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a) \(A=4\sqrt{x^2+1}-2\sqrt{16\left(x^2+1\right)}+5\sqrt{25\left(x^2+1\right).}\)
\(=4\sqrt{x^2+1}-2.4\sqrt{x^2+1}+5.5\sqrt{x^2+1}\)
\(=4\sqrt{x^2+1}-8\sqrt{x^2+1}+25\sqrt{x^2+1}\)
\(=\left(4-8+25\right)\sqrt{x^2+1}\)
\(=21\sqrt{x^2+1}\)
b) \(B=\frac{2}{x+y}\sqrt{\frac{3\left(x+y\right)^2}{4}}\)
\(B=\frac{2}{x+y}.\frac{\sqrt{3}\left(x+y\right)}{2}\)
\(B=\frac{\sqrt{3}\left(x+y\right)}{x+y}\)
\(B=\sqrt{3}\)
Ta có: \(\left(a+\sqrt{a^2+9}\right)\left(b+\sqrt{b^2+9}\right)=9\)
\(\Leftrightarrow\frac{\left(a-\sqrt{a^2+9}\right)\left(a+\sqrt{a^2+9}\right)\left(b+\sqrt{b^2+9}\right)}{a-\sqrt{a^2+9}}=9\)
\(\Leftrightarrow\frac{-9\left(b+\sqrt{b^2+9}\right)}{a-\sqrt{a^2+9}}=9\)
\(\Rightarrow b+\sqrt{b^2+9}=\sqrt{a^2+9}-a\)
Tương tự chỉ ra được: \(a+\sqrt{a^2+9}=\sqrt{b^2+9}-b\)
Cộng vế 2 PT trên lại ta được:
\(a+b+\sqrt{a^2+9}+\sqrt{b^2+9}=\sqrt{a^2+9}+\sqrt{b^2+9}-a-b\)
\(\Leftrightarrow2\left(a+b\right)=0\Rightarrow a=-b\)
Thay vào M ta được:
\(M=2a^4-a^4-6a^2+8a^2-10a+2a+2026\)
\(M=a^4+2a^2-8a+2026\)
\(M=\left(a^4+2a^2-8a+5\right)+2021\)
\(M=\left[\left(a^4-a^3\right)+\left(a^3-a^2\right)+\left(3a^2-3a\right)-\left(5a-5\right)\right]+2021\)
\(M=\left(a-1\right)\left(a^3+a^2+3a-5\right)+2021\)
\(M=\left(a-1\right)^2\left(a^2+2a+5\right)+2021\)\(\ge0+2021=2021\)
Dấu "=" xảy ra khi: a = 1 => b = -1
Vậy Min(M) = 2021 khi a = 1 và b = -1
a, Ta có : \(x=81\Rightarrow\sqrt{x}=9\)
Thay \(\sqrt{x}=9\)vào biểu thức A ta được :
\(A=\frac{2}{9+1}=\frac{2}{10}=\frac{1}{5}\)
b, Ta có : \(P=\frac{B}{A}\)hay\(P=\frac{\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}}{\frac{2}{\sqrt{x}+1}}\)
\(=\frac{1+\sqrt{x}}{x+\sqrt{x}}.\frac{\sqrt{x}+1}{2}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)
c, Ta có \(\frac{1}{2}=\frac{\sqrt{x}}{2\sqrt{x}}\)mà \(\sqrt{x}< \sqrt{x}+1\)
nên \(P>\frac{1}{2}\)
a) \(A=\frac{2}{\sqrt{x}+1}=\frac{2}{\sqrt{81}+1}=\frac{2}{9+1}=\frac{1}{5}\)
b) \(B=\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}\)
\(=\frac{1+\sqrt{x}}{\left(1+\sqrt{x}\right)\sqrt{x}}=\frac{1}{\sqrt{x}}\)
\(\Rightarrow P=\frac{B}{A}=\frac{1}{\sqrt{x}}\div\frac{2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)
c) Ta có: \(P=\frac{\sqrt{x}+1}{2\sqrt{x}}=\frac{1}{2}+\frac{1}{\sqrt{x}}+\frac{1}{2}+0=\frac{1}{2}\)
=> P>1/2
\(A=\left(\frac{x+2}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
\(A=\left[\frac{\left(x+2\right)^2}{4-x^2}+\frac{4x^2}{4-x^2}-\frac{\left(2-x\right)^2}{4-x^2}\right]:\left[\frac{x\left(x-3\right)}{x^2.\left(2-x\right)}\right]\)
\(A=\left[\frac{x^2+4x+4+4x^2-4+4x-x^2}{4-x^2}\right]:\left[\frac{x-3}{x\left(2-x\right)}\right]\)
\(A=\frac{4x^2+8x}{4-x^2}:\frac{x-3}{x\left(2-x\right)}\)
\(A=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(A=\frac{4x^2}{x-3}\)
Đề là:
\(x^{n+3}y^4:x^7y^n\) hay \(x^{n+3}y^4:\left(x^7y^n\right)\)vậy bạn?
