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=\(\sqrt{3+2\sqrt{3}+1}\)+\(\sqrt{3-2\sqrt{3}+1}\)
=\(\sqrt{\left(\sqrt{3}+1\right)^2}\)+\(\sqrt{\left(\sqrt{3}-1\right)^2}\)
=\(\sqrt{3}+1+\sqrt{3}-1\)
=\(2\sqrt{3}\)
k mk nha
\(=\frac{1}{\sqrt{2}-\sqrt{3}}.\sqrt{\frac{\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{3}\right)}{\left(3\sqrt{2}+2\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{3}\right)}}\)
\(=\frac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)}.\left(\frac{\left(3\sqrt{2}-2\sqrt{3}\right)^2}{\sqrt{6}}\right)\)
\(=\frac{\sqrt{3}+\sqrt{2}}{-1}.\left(\frac{30-12\sqrt{6}}{\sqrt{6}}\right)\)
\(=\frac{\sqrt{6}\left(\sqrt{150}-12\right)\left(\sqrt{3}+\sqrt{2}\right)}{-\sqrt{6}}\)
\(=-\left(5\sqrt{6}-12\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=-\left(5\sqrt{18}+5\sqrt{12}-12\sqrt{3}-12\sqrt{2}\right)\)
\(=-\left(15\sqrt{2}+10\sqrt{3}-12\sqrt{3}-12\sqrt{2}\right)\)
\(=-\left(3\sqrt{2}-2\sqrt{3}\right)\)
\(=2\sqrt{3}-3\sqrt{2}\)
VẬY \(VT=2\sqrt{3}-3\sqrt{2}\)
\(\frac{1}{\sqrt{2}-\sqrt{3}}.\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}.\sqrt{\frac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{2-3}.\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}}\)
\(=-\left(\sqrt{2}+\sqrt{3}\right).\left(\sqrt{3}-\sqrt{2}\right).\sqrt{\frac{1}{3-2}}\)
\(=-\left(3-2\right)=-1\)
Câu c nè
Đặt \(3x=a\)
=>\(9x^2=a^2\)
Đăt \(x+2=b\)
=>\(\left(x+2\right)^2=b^2\)
ta có
\(a-b=3x-x-2=2x-2\)
<=>\(2x=a-b+2\)
Khi đó pt đã cho trở thành
\(2+3\sqrt[3]{a^2b}=a-b+3\sqrt[3]{ab^2}\)\(a-b+3\sqrt[3]{ab^2}-3\sqrt[3]{a^2b}=\left(\sqrt[3]{a}\right)^3-3\sqrt[3]{a^2b}+3\sqrt[3]{ab^2}-b^3=0\)
<=>\(\left(\sqrt[3]{a}-\sqrt[3]{b}\right)^3=0\)
<=>\(\sqrt[3]{a}=\sqrt[3]{b}\)
<=>a=b
=>3x=x+2
<=>2x-2=0
<=>x=1
nhớ tick nha
Ok !! chi tiết =))
\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{1+2+3+2\sqrt{2}.\sqrt{1}+2\sqrt{2}.\sqrt{3}+2\sqrt{1}.\sqrt{3}}-\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=1+\sqrt{2}+\sqrt{3}-\sqrt{3}-1\)
\(=\sqrt{2}\)
câu trên không có điều kiện các bạn nhé ! chỉ có thế thôi!
\(=\left(-1\right)\sqrt{\left(\sqrt{3}+\sqrt{1}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}\)
\(=\left(-1\right)\cdot\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)\)
\(=\left(-\sqrt{3}-1\right)+\left(\sqrt{3}-1\right)\)
\(=-2\)