\(\frac{2}{\sqrt{3}-1}-\frac{1}{\sqrt{3}-2}+\frac{12}{\sqrt{3}+3}=\frac{2\sqrt{3}+2}{3-1}-\frac{\sqrt{3}+2}{3-2}+\frac{12\sqrt{3}-36}{3-9}\)

\(=\sqrt{3}+1+\sqrt{3}-2-2\sqrt{3}+6=5\)

\(B=\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\frac{3\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\frac{a-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)

\(=\frac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\frac{11}{a-9}\)

Để B nguyên thì \(\frac{11}{a-9}\inℤ\Leftrightarrow a-9\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

đến đây bạn tự lập bảng xét ước nhé :c chú ý ĐK giùm mình không lại sai :>

\(\left(x^2+2x+1\right)^2+2\left(x^2+2x+1\right)+1=x\)

\(\Leftrightarrow\left(x^2+2x+2\right)^2=x\Leftrightarrow\left|x^2+2x+2\right|=\sqrt{x}\)

Với : x >= 0 => \(x^2+2x+2>0\)

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

\(\Delta=4-4\left(2-\sqrt{x}\right)=4-8+4\sqrt{x}=-4+4\sqrt{x}\)

Để pt có nghiệm khi delta >=0 

\(-4+4\sqrt{x}\ge0\Leftrightarrow1-\sqrt{x}\le0\Leftrightarrow0\le x\le1\)

\(ĐKXĐ:x\le2\)

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

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

\(x^4-4x^2+4=2-x\)

\(x^4-4x^2+x+2=0\)

\(x^4+x^3-3x^2-2x-1x^3-x^2+3x+2=0\)

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

\(\left(x-1\right)\left(x^3+x^2-3x-2\right)=0\)

\(\left(x-1\right)\left(x^3+2x^2-x^2-2x-x-2\right)=0\)

\(\left(x-1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)-\left(x+2\right)\right]=0\)

\(\left(x-1\right)\left(x+2\right)\left(x^2-x-1\right)=0\)

\(TH1:x-1=0< =>x=1\left(KTM\right)\)

mình cũng chưa biết tại sao nó lại không thỏa mãn nữa :v

\(TH2:x+2=0< =>x=-2\left(TM\right)\)

xét \(x^2-x-1=0\)

\(\sqrt{\Delta}=\sqrt{\left(-1\right)^2-4.1.\left(-1\right)}=\sqrt{5}\)

\(\orbr{\begin{cases}x=\frac{1+\sqrt{5}}{2}\left(TM\right)\\x_2=\frac{1-\sqrt{5}}{2}\left(KTM\right)\end{cases}}\)

vậy kl...................

dốt quá bạn thêm đkxđ vào lúc bình cả hai vế lên nha

\(\orbr{\begin{cases}x\le-\sqrt{2}\\x\ge\sqrt{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=1\left(KTM\right)\\x=\frac{1-\sqrt{5}}{2}\left(KTM\right)\end{cases}}\)

\(\sqrt{5}\cdot\sqrt{45}=\sqrt{5\cdot45}=\sqrt{225}=15\)

\(\sqrt{75\cdot48}=\sqrt{75}\cdot\sqrt{48}=5\sqrt{3}\cdot4\sqrt{3}=20\cdot3=60\)

\(\sqrt{2}\cdot\sqrt{162}=\sqrt{2\cdot162}=\sqrt{324}=18\)

\(\sqrt{0,25\cdot196}=\sqrt{0,25}\cdot\sqrt{196}=0,5\cdot14=7\)

\(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}=\sqrt{9\cdot25}=\sqrt{9}\cdot\sqrt{25}=3\cdot5=15\)

\(\sqrt{6,8^2-3,2^2}=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}=\sqrt{3,6\cdot10}=\sqrt{36}=6\)

\(\sqrt{36a^2}=\left|6a\right|=-6a\left(a< 0\right)\)

\(\sqrt{a^4b^6c^2}=\left|a^2b^3c\right|=a^2\left|b^3c\right|\)

\(\sqrt{4\left(a-3\right)^2}=2\left|a-3\right|=2a-6\left(a\ge3\right)\)

\(\sqrt{9\left(b-2\right)^2}=3\left|b-2\right|=6-3b\left(b< 2\right)\)

\(\sqrt{b^2\left(b-1\right)^2}=\left|b\left(b-1\right)\right|=b^2-b\left(b< 0\right)\)

ĐK : x ≥ -3/2

\(\Leftrightarrow x^2+4x+3-\left(2\sqrt{2x+3}-2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-\frac{8x+12-4}{2\left(\sqrt{2x+3}+1\right)}=0\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x+3\right)-\frac{4}{\sqrt{2x+3}+1}\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+3\right)-\frac{4}{\sqrt{2x+3}+1}=0\end{cases}}\)

TH1 : x + 1 = 0 <=> x = -1 (tm) (1)

TH2 : \(\left(x+3\right)-\frac{4}{\sqrt{2x+3}+1}=0\Leftrightarrow\left(x+3\right)\sqrt{2x+3}+x+3=4\)

\(\Leftrightarrow\left(x+3\right)\left(\sqrt{2x+3}-1\right)+2x+2=0\)

\(\Leftrightarrow\left(x+3\right)\frac{2x+2}{\sqrt{2x+3}+1}+2x+2=0\)

\(\Leftrightarrow2\left(x+1\right)\left[\frac{x+3}{\sqrt{2x+3}+1}+1\right]=0\)(*)

Dễ thấy với x ≥ -3/2 thì \(\frac{x+3}{\sqrt{2x+3}+1}+1>0\)

nên (*) <=> x + 1 = 0 <=> x = -1 (tm) (2)

Từ (1) và (2) => pt có nghiệm x = -1

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

\(=\frac{9+4-12}{\sqrt{6}}=\frac{1}{\sqrt{6}}\)

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

\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\)

gọi 3 số đó là a,b,c

a+b+c=100

theo bdt cosi: a+b+c>=\(3\sqrt[3]{abc}\)

\(\Leftrightarrow100\ge3\sqrt[3]{abc}\Leftrightarrow\frac{1000000}{27}\ge abc\)

vậy abc đạt gtln là 1000000/27 hay tích 3 số đó có GTLN là 1000000/27