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

 \(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)

Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)

=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)

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

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

Do đó: N = (-1)²⁰¹⁹ + 3.(-1)²⁰²⁰ - 2.(-1)²⁰²¹ = -1 + 3 + 2 = 4

ai nay dung kinh nghiem la chinh

cau a)

ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)

\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)

khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)

\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)

\(x=\frac{3-1}{1}=2\)

suy ra 

x^3-4x+1=1

A=1^2018

A=1

b)

ta thay

\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)

khi do 

\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)

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

x=2

thay vao

x^3+3x-14=0

B=0^2018

B=0

a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)

\(\Rightarrow x+\frac{1}{4x}=a^2-1\)

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

e/ ĐKXĐ: ...

\(\Leftrightarrow x^2-1+2x\sqrt{\frac{x^2-1}{x}}=3x\)

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

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

Đặt \(\sqrt{\frac{x^2-1}{x}}=a\ge0\)

\(a^2+2a=3\Leftrightarrow a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\left(l\right)\end{matrix}\right.\)

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

f/ ĐKXĐ: ...

\(\Leftrightarrow x^2-6+x\sqrt{\frac{x^2-6}{x}}-6x=0\)

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{x^2-6}{x}+\sqrt{\frac{x^2-6}{x}}-6=0\)

Đặt \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)

\(a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)