\(B=\dfrac{\left(a+3\right)^2}{2a^2+6a}\cdot\dfrac{1-6a-18}{a^2-9}\\ a,ĐK:a\ne0;a\ne\pm3\\ b,B=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}\cdot\dfrac{-17-6a}{\left(a-3\right)\left(a+3\right)}=\dfrac{-17-6a}{2a\left(a-3\right)}\\ c,B=0\Leftrightarrow-17-6a=0\Leftrightarrow a=-\dfrac{17}{6}\left(tm\right)\\ d,B=1\Leftrightarrow-17-6a=2a^2-6a\\ \Leftrightarrow2a^2=-17\Leftrightarrow a\in\varnothing\)

Lời giải:

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

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

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

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

...

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.

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

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

\(B=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)

\(=2+\dfrac{1}{\sqrt{a}}\cdot\dfrac{2a+2}{\sqrt{a}+1}\)

\(=\dfrac{2a+2\sqrt{a}+2a+2}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{4a+2\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+1\right)}\)

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

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

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

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

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

Vậy B=0

Đk:\(x>0;x\ne1\)

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

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

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

\(=\dfrac{1}{\sqrt{x}-1}\)

\(B=\dfrac{1}{2}\Leftrightarrow\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=2\)\(\Leftrightarrow x=9\) (tm)

Vậy..

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

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

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

\(B=\dfrac{1}{\sqrt{x}-1}\)

b) Với \(B=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=2\)

\(\Leftrightarrow\sqrt{x}=3\)

\(\Leftrightarrow x=9\)

Vậy...

Chúc bạn học tốt

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

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

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

\(=\dfrac{1}{\sqrt{x}-1}\)

b, Thay B = 1/2 vào ta được :\(\dfrac{1}{2}=\dfrac{1}{\sqrt{x}-1}\)

\(\Leftrightarrow\sqrt{x}=3\)

\(\Leftrightarrow x=9\)

Vậy ...

`B=(1/(3-sqrtx)-1/(3+sqrtx))*(3+sqrtx)/sqrtx(x>=0,x ne 9)`

`B=((3+sqrtx)/(9-x)-(3-sqrtx)/(9-x))*(3+sqrtx)/sqrtx`

`B=((3+sqrtx-3+sqrtx)/(9-x))*(3+sqrtx)/sqrtx`

`B=(2sqrtx)/((3-sqrtx)(3+sqrtx))*(3+sqrtx)/sqrtx`

`B=2/(3-sqrtx)`

`B>1/2`

`<=>2/(3-sqrtx)-1/2>0`

`<=>(4-3+sqrtx)/[2(3-sqrtx)]>0`

`<=>(sqrtx+1)/(2(3-sqrtx))>0`

Mà `sqrtx+1>=1>0`

`<=>2(3-sqrtx)>0`

`<=>3-sqrtx>0`

`<=>sqrtx<3`

`<=>x<9`

Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn.

\(a,A=\left(\cos^220^0+\cos^270^0\right)+\left(\cos^240^0+\cos^250^0\right)\\ A=\left(\cos^220^0+\sin^220^0\right)+\left(\cos^240^0+\sin^240^0\right)=1+1=2\\ b,B=\left(\cos^2\alpha\right)^3+\left(\sin^2\alpha\right)^3+3\sin^2\alpha\cdot\cos^2\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)\\ B=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)