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\(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\left(đk:x\ge2;y\ge3;z\ge5\right)\)
\(< =>\left(x-2\right)-2\sqrt{x-2}+1+\left(y-3\right)-4\sqrt{y-3}+4+\left(z-5\right)-6\sqrt{z-5}+9=0\)
\(< =>\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Do \(\left(\sqrt{x-2}-1\right)^2\ge0;\left(\sqrt{y-3}-2\right)^2\ge0;\left(\sqrt{z-5}-3\right)^2\ge0\)
Cộng theo vế ta được \(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2\ge0\)
Mà \(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Dấu "=" xảy ra khi và chỉ khi x = 3 ; y = 7 ; z = 14 ( tmđk )
Vậy ...
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\(-7xy\sqrt{\frac{16}{xy}}\)
\(-7xy\frac{4\sqrt{xy}}{xy}\)
\(-28\sqrt{xy}\)
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\(15.a,\sqrt{3}\left(\sqrt{4}+2\sqrt{9}\right)\frac{\sqrt{3}}{2}-\sqrt{150}\)
\(\left(2+2.3\right)\frac{3}{2}-\sqrt{150}\)
\(12-\sqrt{150}\)
\(6\left(2-\sqrt{25}\right)=6\left(-3\right)=-18\)
\(b,\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)
\(\sqrt{196}-\sqrt{84}-7+2\sqrt{21}\)
\(14-\sqrt{21}\left(\sqrt{4}+2\right)-7\)
\(7-\sqrt{21}.4\)
\(7-\sqrt{84}\)
\(c,\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)\)
\(1+\sqrt{2}-\sqrt{3}+\sqrt{2}+2-\sqrt{6}+\sqrt{3}+\sqrt{6}-3\)
\(=2\sqrt{2}\)
\(d,\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)^2-\sqrt{3}+\sqrt{2}\)
\(\sqrt{3}\left(2-\sqrt{6}+3\right)-\sqrt{3}+\sqrt{2}\)
\(\sqrt{3}\left(5-\sqrt{6}-1\right)+\sqrt{2}\)
\(\sqrt{3}\left(4-\sqrt{6}\right)+\sqrt{2}\)
\(4\sqrt{3}-\sqrt{18}+\sqrt{2}\)
\(\sqrt{2}\left(2\sqrt{3}-3+1\right)\)
\(\sqrt{2}\left(2\sqrt{3}-2\right)\)
\(2\sqrt{6}-\sqrt{6}=\sqrt{6}\)