6/ Đặt \(\hept{\begin{cases}\sqrt[4]{x}=a\\\sqrt[4]{2-x}=b\end{cases}}\)

\(\Rightarrow b^4+a^4=2\)

Từ đó ta có: a + b = 2

Ta có: \(a^4+b^2\ge\frac{\left(a^2+b^2\right)^2}{2}\ge\frac{\left(a+b\right)^4}{8}=\frac{16}{8}=2\)

Dấu = xảy ra khi a = b = 1

=> x = 1

\(x-1\)

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

\(x-4\sqrt{x}+4\)

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

a) = a^10 - a + a^5 - a^2 + a^2 + a + 1

= a(a^9 - 1) + a^2(a^3 - 1) + (a^2 + a + 1)

= a.(a^3-1)(a^6 + a^3 + 1) + a^2(a-1)(a^2+a+1) + (a^2 + a + 1)

= a.(a-1)(a^2 + a + 1)(a^6 + a^3 + 1) + a^2(a-1)(a^2+a+1) + (a^2 + a + 1)

= (a^2 + a + 1)[(a.(a-1)(a^6 + a^3 + 1) + a^2 + 1]

b) x^5 - x^4 - 1 = x^5 - x^4 + x^3 - x^3 + x^2 - x - x^2 + x - 1

= x^3(x^2 - x + 1) - x(x^2 - x + 1) - (x^2 - x + 1)

= (x^2 - x + 1)(x^3 - x - 1)

a) \(a^{10}+a^5+1\)

\(=\left(a^{10}-a^9+a^7-a^6+a^5-a^3+a^2\right)\)

\(+\left(a^9-a^8+a^6-a^5+a^4-a^2+a\right)\)

\(+\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(=a^2\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(+a\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(+\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

\(=\left(a^2+a+1\right)\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)

1. \(x^3-6x^2+10x-4=0\)

<=> \(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)

<=>  \(\left(x-2\right)\left(x^2-4x+2\right)=0\)

<=> \(\orbr{\begin{cases}x=2\\x^2-4x+2=0\left(1\right)\end{cases}}\)

Giải pt (1): \(\Delta=\left(-4\right)^2-4.2=8>0\)

=> pt (1) có 2 nghiệm: \(x_1=\frac{4+\sqrt{8}}{2}=2+\sqrt{2}\)

\(x_2=\frac{4-\sqrt{8}}{2}=2-\sqrt{2}\)

1) Ta có: \(x^3-6x^2+10x-4=0\)

       \(\Leftrightarrow\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)

       \(\Leftrightarrow x^2\left(x-2\right)-4x\left(x-2\right)+2\left(x-2\right)=0\)

       \(\Leftrightarrow\left(x^2-4x+2\right)\left(x-2\right)=0\)

   + \(x-2=0\)\(\Leftrightarrow\)\(x=2\)\(\left(TM\right)\)

   + \(x^2-4x+2=0\)\(\Leftrightarrow\)\(\left(x^2-4x+4\right)-2=0\)

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

                                             \(\Leftrightarrow\)\(x-2=\pm\sqrt{2}\)

                                             \(\Leftrightarrow\)\(\orbr{\begin{cases}x=2+\sqrt{2}\approx3,4142\left(TM\right)\\x=2-\sqrt{2}\approx0,5858\left(TM\right)\end{cases}}\)

Vậy \(S=\left\{0,5858;2;3,4142\right\}\)

Câu 2b đề là tìm x chứ nhỉ???

b) \(\sqrt{x^2-4}+\sqrt{x-2}=0\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{x^2-4}\ge0\\\sqrt{x-2}\ge0\end{matrix}\right.\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x-2}=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2-4=0\\x-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\pm2\\x=2\end{matrix}\right.\) <=> x = 2

Vậy x = 2

bài 2 câu b) đề sai rồi bạn

còn bài 1 câu b) mình cảm thấy sai sai

Sửa đề :

a) \(A=\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{4}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-5}{x-\sqrt{x}-2}\right)\)

\(\Leftrightarrow A=\frac{x-\sqrt{x}+4\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-4-x+\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

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

\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}\)

b) \(A=4\)

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

\(\Leftrightarrow x+3\sqrt{x}+4=4\sqrt{x}+4\)

\(\Leftrightarrow x-\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy \(A=4\Leftrightarrow x\in\left\{0;1\right\}\)

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

\(\Leftrightarrow x^3=18+3x\)

Tương tự co:

\(y^3=6+3y\)

\(\Rightarrow P=18+3x+6+3y-3\left(x+y\right)+2019=2043\)

a: \(A=10\sqrt{x}-10\sqrt{x}-\dfrac{4}{3}\cdot\dfrac{x\sqrt{x}}{2}\)

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

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

\(=-\left(5-4\right)=-1\)