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\(P=4a+\frac{9}{a-3}+6\)
\(=4\left(a-3\right)+\frac{9}{a-3}+18\)
Vì a>3
=> a-3>0
=> \(\frac{9}{a-3}>0\)
Áp dụng bđt Cô-sy cho 2 số ko âm ta có:
\(4\left(a-3\right)+\frac{9}{a-3}\ge2\sqrt{4\left(a-3\right)\cdot\frac{9}{a-3}}=2\sqrt{36}=12\)
\(\Rightarrow4\left(a-3\right)+\frac{9}{a-3}+18\ge30\)
Dấu = xảy ra
\(\Leftrightarrow4\left(a-3\right)=\frac{9}{a-3}\)
\(\Leftrightarrow4\left(a-3\right)^2=9\)
\(\Leftrightarrow\left(a-3\right)^2=\frac{9}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}a-3=\frac{3}{2}\\a-3=-\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a=\frac{9}{2}\left(tm\right)\\a=\frac{3}{2}\left(kotm\right)\end{cases}}\)
Vậy ...
\(B=\left(\frac{x}{2}+y\right)^3-6\left(\frac{x}{2}+y\right)^2.z+6\left(x+2y\right)z^2-8z^3\)
\(=\left(\frac{x}{2}+y\right)^3-3.\left(\frac{x}{2}+y\right)^2.2z+3.\left(\frac{x}{2}+y\right).\left(2z\right)^2-\left(2z\right)^3\)
\(=\left(\frac{x}{2}+y-2z\right)^3\)
\(C=\left(m-n\right)^3+15\left(m-n\right)^2.\left(m-p\right)-75\left(n-m\right)\left(p-m\right)^2-125\left(p-m\right)^3\)
\(=\left(m-n\right)^3+3.\left(m-n\right).\left[5\left(m-p\right)\right]+3.\left(m-n\right).\left[5\left(m-p\right)\right]^2+\left[5\left(m-p\right)\right]^3\)
\(=\left(m-n+5m-5p\right)^3=\left(6m-n-5p\right)^3\)
\(a,y^4-14y^2+49\)
\(\left(y^2-7\right)^2\)
\(b,x^2-2\)
\(x^2-\left(\sqrt{2}\right)^2=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)
\(c,y^2-13\)
\(y^2-\left(\sqrt{13}\right)^2=\left(y-\sqrt{13}\right)\left(y+\sqrt{13}\right)\)
\(d,-4x^2+9y^2\)
\(\left(3y\right)^2-\left(2x\right)^2\)
\(\left(3y-2x\right)\left(3y+2x\right)\)
\(\left(-2+x^2\right)\left(-2+x^2\right)\left(-2+x^2\right)\left(-2+x^2\right)\left(-2+x^2\right)=1\)
\(\left(x^2-2\right)^5=1\)
\(x^2-2=1\)
\(x^2=3\)
\(\orbr{\begin{cases}x=\sqrt{3}\\x=-\sqrt{3}\end{cases}}\)
\(x^2+2\sqrt{3}x+3=0\)
\(x^2+2\sqrt{3}x+\sqrt{3}^2=0\)
\(\left(x+\sqrt{3}\right)^2=0\)
\(x+\sqrt{3}=0\)
\(x=-\sqrt{3}\)