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bài 2
ta có \(\left(\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\right)^2\)
\(=\left(\sqrt{a}.\sqrt{\frac{8a^2+1}{a}}+\sqrt{b}.\sqrt{\frac{8b^2+1}{b}}+\sqrt{c}.\sqrt{\frac{8c^2+1}{c}}\right)^2\)\(=\left(A\right)\)
Áp dụng bất đẳng thức Bunhiacopxki ta có;
\(\left(A\right)\le\left(a+b+c\right)\left(8a+\frac{1}{a}+8b+\frac{1}{b}+8c+\frac{8}{c}\right)\)
\(=\left(a+b+c\right)\left(9a+9b+9c\right)=9\left(a+b+c\right)^2\)
\(\Rightarrow3\left(a+b+c\right)\ge\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\)(đpcm)
Dấu \(=\)xảy ra khi \(a=b=c=1\)
5/ĐK: \(\left[{}\begin{matrix}x\le-1\\x\ge5\end{matrix}\right.\)
PT \(\Leftrightarrow2\left(x^2-4x-6\right)+\sqrt{x^2-4x-5}-1=0\)
\(\Leftrightarrow\left(x^2-4x-6\right)\left(2+\frac{1}{\sqrt{x^2-4x-5}+1}\right)=0\)
\(\Leftrightarrow x^2-4x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{10}\\x=2-\sqrt{10}\end{matrix}\right.\)
Vậy..
ráng làm nốt rồi đi ngủ thoyy
1.
a) ĐK: \(x\ge2\)
\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}-\sqrt{x-2}-\sqrt{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-2}=\sqrt{x+3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\varnothing\end{matrix}\right.\)
Vậy...
b) \(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)
\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-x^2+2x-1\)
\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=\left(2x+1\right)^2+\left(x+8\right)-\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x-1\right)\sqrt{x+8}+\left(x+8\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-\sqrt{x+8}-x+1\right)\left(2x+1-\sqrt{x+8}+x-1\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x+8}+2\right)\left(3x-\sqrt{x+8}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{x+8}\\3x=\sqrt{x+8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)\(\Leftrightarrow x=1\)
Vậy...
c) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
Nhân cả 2 vế với \(\sqrt{2}\) ta được :
\(pt\Leftrightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)
Ta có : \(\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)
\(=\left|\sqrt{2x-1}+1\right|+\left|1-\sqrt{2x-1}\right|\ge\left|\sqrt{2x-1}+1+1-\sqrt{2x-1}\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le1\)
2) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\frac{1}{x+y+z}=1\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-x-y-z}{z\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}\)
\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\cdot\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)\left(xz+yz+z^2+xy\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
TH1: \(x=-y\Leftrightarrow x^{29}=-y^{29}\Leftrightarrow x^{29}+y^{29}=0\)
Khi đó \(B=0\cdot\left(x^{11}+y^{11}\right)\cdot\left(x^{2013}+y^{2013}\right)=0\)
Tương tự 2 trường hợp còn lại ta đều được \(B=0\)
Vậy \(B=0\)
Ta có : \(\left(x+\sqrt{x^2+2017}\right)\left(-x+\sqrt{x^2+2017}\right)=2017\left(1\right)\)
\(\left(y+\sqrt{y^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\left(2\right)\)
nhân theo vế của ( 1 ) ; ( 2 ) , ta có :
\(2017\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017^2\)
\(\Rightarrow\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\)
rồi bạn nhân ra , kết hợp với việc nhân biểu thức ở phần trên xong cộng từng vế , cuối cùng ta đc :
\(xy+\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017\)
\(\Leftrightarrow\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017-xy\)
\(\Leftrightarrow x^2y^2+2017\left(x^2+y^2\right)+2017^2=2017^2-2\cdot2017xy+x^2y^2\)
\(\Rightarrow x^2+y^2=-2xy\Rightarrow\left(x+y\right)^2=0\Rightarrow x=-y\)
A = 2017
( phần trên mk lười nên không nhân ra, bạn giúp mk nhân ra nha :) )
2/ \(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{4\sqrt{x-2011}-4}{x-2011}+\frac{4\sqrt{y-2012}-4}{y-2012}+\frac{4\sqrt{z-2013}-4}{z-2013}=3\)
\(\Leftrightarrow\left(1-\frac{4\sqrt{x-2011}-4}{x-2011}\right)+\left(1-\frac{4\sqrt{y-2012}-4}{y-2012}\right)+\left(1-\frac{4\sqrt{z-2013}-4}{z-2013}\right)=0\)
\(\Leftrightarrow\left(\frac{x-2011-4\sqrt{x-2011}+4}{x-2011}\right)+\left(\frac{y-2012-4\sqrt{y-2012}+4}{y-2012}\right)+\left(\frac{z-2013-4\sqrt{z-2013}+4}{z-2013}\right)=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2011}-2\right)^2}{x-2011}+\frac{\left(\sqrt{y-2012}-2\right)^2}{y-2012}+\frac{\left(\sqrt{z-2013}-2\right)^2}{z-2013}=0\)
Dấu = xảy ra khi \(\sqrt{x-2011}=2;\sqrt{y-2012}=2;\sqrt{z-2013}=2\)
\(\Leftrightarrow x=2015;y=2016;z=2017\)
A
Áp dụng BĐT cosi ta có
\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)
\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)
Khi đó
\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)
MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)
B
Áp dụng BĐT cosi ta có :
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)
Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)
=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)
\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z
câu 1 sai đề
\(\sqrt{x}+1chứkophải\sqrt{x+1}\)