Bài 3

\(A=\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\sqrt{3+\sqrt{5}}\cdot\sqrt{2}\left(\sqrt{5}-1\right)\)

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

\(=2\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=2\cdot4=8\left(đpcm\right)\)

\(B=\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{2-\sqrt{3}}\right)=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)

\(=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\left(đpcm\right)\)

Bài 4

\(P=\frac{3\sqrt{10}+\sqrt{20}-3\sqrt{6}-\sqrt{12}}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{10}\left(\sqrt{2}+1\right)-\sqrt{6}\left(\sqrt{2}+1\right)}{\sqrt{5}-\sqrt{3}}\)

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

\(Q=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+2\sqrt{2}+2+2}{\sqrt{2}+\sqrt{3}+2}\)

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

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

\(\left|3-\sqrt{5}\right|+\left|3+\sqrt{5}\right|+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

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

\(6+2\sqrt{9-5}\)

\(6+2\sqrt{4}\)

\(=10\)

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

\(=4+2\sqrt{3}-2=2+2\sqrt{3}\)

\(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)

\(1+\sqrt{3}+\sqrt{2}+\sqrt{3}+3+\sqrt{6}-\sqrt{2}-\sqrt{6}-2\)

\(2+2\sqrt{3}\)

\(a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}\right)^3-\left(\sqrt{b}\right)^3=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\)

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

Điểm rơi đạt tại \(a=b=c=1\)

Áp dụng BĐT Cô-si có : \(\frac{a^3}{b}+ab\ge2\sqrt{\frac{a^4b}{b}}=2a^2\)

Xây dựng các đánh giá tương tự và cộng theo vế ta được :

\(LHS+ab+bc+ca\ge2a^2+2b^2+2c^2\)

\(< =>LHS\ge2a^2+2b^2+2c^2-ab-bc-ca\)(*)

Sử dụng BĐT phụ \(a^2+b^2+c^2\ge ab+bc+ca\)khi đó : 

\(2a^2+2b^2+2c^2-ab-bc-ca\ge ab+bc+ca=3\)(**)

Từ (*) và (**) suy ra điều phải chứng minh nhéeeeeeee

Áp dụng BĐT AM-GM cho 2 số thực dương có : \(\frac{a^3}{b}+ab\ge2a^2\)

Tương tự và cộng theo vế ta được :\(VT+ab+bc+ca\ge2\left(a^2+b^2+c^2\right)\)

\(< =>VT\ge2\left(a^2+b^2+c^2\right)-ab-bc-ca\)(1)

Ta cần cm BĐT phụ sau \(a^2+b^2+c^2\ge ab+bc+ca\)(2)

\(< =>\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)(đúng)

Từ (1) và (2) suy ra \(VT\ge2\left(a^2+b^2+c^2\right)-ab-bc-ca\ge ab+bc+ca=3\)

Dấu "=" xảy ra \(< =>a=b=c=1\)

xong rồi nhé

Áp dung BĐT Cô-si, ta có:

\(\frac{a^3}{b}+ab\ge2\sqrt{\frac{a^3}{b}.ab}=2a^2\)

\(\frac{b^3}{c}+bc\ge2\sqrt{\frac{b^3}{c}.bc}=2b^2\)

\(\frac{c^3}{a}+ac\ge2\sqrt{\frac{c^3}{a}.ac}=2c^2\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+ab+bc+ca\ge2\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\ge2\left(ab+bc+ca\right)-\left(ab+bc+ca\right)\)

\(\Leftrightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca=3\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

Áp dụng bất đẳng thức Bu-nhia-cốp-xki ta được:

\(\left(x-2+4-x\right)\left(1+9\right)\ge\left(\sqrt{x-2}+3\sqrt{4-x}\right)^2\).

\(\Leftrightarrow20\ge P^2\Leftrightarrow-\sqrt{20}\le P\le\sqrt{20}.\)

Dấu bằng bạn tự tìm dấu bằng xảy ra của BĐT Bunhiacopxki nha, trên mạng có nhiều.

ờ .....ừm ko biết nhiều cho lắm

\(a,\sqrt{1-3x}\)

\(< =>1-3x\ge0\)

\(3x\le1\)

\(x\le\frac{1}{3}\)

\(b,-3< 0\)

\(< =>2x-5\ne0;2x-5\le0< =>2x-5< 0\)

\(x< \frac{5}{2}\)

\(c,\sqrt{3x+2}+\sqrt{-2x+3}\)

\(\hept{\begin{cases}3x+2\ge0\\-2x+3\ge0\end{cases}}\)

\(\hept{\begin{cases}x\ge-\frac{2}{3}\\x\le\frac{3}{2}\end{cases}}\)

\(< =>-\frac{2}{3}\le x\le\frac{3}{2}\)

\(d,\frac{x-5}{\sqrt{-4x}}\)

\(\sqrt{-4x}\ge0;\sqrt{-4x}\ne0< =>\sqrt{-4x}>0\)

\(-4x>0\)

\(x< 0\)

\(e,\sqrt{x-2}+\frac{1}{x-3}\)

\(\sqrt{x-2}\ge0;x-3\ne0\)

\(x\ge2;x\ne3\)

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

\(\sqrt{-\left(x-2\right)^2}\ge0\)

\(-\left|x-2\right|\ge0\)

\(-\left|x-2\right|\le0\)

lên chỉ có 1 nghiệm duy nhất là 

\(x-2=0< =>x=2\)

\(g,\sqrt{\frac{-2x^2}{3x+2}}\)

\(-2x^2\le0\)

\(\sqrt{\frac{-2x^2}{3x+2}}\ge0< =>3x+2\le0;3x+2\ne0\)

\(x\le-\frac{2}{3};x\ne-\frac{2}{3}< =>x< -\frac{2}{3}\)

a)\(\sqrt{1-3x}\)có nghĩa \(\Leftrightarrow\sqrt{1-3x}\ge0\)

\(\Leftrightarrow1-3x\ge0\)

\(\Leftrightarrow-3x\ge-1\)

\(\Leftrightarrow x\ge\frac{1}{3}\)

b)\(\sqrt{\frac{-3}{2x-5}}\)có nghĩa \(\Leftrightarrow\sqrt{\frac{-3}{2x-5}}\ge0\)

\(\Leftrightarrow\frac{-3}{2x-5}\ge0\)

\(\Leftrightarrow2x-5>0\)

\(\Leftrightarrow2x>5\)

\(\Leftrightarrow x>\frac{5}{2}\)

c)\(\sqrt{3x+2}+\sqrt{-2x+3}\)có nghĩa \(\sqrt{3x+2}+\sqrt{-2x+3}\ge0\)

\(\Leftrightarrow3x+2-2x+3\ge0\)

\(\Leftrightarrow x+5\ge0\)

\(\Leftrightarrow x\ge-5\)

d)\(\frac{x-5}{\sqrt{-4x}}\)có nghĩa \(\Leftrightarrow\frac{x-5}{\sqrt{-4x}}\ge0\)

\(\Leftrightarrow\frac{x-5}{\sqrt{-\left(2x\right)^2}}\ge0\)

\(\Leftrightarrow\frac{x-5}{-2x}\ge0\)

\(\Leftrightarrow-2x>0\)

\(\Leftrightarrow x>2\)(Câu này không chắc làm đúng không, chắc sai goi)

f)\(\sqrt{-x^2+4x-4}\)có nghĩa \(\Leftrightarrow\sqrt{-x^2+4x-4}\ge0\)

\(\Leftrightarrow-x^2+4x-4\ge0\)

\(\Leftrightarrow-\left(x-2\right)^2\ge0\)

không có z thỏa mãn

g)\(\sqrt{\frac{-2x^2}{3x+2}}\)có nghĩa \(\Leftrightarrow\sqrt{\frac{-2x^2}{3x+2}}\ge0\)

\(\Leftrightarrow\frac{-2x^2}{3x+2}\ge0\)

\(\Leftrightarrow3x+2>0\)

\(\Leftrightarrow3x>-2\)

\(\Leftrightarrow x>\frac{-2}{3}\)

@Cừu

√96 - 6√2/3 + 3/(3+√6) - √10-4√6

= 4√6 - 2√2 + 3(3 - √6)/(3 - √6)(3 + √6) - √10 - 4√6

= -2√2 + 3(3 - √6)/(9 - 6) - √10

= -2√2 + 3 - √6 - √10