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B1:
\(\Leftrightarrow5a-5b\sqrt{2}-4a-4b\sqrt{2}+18\sqrt{2}\left(a^2-2b^2\right)=3\left(a^2-2b^2\right)\)
\(\Leftrightarrow5a-5b\sqrt{2}-4a-4b\sqrt{2}+18a^2\sqrt{2}-36b^2\sqrt{2}=3a^2-6b^2\)
\(\Leftrightarrow18a^2\sqrt{2}-36b^2\sqrt{2}-9b\sqrt{2}=3a^2-6b^2-a\)
\(\Leftrightarrow\left(18a^2-36b^2-9b\right)\sqrt{2}=3a^2-6b^2-a\)
Nếu \(18a^2-36b^2-9b\ne0\Rightarrow\sqrt{2}=\frac{3a^2-6b^2-a}{18a^2-36b^2-9b}\)
Vì a,b nguyên nên \(\frac{3a^2-6b^2-a}{18a^2-36b^2-9b}\in Q\Rightarrow\sqrt{2}\in Q\)=> Vô lý vì \(\sqrt{2}\)là số vô tỉ.
Vậy ta có: \(18a^2-36b^2-9b=0\Rightarrow\hept{\begin{cases}18a^2-36b^2-9b=0\\3a^2-6b^2-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3a^2-6b^2=\frac{3}{2}b\\3a^2-6b^2=a\end{cases}\Leftrightarrow a=\frac{3}{2}b}\)
Thay \(a=\frac{3}{2}b\)vào \(3a^2-6b^2-a=0\)ta có:
\(3.\frac{9}{4}b^2-6b^2-\frac{3}{2}b=0\Leftrightarrow27b^2-24b^2-6b=0\Leftrightarrow3b\left(b-2\right)=0\)
Ta có: b=0(loại) ; b=2(thoả mãn) . Vậy a=3. KL:...
B2: \(GT\Rightarrow\left[\left(a+b\right)^2-2\left(ab+1\right)\right]\left(a+b\right)^2+\left(1+ab\right)^2=0\)
\(\Leftrightarrow\left(a+b\right)^4-2\left(a+b\right)^2\left(1+ab\right)+\left(1+ab\right)^2=0\)
\(\Leftrightarrow\left[\left(a+b\right)^2-\left(1+ab\right)\right]^2=0\Rightarrow\left(a+b\right)^2-\left(1+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)^2=1+ab\Leftrightarrow\left|a+b\right|=\sqrt{1+ab}\in Q\)( vì a,b thuộc Q)
KL:....
Đặt: \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\)
\(\Leftrightarrow xy+yz+zx=1\)
Ta có:
\(S=\frac{\frac{1}{x}}{\sqrt{\frac{1}{y}.\frac{1}{z}\left(1+\frac{1}{x^2}\right)}}+\frac{\frac{1}{y}}{\sqrt{\frac{1}{z}.\frac{1}{x}\left(1+\frac{1}{y^2}\right)}}+\frac{\frac{1}{z}}{\sqrt{\frac{1}{x}.\frac{1}{y}\left(1+\frac{1}{z^2}\right)}}\)
\(=\sqrt{\frac{yz}{1+x^2}}+\sqrt{\frac{zx}{1+y^2}}+\sqrt{\frac{xy}{1+z^2}}\)
\(=\sqrt{\frac{yz}{xy+yz+zx+x^2}}+\sqrt{\frac{zx}{xy+yz+zx+y^2}}+\sqrt{\frac{xy}{xy+yz+zx+z^2}}\)
\(=\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\frac{zx}{\left(y+x\right)\left(y+z\right)}}+\sqrt{\frac{xy}{\left(z+x\right)\left(z+y\right)}}\)
\(\le\frac{1}{2}.\left(\frac{y}{x+y}+\frac{z}{x+z}+\frac{z}{y+z}+\frac{x}{x+y}+\frac{x}{z+x}+\frac{y}{z+y}\right)\)
\(=\frac{1}{2}.\left(1+1+1\right)=\frac{3}{2}\)
Dấu = xảy ra khi \(x=y=z=\sqrt{3}\)
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
a2(b+c)2+5bc+b2(a+c)2+5ac≥4a29(b+c)2+4b29(a+c)2=49(a2(1−a)2+b2(1−b)2)(vì a+b+c=1)
a2(1−a)2−9a−24=(2−x)(3x−1)24(1−a)2≥0(vì )<a<1)
⇒a2(1−a)2≥9a−24
tương tự: b2(1−b)2≥9b−24
⇒P⩾49(9a−24+9b−24)−3(a+b)24=(a+b)−94−3(a+b)24.
đặt t=a+b(0<t<1)⇒P≥F(t)=−3t24+t−94(∗)
Xét hàm (∗) được: MinF(t)=F(23)=−19
⇒MinP=MinF(t)=−19.dấu "=" xảy ra khi a=b=c=13