Ta có: \(a=\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}=\frac{\sqrt{3}}{\sqrt{5}}+\frac{\sqrt{5}}{\sqrt{3}}=\frac{8\sqrt{15}}{15}\)

=> \(a^2=\frac{64}{15}\)

=> \(M=\sqrt{15a^2-8a\sqrt{15}+16}=\sqrt{15.\frac{64}{15}-8.\frac{8\sqrt{15}}{15}.\sqrt{15}+16}\)

\(M=\sqrt{64-64+16}=4\)

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

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

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

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

\(C^2=2x^2-2\sqrt{x^4-4\left(x^2-1\right)}\)

\(C^2=2x^2-2\sqrt{x^4-4x^2+4}\)

\(C=\sqrt{2x^2-2\sqrt{x^4-4x^2+4}}\) 

Thay: \(x=\sqrt{5}\) vào C, ta có:

\(C=\sqrt{2\sqrt{5}^2-2\sqrt{\sqrt{5}^4-4\sqrt{5}^2+4}}\)

\(C=\sqrt{10-2\sqrt{25-20+4}}\)

\(C=\sqrt{10-2\sqrt{9}}\)

\(C=\sqrt{10-6}\)

\(C=\orbr{\begin{cases}-2\\2\end{cases}}\)

Mà theo bài ra: \(\sqrt{x^2+2\sqrt{x^2-1}}>\sqrt{x^2-2\sqrt{x^2-1}}\)

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

\(\Rightarrow C=2\)

Đề câu a là \(4\sqrt{5}a\) hay \(4\sqrt{5a}\) . Thấy \(4\sqrt{5}a\) đúng hơn
 

\(A=\sqrt{5a^2-4\sqrt{5}a+4}\) 

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

\(A=\sqrt{5}a-2\)

Thay \(a=\sqrt{5}+\frac{1}{\sqrt{5}}\) vào A, có:

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

\(A=5+1-2\) 

\(A=4\)

Giải

Ta có: \(\sqrt{\dfrac{5}{3}}+\sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{5}}{\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{5}}=\dfrac{8}{\sqrt{15}}\)

Vậy M = \(\sqrt{15\left(\dfrac{8}{15}\right)^2-8.\dfrac{8}{\sqrt{15}}.\sqrt{15}+16}\)

= \(\sqrt{8^2-8^2+16}=\sqrt{16}=4\)

\(M=\sqrt{15a^2-8a\sqrt{15}+16}=\sqrt{\left(\sqrt{15}a-4\right)^2}\)

\(=\sqrt{15}a-4=\sqrt{15}\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)-4\)

\(=\left(3+5\right)-4=4\)

2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)

b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\)  thì 

     \(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)

         \(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)

        \(=\sqrt{4^2}\)

         \(=4\)

ĐK: \(x-9\ne0\Rightarrow x\ne9\)

\(\sqrt{x}\ge0\Rightarrow x\ge0\)

\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)

\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)

ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)

\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

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

\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)

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

2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)

\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)

\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)

a,\(x\ge0,x\ne49\)

a/ \(\sqrt{4a^4-12a^2+9}-\sqrt{a^4-8a^2+16}\)

= \(\sqrt{\left(2a^2-3\right)^2}-\sqrt{\left(a^2-4\right)^2}\)

= \(|2a^2-3|-|a^2-4|\)

= \(2a^2-3+a^2-4\)
= \(3a^2-7\)

Thay a=\(\sqrt{3}\).Ta có:

\(3.\left(\sqrt{3}\right)^2-7\)

= 3.3-7=2

b/ \(\sqrt{10a^2-12a\sqrt{10}+36}\)

= \(\sqrt{\left(a\sqrt{10}\right)^2-2.a\sqrt{10}.6+6^2}\)

= \(\sqrt{\left(a\sqrt{10}-6\right)^2}\)

= \(|a\sqrt{10}-6|\)

=  \(-a\sqrt{10}+6\)

Thay  a= \(\sqrt{\frac{5}{2}}-\sqrt{\frac{2}{5}}\)=\(\frac{3}{\sqrt{10}}\),Ta có:

\(-\frac{3}{\sqrt{10}}.\sqrt{10}+6\)

= -3+6 =3

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