Lời giải:

Áp dụng công thức: $\cos 2x=\cos ^2x-\sin ^2x=1-2\sin ^2x=2\cos ^2x-1$ ta có:

\(\frac{6+2\cos 4a}{1-\cos 4a}=\frac{6+2(2\cos ^22a-1)}{2\sin ^22a}=\frac{2+2\cos ^22a}{\sin ^22a}=\frac{2+2(\cos ^2a-\sin ^2a)^2}{4\sin ^2a\cos ^2a}\)

\(=\frac{1+(\sin ^2a-\cos ^2a)^2}{2\sin ^2a\cos ^2a}=\frac{(\sin ^2a+\cos ^2a)^2+(\sin ^2a-\cos ^2a)^2}{2\sin ^2a\cos ^2a}=\frac{2(\sin ^4a+\cos ^4a)}{2\sin ^2a\cos ^2a}=\frac{\sin ^4a+\cos ^4a}{\sin ^2a\cos ^2a}\)

\(=\frac{\sin ^2a}{\cos ^2a}+\frac{\cos ^2a}{\sin ^2a}=\tan ^2a+\cot ^2a\) (đpcm)

\(\frac{4tana\left(1-tan^2a\right)}{\left(1+tan^2a\right)^2}=\frac{4\frac{sina}{cosa}\left(\frac{cos^2a-sin^2a}{cos^2a}\right)}{\left(\frac{sin^2a+cos^2a}{cos^2a}\right)^2}=4sina.cosa.cos2a\)

\(=2sin2a.cos2a=sin4a\)

Nếu chứng minh $\sqrt{x}+\sqrt{x+1}=1$ thì không có đủ cơ sở để cm bạn nhé. Bạn viết lại đề hoặc bổ sung thêm điều kiện để mọi người trợ giúp tốt hơn.

b) Ta có : a\(^2\)+ b\(^2\)+ c\(^2\) =ab+bc+ca

=> 2(a\(^2\)+b\(^2\)+c\(^2\))= 2(ab+bc+ca)

<=>2a\(^2\)+2b\(^2\)+2c\(^2\)=2ab+2bc+2ca

<=> 2a\(^2\)+2b\(^2\)+2c\(^2\)-2ab-2bc-2ca=0

<=> a\(^2\)+a\(^2\)+b\(^2\)+b\(^2\)+c\(^2\)+c\(^2\)-2ab-2bc=2ca=0

<=> (a\(^2\)-2ab+b\(^2\))+(b\(^2\)-2bc+b\(^2\))+(a\(^2\)-2ca+c\(^2\))

<=> (a-b)\(^2\)+(b-c)\(^2\)+(a-c)\(^2\) =a

<=> hoặc a-b=0 hoặc b-c=o hoặc a-c=o <=>a=b hoặc b=c hoặc a=c

=>a=b=c (đpcm)

a) Theo đề bài: \(a^2+b^2=ab\)

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

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

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

Vì \(\left(a-b\right)^2\ge0\)  để \(\left(a-b\right)^2+ab=0\) <=> \(\left(a-b\right)^2=ab=0\)

(a-b)²=0 <=> a-b=0 <=> a=b (đpcm)

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

=>\(2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)

=>\(2a^2+2b^2+2c^2=2ab+2bc+2ac\)

=>\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

Vì \(\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{cases}\) để \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

<=>\(\left(a-b\right)^2=\left(b-c\right)^2=\left(a-c\right)^2=0\)

<=>a-b=b-c=a-c=0

<=>a=b=c (đpcm)

\(\left(x+y+z\right)^2-x^2-y^2-z^2\)

\(=x^2+y^2+z^2+2xy+2yz+2zx-x^2-y^2-z^2\)

\(=2xy+2yz+2zx\)

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

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

\(VT=x^2+y^2+z^2+2xy+2yz+2xz-x^2-y^2-z^2\)

\(VT=2xy+2yz+2xz\)

\(VT=2\left(xy+yz+xz\right)\)

\(VT=VP\left(đpcm\right)\)

* VT: vế trái
  VP: vế phải

(x+y+z)²-x²-y²-z²   (1)

Ta co : 2(xy+yz+xz)

=2xy+2yz+2xz

=2xy+2yz+2xz+x²+y²+z²-x²-y²-z²

=(x+y+z)²-x²-y²-z²

Tu (1) suy ra dpcm 

\(\frac{a^2+2ab+b^2}{4}\ge ab\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\)

vì BĐT cuối đúng nên BĐT đầu đúng

Cảm ơn bạn nhiều nha....