng)

b,\(a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge0\) 

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) luôn đúng