Ta có: 

\(\sqrt{\dfrac{7+3\sqrt{5}}{2}}\)

\(=\sqrt{\dfrac{2\cdot\left(7+3\sqrt{5}\right)}{2\cdot2}}\)

\(=\sqrt{\dfrac{14+6\sqrt{5}}{4}}\)

\(=\sqrt{\dfrac{\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot3-3^2}{2^2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{5}+3\right)^2}{2^2}}\)

\(=\dfrac{3+\sqrt{5}}{2}\)

Mà: \(\dfrac{3+\sqrt{5}}{2}=a+b\sqrt{5}\)

Nên:  \(\dfrac{3+\sqrt{5}}{2}=\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}=\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}\)

Vậy: \(a=\dfrac{3}{2};b=\dfrac{1}{2}\)

\(\Rightarrow a+b=\dfrac{3}{2}+\dfrac{1}{2}=\dfrac{4}{2}=2\)

\(\sqrt{\dfrac{7+3\sqrt{5}}{2}}=\sqrt{\dfrac{14+6\sqrt{5}}{4}}=\sqrt{\left(\dfrac{3+\sqrt{5}}{2}\right)^2}\)

\(=\dfrac{3+\sqrt{5}}{2}\)

=>a=3/2; b=1/2

a+b=3/2+1/2=2

a: R-3=(x^2+x-1-3x)/x=(x-1)^2/x

Nếu x>0 thì R-3>0

=>R>3

Nếu x<0 thì R-3<0

=>R<3

c: Để R>4 thì R-4>0

=>\(\dfrac{x^2+x+1-4x}{x}>0\)

=>\(\dfrac{x^2-3x+1}{x}>0\)

TH1: x>0 và x^2-3x+1>0

=>x>0 và \(\left[{}\begin{matrix}x< \dfrac{3-\sqrt{5}}{2}\\x>\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow x>\dfrac{3+\sqrt{5}}{2}\)

mà x nguyên

nên x>3

TH2: x<0 và x^2-3x+1<0

=>x<0 và \(\dfrac{3-\sqrt{5}}{2}< x< \dfrac{3+\sqrt{5}}{2}\)(loại)

ai lm tr mình cho tích

a) Có:

 \(a+b+c=0\\\Leftrightarrow\left(a+b+c\right)^2=0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\\ \Leftrightarrow2ab+2bc+2ca=-1\\ \Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\\ \Leftrightarrow\left(ab+bc+ca\right)^2=\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}-0=\dfrac{1}{4} \)

câu (b) cho đa thức P (x) = cái gì?

Bài 3:

a: \(\Leftrightarrow8n^2+4n-8n-4+5⋮2n+1\)

\(\Leftrightarrow2n+1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{0;-1;2;-3\right\}\)

b: \(\Leftrightarrow4n^3-2n^2-6n+3+2⋮2n-1\)

\(\Leftrightarrow2n-1\in\left\{1;-1\right\}\)

hay \(n\in\left\{1;0\right\}\)

\(R=\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{3x-5\sqrt{x}}{4-x}\right):\left(\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}-1\right)\left(ĐK:x\ge0,x\ne4\right)\\ =\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{3x-5\sqrt{x}}{\sqrt{x}^2-2^2}\right):\dfrac{2\sqrt{x}-1-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)

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

\(=\dfrac{7x-9\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

Bạn xem lại đề nhé, rút gọn thường ra kết quả rất đẹp chứ không dài như kết quả này đâu ạ.

Giúp với ạ mình cảm ơn ai làm được mình cho 100sao 

b, \(A=\dfrac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}=\dfrac{2^{12}.3^5-\left(2^2\right)^6.3^4}{2^{12}.3^6+\left(2^3\right)^4.3^5}=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}=\dfrac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}=\dfrac{2}{3.4}=\dfrac{1}{6}\)

a. Vì | x | = 1,5

\(\Rightarrow\left[{}\begin{matrix}x=1,5\\x=-1,5\end{matrix}\right.\)

Với x = 1,5 ; y = -0,75 thì :

P = 1,5 - 4 . 1,5 . (-0,75) + ( -0,75 )

P = 1,5 - ( -4,5 ) -0,75

P = 6 - 0,75

P = 5,25

Với x = -1,5 ; y = -0,75 thì

P = -1,5 - 4 . ( -1,5 ) .( -0,75 ) + ( -0,75 )

P = -1,5 - 4,5 -0,75

P = -6 - 0,75

P = -6,75

\(\left|x-m\right|=25\Leftrightarrow\left[{}\begin{matrix}x=m+25\\x=m-25\end{matrix}\right.\)

\(\left|x\right|\ge2020\Leftrightarrow\left[{}\begin{matrix}x\ge2020\\x\le-2020\end{matrix}\right.\)

+) \(x=m+25\)

Để \(A\cap B=\varnothing\) \(\Leftrightarrow\left\{{}\begin{matrix}m+25>-2020\\m+25< 2020\end{matrix}\right.\)\(\Leftrightarrow-2045< m< 1995\)

+) \(x=m-25\)

Để \(A\cap B=\varnothing\)  \(\Leftrightarrow\left\{{}\begin{matrix}m-25>-2020\\m-25< 2020\end{matrix}\right.\)\(\Leftrightarrow-1995< m< 2045\)

1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)

\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)

c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)

Vậy \(x>4\)thì \(R>0\)

2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)

Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)

3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)

\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)

b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)

\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)