\(a^3+b^3=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}\)

\(=\sqrt{6}-\sqrt{2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}=0\)

\(\Rightarrow a=-b\Rightarrow a^5+b^5=0\)

Câu 1: 

Ta có: \(\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)

\(=6x^2+9x+14x+21-\left(6x^2+33x-10x-55\right)\)

\(=6x^2+23x+21-6x^2-23x+55\)

=76

`3b)\sqrt{25x-25}-15/2\sqrt{(x-2)/9}=6+3/2\sqrt{x-1}`

ĐK:`x>=1`

`pt<=>sqrt{25(x-1)}-15/2*1/3sqrt{x-1}-3/2sqrt{x-1}=6`

`<=>5sqrt{x-1}-5/2sqrt{x-1}-3/2sqrt{x-1}=6`

`<=>5sqrt{x-1}-4sqrt{x-1}=6`

`<=>sqrt{x-1}=6`

`<=>x-1=36`

`<=>x=37(tmddk)`

Vậy `S={37}`

3b) \(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\dfrac{3}{2}\sqrt{x-1}\left(x\ge1\right)\)

\(\Leftrightarrow\sqrt{25\left(x-1\right)}-\dfrac{15}{2}\sqrt{\dfrac{1}{9}.\left(x-1\right)}-\dfrac{3}{2}\sqrt{x-1}=6\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}.\dfrac{1}{3}\sqrt{x-1}-\dfrac{3}{2}\sqrt{x-1}=6\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{x-1}=6\)

\(\Leftrightarrow\sqrt{x-1}=6\Rightarrow x-1=36\Rightarrow x=37\)

Bài 3: 

c) ĐKXĐ: \(x\ge0\)

Ta có: \(\sqrt{4+5\sqrt{x}}=3\)

\(\Leftrightarrow5\sqrt{x}+4=9\)

\(\Leftrightarrow5\sqrt{x}=5\)

\(\Leftrightarrow x=1\)

Vậy: S={1}

d) Ta có: \(x-\sqrt{4x-20}=20\)

\(\Leftrightarrow\sqrt{4x-20}=x-20\)

\(\Leftrightarrow\left(x-20\right)^2=4x-20\)

\(\Leftrightarrow x^2-40x+400-4x+20=0\)

\(\Leftrightarrow x^2-44x+420=0\)

\(\Delta=\left(-44\right)^2-4\cdot1\cdot420=256\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{44-16}{2}=\dfrac{28}{2}=14\\x_2=\dfrac{44+16}{2}=\dfrac{60}{2}=30\end{matrix}\right.\)

Vậy: S={14;30}

1:

1: Để hàm số đồng biến thì m>0

2: Khi m=2 thì y=2x+1

Tọa độ giao là;

2x+1=x+3 và y=x+3

=>x=2 và y=5

Ta có:x²+y²=25➝(x+y)²-2xy=25➝(x+y)²=1➝x+y=1.Đến đây bạn tự làm.

a: =>x=y+11

xy=60

\(\Leftrightarrow y^2+11y-60=0\)

\(\Leftrightarrow\left(y+15\right)\left(y-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=-15\\y=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=15\end{matrix}\right.\)