Bài 2: 

\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

Ta có: \(P=x^2-2x+2020\)

\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)

\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)

=2026

Bài 1: 

\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)

=-6

b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)

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

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

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

Đặt a = \(\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}+\sqrt{2}-\sqrt{\frac{5+\sqrt{5}}{2}}}\)

\(a^2=4+2\sqrt{4-\frac{5+\sqrt{5}}{2}}=4+\sqrt{6-2\sqrt{5}}\)

\(=4+\sqrt{\left(\sqrt{5}-1\right)^2}=3+\sqrt{5}\Rightarrow a=\sqrt{3}+\sqrt{5}\)

\(\Rightarrow\)\(x=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-1\)

\(=\sqrt{\frac{6+2\sqrt{5}}{2}}-\sqrt{\frac{6-2\sqrt{5}}{2}}-1=\frac{\sqrt{5}+1}{\sqrt{2}}-\frac{\sqrt{5}-1}{\sqrt{2}}-1\)

\(=\sqrt{2}-1\Rightarrow x=\sqrt{2}-1\Rightarrow x=x^2+2x-1=0\)

\(B=2x^3+3x^2-4x+2\)

\(B=2x\left(x^2+2x-1\right)-\left(x^2+2x-1\right)+1=1\)

Tham khao:

2,Biết x+y=5x+y=5 và x+y+x2y+xy2=24x+y+x2y+xy2=24 Giá trị của biểu thức x3+y3x3+y3 là

3,Nếu đa thức x2+px2+qx2+px2+q chia hết cho đa thức x2−2x−3x2−2x−3 thì khi đó giá trị của

2) x+y+x2y+xy2=24⇔x+y+xy(x+y)=24⇔5+5xy=24⇔xy=24−55=3,8x+y+x2y+xy2=24⇔x+y+xy(x+y)=24⇔5+5xy=24⇔xy=24−55=3,8

(x+y)=5⇔x2+2xy+y2=25⇔x2+y2=25−2xy=17,4(x+y)=5⇔x2+2xy+y2=25⇔x2+y2=25−2xy=17,4

x3+y3=(x+y)(x2−xy+y2)=5(17,4−3,8)=68

3) x4−2x−3=(x+1)⋅(x−3)x4−2x−3=(x+1)⋅(x−3)

Để đa thức x4+px2+q⋮x2−2x−3x4+px2+q⋮x2−2x−3 => Có hai nghiệm của x là x = -1 hoặc x = 3.

+) Xét x = -1 : x4+px2+q=0⇒(−1)4+p⋅(−1)2+q=0x4+px2+q=0⇒(−1)4+p⋅(−1)2+q=0

⇒1+p+q=0→q=−1−p⇒1+p+q=0→q=−1−p (1)

+) Xét x = 3 : x4+px2+q=0⇒34+p⋅32+q=0x4+px2+q=0⇒34+p⋅32+q=0

⇒81+p⋅9+q=0⇒81+p⋅9+q=0 (2)

Thế (1) vào (2) ta có : 81+9⋅p−1−p=081+9⋅p−1−p=0

⇔80+8p=0⇔80+8p=0

⇔p=−10⇔p=−10

Vậy giá trị của p là -10.

Đặt a\(=\sqrt{2-\sqrt{\frac{5-\sqrt{5}}{2}-\sqrt{2}-\sqrt{\frac{5+\sqrt{5}}{2}}}}\)

\(a^2=4+2\sqrt{4-\frac{5+\sqrt{5}}{2}=4+\sqrt{6-2\sqrt{5}}}\)

\(=4-\sqrt{\left(\sqrt{5-1}\right)^2=3+\sqrt{5}\Rightarrow a=\sqrt{3-\sqrt{5}}}\)

\(\Rightarrow x=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-1\)

\(=\sqrt{\frac{6-2\sqrt{5}}{2}}-\sqrt{\frac{6-2\sqrt{5}}{2}}-1=\frac{\sqrt{5+1}}{\sqrt{2}}-\frac{\sqrt{5}-1}{\sqrt{2}}-1\)

\(=\sqrt{2}-1\Rightarrow x=\sqrt{2}-1\Rightarrow x=x^2-2x-1=0\)

\(B=2x^3-3x^2-4x+2\)

\(B=2x\left(x^2-2x-1\right)-\left(x^2-2x-1\right)+1=1\)

P/s : Mình tự làm không phải mình copy đâu

1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)

\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)

\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)

\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)

\(=-8\sqrt{3}\)

2) \(A=\sqrt{12-4x}\) có nghĩa khi:

\(12-4x\ge0\)

\(\Leftrightarrow4x\le12\)

\(\Leftrightarrow x\le\dfrac{12}{4}\)

\(\Leftrightarrow x\le3\)

3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)

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

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

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

a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

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

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

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

\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\\ \Leftrightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow x^3=18+3x\sqrt[3]{81-80}=18-3x\\ \Leftrightarrow x^3-3x=18\\ y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\\ \Leftrightarrow y^3=6+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\right)\\ \Leftrightarrow y^3=6+3y\sqrt[3]{9-8}=6+3y\\ \Leftrightarrow y^3-3y=6\\ \Leftrightarrow P=x^3+y^3-3\left(x+y\right)+1993\\ P=x^3+y^3-3x-3y+1993=18+6+1993=2017\)

Áp dụng: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab\left(a+b\right)\)

\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)

\(=18+3\sqrt[3]{81-80}.x=18+3x\)

\(y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\)

\(\Rightarrow y^3=3-2\sqrt{2}+3+2\sqrt{2}+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\right)\)

\(=6+3\sqrt[3]{9-8}y=6+3y\)

\(P=x^3+y^3-3\left(x+y\right)+1993\)

\(=18+3x+6+3y-3x-3y+1993=2017\)