\(Q^3=\dfrac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}+b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}+3Q\sqrt[3]{\dfrac{\left(b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}\right)\left(b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}\right)}{4}}\)

\(Q^3=\dfrac{2b^3-6b}{2}+3Q\sqrt[3]{\dfrac{\left(b^3-3b\right)^2-\left(b^2-1\right)^2\left(b^2-4\right)}{4}}\\ Q^3=b^3-3b+3Q\sqrt[3]{\dfrac{b^6-6b^4+9b^2-b^6+6b^4-9b^2+4}{4}}\\ Q^3=b^3-3b+3Q\sqrt[3]{\dfrac{4}{4}}=b^3-3b+3Q\\ \Leftrightarrow Q^3-3Q=b^3-3b\\ \Leftrightarrow Q\left(Q^2-3\right)=b\left(b^2-3\right)\)

\(\Leftrightarrow Q=b=\sqrt[3]{2020}\) (hmm ko chắc)

Akai Haruma

Bài 1:
Xét tử số:

\(\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}=\sqrt{3^2+5+2.3\sqrt{5}}-\sqrt{3^2+5-2.3\sqrt{5}}\)

\(=\sqrt{(3+\sqrt{5})^2}-\sqrt{(3-\sqrt{5})^2}=3+\sqrt{5}-(3-\sqrt{5})=2\sqrt{5}\)

Xét mẫu số:
\(\sqrt{(\sqrt{5}+1)\sqrt{6-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{5+1-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{(\sqrt{5}-1)^2}}\)

\(=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}=\sqrt{4}=2\)

Do đó: $A=\frac{2\sqrt{5}}{2}=\sqrt{5}$

Bài 2:

a)

$B=(\sqrt[3]{2}+1)^3(\sqrt[3]{2}-1)^3$
$=[(\sqrt[3]{2}+1)(\sqrt[3]{2}-1)]^3$
$=(\sqrt[3]{4}-1)^3$

$=3-3\sqrt[3]{16}+3\sqrt[3]{4}$

b)

Với $a,b$ đã cho ta đặt $\sqrt[3]{2}=x$. Khi đó:

\(a=\frac{6}{2x-2+\frac{2}{x}}=\frac{3x}{x^2-x+1}=\frac{3x(x+1)}{x^3+1}=\frac{3x(x+1)}{2+1}=x(x+1)\)

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

Khi đó:

$C=a^3b-ab^3=ab(a^2-b^2)=ab(a-b)(a+b)$

$=x^2(x^2-1)(2x)(2x^2)=4x^5(x^2-1)=8\sqrt[3]{4}(\sqrt[3]{4}-1)$

Bài 3:

Ta biết rằng $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}>0$ với mọi $x\in\mathbb{R}$

Do đó:

$|x^2-x+1|-|x-2|=6$

$\Leftrightarrow x^2-x+1-|x-2|=6(*)$

Nếu $x\geq 2$ thì $(*)\Leftrightarrow x^2-x+1-(x-2)=6$

$\Leftrightarrow x^2-2x-3=0$

$\Leftrightarrow (x-3)(x+1)=0$

$\Leftrightarrow x=3$ (do $x\geq 2$)

Nếu $x< 2$ thì $(*)\Leftrightarrow x^2-x+1-(2-x)=6$

$\Leftrightarrow x^2-7=0$

$\Rightarrow x=-\sqrt{7}$ (do $x< 2$)

Vậy........

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)

\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)

\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)

Từ ( 1 ) và ( 2 ) có đpcm

b)

)\(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)

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

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

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

=\(\frac{4\sqrt{5}}{4-5}\)

=\(\frac{4\sqrt{5}}{-1}\)

\(-4\sqrt{5}\)

ko khó nhưng mà bn đăng từng câu 1 hộ mk mk giải giúp cho

gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)

=> Thay vào thì     \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)

\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)

Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào

=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)

=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)

=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\) 

Đặt: \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)

=>     \(P=\frac{xy}{z^2+3xy}+\frac{yz}{x^2+3yz}+\frac{zx}{y^2+3zx}\)

=>     \(3P=\frac{3xy}{z^2+3xy}+\frac{3yz}{x^2+3yz}+\frac{3zx}{y^2+3zx}=1-\frac{z^2}{z^2+3xy}+1-\frac{x^2}{x^2+3yz}+1-\frac{y^2}{y^2+3zx}\)

Ta sẽ CM: \(3P\le\frac{9}{4}\)<=> Cần CM: \(\frac{x^2}{x^2+3yz}+\frac{y^2}{y^2+3zx}+\frac{z^2}{z^2+3xy}\ge\frac{3}{4}\)

Có:    \(VT\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)

Ta sẽ CM: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{3}{4}\)

<=> \(4\left(x+y+z\right)^2\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(4\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(x^2+y^2+z^2\ge xy+yz+zx\)

Mà đây lại là 1 BĐT luôn đúng => \(3P\le\frac{9}{4}\)=> \(P\le\frac{3}{4}\)

Vậy P max \(=\frac{3}{4}\)<=> \(a=b=c\)