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14 tháng 5 2021

Em gửi ảnh ạ !

14 tháng 5 2021

Em gửi ảnh trên ạ !!!!!

14 tháng 5 2021

1) Khi x = 49 thì:

\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)

2) Ta có:

\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)

\(B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)

Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)

\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)

\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)

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

Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)

\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)

Vậy x = 4

19 tháng 1 2022

a) \(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}\)

\(=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}=-5\sqrt{x-1}\)

b) \(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)

\(=5\sqrt{y+4}+6\sqrt{y+4}-18\sqrt{y+4}=-7\sqrt{y+4}\)

c) \(P=\sqrt{y-2}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)

\(=\sqrt{y-2}-24\sqrt{y-2}+28\sqrt{y-2}=5\sqrt{y-2}\)

a) \(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}.\)

\(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}\)

\(=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}\)

\(=-5\sqrt{x-1}\)

b) \(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)

\(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)

\(=5\sqrt{y+4}+6\sqrt{y+4}\)

\(=-7\sqrt{y+4}\)

c) \(P=\sqrt{\left(y-2\right)}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)

\(P=\sqrt{\left(y-2\right)}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)

\(=\sqrt{y-2}-24\sqrt{y-2}+28\sqrt{y-2}\)

\(=5\sqrt{y-2}\)

30 tháng 6 2017

Áp dụng BĐT Mincopxki và AM-GM ta có:

\(P=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{9}{x+y+z}\right)^2}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\dfrac{81}{\left(x+y+z\right)^2}}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\dfrac{1}{\left(x+y+z\right)^2}+\dfrac{80}{\left(x+y+z\right)^2}}\)

\(\ge\sqrt{2\sqrt{\left(x+y+z\right)^2\cdot\dfrac{1}{\left(x+y+z\right)^2}}+80}\)

\(\ge\sqrt{2+80}=\sqrt{82}\)

Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{3}\)