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1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4
--> Pmin=4 khi x=4
2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1
=> M=2x2-8x+\(\sqrt{x^2-4x+5}\)+6=2(t2-5)+t+6
<=> M=2t2+t-4\(\ge\)2.12+1-4=-1
Mmin=-1 khi t=1 hay x=2
a) \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}=27-4\sqrt{3x}\)
b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28=3\sqrt{2x}+2\sqrt{8x}+28=3\sqrt{2x}+4\sqrt{2x}+28=7\sqrt{2x}+28\)
c) \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{\left(x-y\right)\left(x+y\right)}.\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{\sqrt{6}}{x-y}\)
d) \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4x+4a^2\right)}=\frac{2}{2a-1}\sqrt{5a^2\left(2a-1\right)^2}=\frac{2}{2a-1}.\sqrt{5}\left|a\left(2a-1\right)\right|=2a\sqrt{5}\)
Thiếu ĐKXĐ : ..............
a) Ta có: \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}\)
\(=27-4\sqrt{3x}\)
b) Ta có: \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28\)
\(=3\sqrt{2x}-5.2\sqrt{2x}+7.2\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+14\sqrt{2x}+28\)
\(=7\sqrt{2x}+28\)
c) Ta có: \(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{4}{\left(x-y\right)^2.\left(x+y\right)^2}.\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{2.3}{\left(x-y\right)^2}}\)
\(=\frac{1}{x-y}.\sqrt{6}\)
d) Ta có: \(\frac{2}{2a-1}.\sqrt{5a^2.\left(1-4a+4a^2\right)}\)
\(=\sqrt{\frac{4}{\left(2a-1\right)^2}.5a^2.\left(2a-1\right)^2}\)
\(=2a.\sqrt{5}\)
\(P\ge\frac{1}{2}\left(x+y\right)^2+\frac{32}{x+y+2}=\frac{1}{2}\left[\left(x+y\right)^2+4\right]+\frac{32}{x+y+2}-2\)
\(P\ge\frac{1}{4}\left(x+y+2\right)^2+\frac{32}{x+y+2}-2\)
\(P\ge\frac{1}{4}\left(x+y+2\right)^2+\frac{16}{x+y+2}+\frac{16}{x+y+2}-2\)
\(P\ge3\sqrt[3]{\frac{16^2\left(x+y+2\right)^2}{4\left(x+y+2\right)^2}}-2=10\)
\(P_{min}=10\) khi \(x=y=1\)
\(A=\left(\frac{2xy}{x^2+y^2}\right)^2+\frac{x^4+y^4+2\left(xy\right)^2}{\left(xy\right)^2}-2=4\left(\frac{xy}{x^2+y^2}\right)^2+\left(\frac{x^2+y^2}{xy}\right)^2-2\)
\(=\left(\frac{2xy}{x^2+y^2}\right)^2+\left(\frac{x^2+y^2}{2xy}\right)^2+3\left(\frac{x^2+y^2}{2xy}\right)^2-2\)
\(\ge2\sqrt{\left(\frac{2xy}{x^2+y^2}\right)^2.\left(\frac{x^2+y^2}{2xy}\right)^2}+3\left(\frac{2xy}{2xy}\right)^2-2=3\)
\(y=\frac{1}{6}\left(\frac{4\left(x+1\right)^2+9}{x+1}\right)=\frac{1}{6}\left(4\left(x+1\right)+\frac{9}{x+1}\right)\)
\(y\ge\frac{1}{6}.2\sqrt{\frac{4\left(x+1\right).9}{x+1}}=2\)
\(y_{min}=2\) khi \(\left(x+1\right)^2=\frac{9}{4}\Rightarrow x=\frac{1}{2}\)