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Bài 3: \(A=\frac{\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Đặt a+b=x;b+c=y;c+a=z
\(A=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\ge\frac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{xyz}=\frac{8xyz}{xyz}=8\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
Bài 4: \(A=\frac{9x}{2-x}+\frac{2}{x}=\frac{9x-18}{2-x}+\frac{18}{2-x}+\frac{2}{x}\ge-9+\frac{\left(\sqrt{18}+\sqrt{2}\right)^2}{2-x+x}=-9+\frac{32}{2}=7\)
Dấu = xảy ra khi\(\frac{\sqrt{18}}{2-x}=\frac{\sqrt{2}}{x}\Rightarrow x=\frac{1}{2}\)
Câu a :
\(x-5\sqrt{x}-14=0\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=0\\\sqrt{x}-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=49\end{matrix}\right.\)
Vậy \(S=\left\{49\right\}\)
Câu b :
\(\left(x^2+x+1\right)\left(x^2+x+2\right)=2\)
Đặt \(x^2+x+1=t\)
\(\Leftrightarrow t\left(t+1\right)=2\)
\(\Leftrightarrow t^2+t-2=0\)
\(\Leftrightarrow\left(t-1\right)\left(t+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-1=0\\t+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-2\end{matrix}\right.\)
Với \(t=1\) thì :
\(x^2+x+1=1\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Với \(t=-2\) thì :
\(x^2+x+1=-2\)
\(\Leftrightarrow x^2+x+3=0\) ( pt vô nghiệm )
Vậy \(S=\left\{-1;0\right\}\)
a: Sửa đề: \(\sqrt{9\left(a-1\right)^2}\)
\(=3|a-1|=3(a-1)=3a-3\)
b: \(=6\cdot\left|a-3\right|=6\left(3-a\right)=18-6a\)
c: \(=a\left|a+2\right|\)
d: \(=\left|a\right|\cdot\left|a-1\right|=a\left(a-1\right)\)
a: \(A=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\cdot\sqrt{6-2\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)\)
\(=18-6\sqrt{5}+6\sqrt{5}-10=8\)
b: \(B=\left(\sqrt{5}+\sqrt{3}\right)\cdot\sqrt{2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\left(5-3\right)=2\cdot2=4\)
\(x^2+4x=23-10\sqrt{2}\Leftrightarrow x^2+4x+4=\left(x+2\right)^2=27-10\sqrt{2}=25-10\sqrt{2}+2=5^2-5.2\sqrt{2}+\left(\sqrt{2}\right)^2=\left(5-\sqrt{2}\right)^2=\left(\sqrt{2}-5\right)^2\Leftrightarrow\left[{}\begin{matrix}x+2=5-\sqrt{2}\\x+2=\sqrt{2}-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3-\sqrt{2}\\x=\sqrt{2}-7\end{matrix}\right.\)
\(b,+,x>\frac{1}{2}\Rightarrow2x>1\Rightarrow1-2x< 0\Rightarrow\left|1-2x\right|=-\left(1-2x\right)=2x-1\Rightarrow\left(2x-1\right)^2=\left(2x-1\right)\Leftrightarrow\left(2x-1\right)\left(2x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(loai\right)\\x=1\left(thoaman\right)\end{matrix}\right.\)\(+,x\le\frac{1}{2}\Rightarrow2x\le1\Rightarrow1-2x\ge0\Rightarrow\left|1-2x\right|=1-2x\Rightarrow\left(2x-1\right)^2=1-2x\Leftrightarrow2x\left(2x-1\right)=0\Leftrightarrow x\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(thoaman\right)\\x=\frac{1}{2}\left(thoaman\right)\end{matrix}\right..\)
\(c,Taco:\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(2x+1\right)^2\ge0\end{matrix}\right.mà:\left(x-2\right)^2+\left(2x+1\right)^2=0nên:\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(2x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-\frac{1}{2}\end{matrix}\right.\left(voli\right).Nên:x\in\varnothing\)
Bài 2:
a: \(A=\left|5x+1\right|-\dfrac{3}{8}>=-\dfrac{3}{8}\)
Dấu '=' xảy ra khi x=-1/5
b: \(B=\left|-\dfrac{1}{6}x+2\right|+0.25>=0.25\)
Dấu '=' xảy ra khi x=12
Bài 3:
a: \(A=2018-\left|x+2019\right|< =2018\)
Dấu '=' xảy ra khi x=-2019
b: \(=-10-\left|2x-\dfrac{1}{1009}\right|< =-10\)
Dấu '=' xảy ra khi x=1/2018
a) Ta có: \(A=\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\left|\sqrt{5}-1\right|}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)(Vì \(\sqrt{5}>1\))
\(=\left(2\sqrt{4+\sqrt{5}-1}\right)\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\)
\(=2\cdot\sqrt{3+\sqrt{5}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{5+2\cdot\sqrt{5}\cdot1+1}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\left|\sqrt{5}+1\right|\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\left(\sqrt{5}+1\right)\)
\(=2\cdot\left(5-1\right)\)
\(=2\cdot4=8\)
b) Ta có: \(B=\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}+\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\cdot\left(1-\frac{2}{a+1}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}-1\right)^2+\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)}\right)\cdot\left(\frac{a+1-2}{a+1}\right)^2\)
\(=\frac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)}\cdot\frac{\left(a-1\right)^2}{\left(a+1\right)^2}\)
\(=\frac{2a+2}{\left(a-1\right)}\cdot\frac{\left(a-1\right)^2}{\left(a+1\right)^2}\)
\(=\frac{2\left(a+1\right)\cdot\left(a-1\right)}{\left(a+1\right)^2}\)
\(=\frac{2a-2}{a+1}\)
\(\left(a+10\right)+\left(a^2+10a\right)^2+2\left(a+10\right)^2=0\)
Có: \(\left\{{}\begin{matrix}\left(a^2+10a\right)^2\ge0\forall a\\2\left(a+10\right)^2\ge0\forall a\end{matrix}\right.\)
Để bt = 0 => \(\left\{{}\begin{matrix}\left(a^2+10a\right)^2=0\\2\left(a+10\right)^2=0\end{matrix}\right.\)\(\Rightarrow a=-10\)
Thay a = -10 vào a + 10 có: -10 + 10 = 0
(tm)
Vậy a = -10
còn 1 nghiệm nữa mà