Tiếc quá 

mình chưa học đến

bik thì giúp cho

xin lỗi bn mik mới học lớp 6 thôi

Xét \(y=\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}+\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}\Rightarrow y^2=4+2\sqrt{4-\frac{5+\sqrt{5}}{2}}\)

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

Suy ra \(\Rightarrow x=3+\sqrt{5}-\sqrt{3-\sqrt{5}}-1=2+\sqrt{5}-\sqrt{3-\sqrt{5}}\)

............................................................

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

câu 2

\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)

câu 1

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

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

\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)

\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)

a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)

\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)

\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

\(=-\left(2-5\right)\)

\(=-\left(-3\right)\)

\(=3\)

b) Ta có:

\(x^2-x\sqrt{3}+1\) 

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

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

Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên

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

Dấu "=" xảy ra:

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

\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)

Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)

a)

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