\(\left(1+tan^2a\right)\left(1-sin^2a\right)-\left(1+cot^2a\right)\left(1-cos^2a\right)\)

\(=\left(1+\dfrac{sin^2a}{cos^2a}\right).cos^2a-\left(1+\dfrac{cos^2a}{sin^2a}\right).sin^2a\)

\(=cos^2a+sin^2a-sin^2a-cos^2a=\)\(0\)

Vậy B=0

Lời giải:
$B=\frac{1}{4}.(-2).3.(x^3.x^3)(y.y^5.y).z^3$

$=\frac{-3}{2}x^6y^7z^3$
Bậc của $B$: $6+7+3=16$

`1/4 x^3y * (-2)x^3y^5 * 3yz^3`

`=[1/4 *(-2) * 3] *(x^3*x^3) *(y*y^5*y) *z^3`

`= -3/2 x^6y^7z^3`

Bậc của đơn thức : `16`

Lời giải:

\(B=\frac{3}{x-1}\sqrt{\frac{(x-1)^2}{(3x)^2}}=\frac{3}{x-1}|\frac{x-1}{3x}|\)

\(=\frac{3}{x-1}.\frac{1-x}{3x}=\frac{-1}{x}\)

\(B=\dfrac{3}{x-1}.\sqrt{\dfrac{x^2-2x+1}{9x^2}}=\dfrac{3}{x-1}.\sqrt{\left(\dfrac{x-1}{3x}\right)^2}\)

\(=\dfrac{3}{x-1}.\left|\dfrac{x-1}{3x}\right|=\dfrac{3}{x-1}.\dfrac{1-x}{3x}=-\dfrac{1}{x}\)

...

đk : x >= 0 ; x khác 1 

\(B=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}=\dfrac{4\sqrt{x}}{x-1}\)

B xác định \(< =>\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}=\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{x-1}=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}=\dfrac{4\sqrt{x}}{x-1}\)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\left(luôn.đúng\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

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

\(B=x^5y^2+\dfrac{1}{2}x^5y^2-6xy+1=\dfrac{3}{2}x^5y^2-6xy+1\)

\(B=-\dfrac{1}{7}x^2y+x^5y^2-xy+\dfrac{1}{2}x^5y^2-5xy+\dfrac{1}{7}x^2y+2021^0\\ =\left(-\dfrac{1}{7}x^2y+\dfrac{1}{7}x^2y\right)+\left(x^5y^2+\dfrac{1}{2}x^5y^2\right)-\left(xy+5xy\right)+1\\ =0+\dfrac{3}{2}x^5y^2-6xy+1\\ =\dfrac{3}{2}x^5y^2-6xy+1\)

ĐKXĐ: \(x\ge0;x\ne9\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{3}{x-9}\right):\dfrac{1}{\sqrt{x}-3}\)

\(=\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{1}{\sqrt{x}-3}\)

\(=\dfrac{\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{1}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)

\(B=\dfrac{\sqrt{x}-3+3}{x-9}\cdot\left(\sqrt{x}-3\right)=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1}{x-1}=\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x-1}{x+1}\)

\(=\dfrac{2}{x-1}.\dfrac{x-1}{x+1}=\dfrac{2}{x+1}\)

Để \(B< 1\Rightarrow\dfrac{2}{x+1}< 1\Rightarrow1-\dfrac{2}{x+1}>0\Rightarrow\dfrac{x-1}{x+1}>0\)

mà \(x+1>0\left(x\ge0\right)\Rightarrow x-1>0\Rightarrow x>1\)

a) Ta có: \(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1}{x-1}\)

\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x+1}\)

\(=\dfrac{2}{x+1}\)