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1) \(1019x^2+18y^4+1007z^2\)
\(=\left(15x^2+15y^4\right)+\left(3y^4+3z^2\right)+\left(1004x^2+1004z^2\right)\)
\(\ge2\sqrt{15x^2.15y^4}+2\sqrt{3y^4.3z^2}+2\sqrt{1004x^2.1004z^2}=30xy^2+6y^2z+2008xz\left(đpcm\right)\)
c: =>3x^2+3y^2=39 và 3x^2-2y^2=-6
=>5y^2=45 và x^2=13-y^2
=>y^2=9 và x^2=4
=>\(\left\{{}\begin{matrix}x\in\left\{2;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{x}=5\\\sqrt{x}-\sqrt{y}=-\dfrac{11}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\\sqrt{y}=1+\dfrac{11}{2}=\dfrac{13}{2}\end{matrix}\right.\)
=>x=1 và y=169/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4-3=1\\-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9-2=7\end{matrix}\right.\)
=>x+1=11/9 và y+4=-11/19
=>x=2/9 và y=-87/19
3.
\(•x=3+\sqrt{2}\\ x^2=\left(3+\sqrt{2}\right)^2\\ x^2=9+2.3.\sqrt{2}+2\\ x^2=11+6\sqrt{2}\\• y=\sqrt{11+6\sqrt{2}}\\ y^2=\left(\sqrt{11+6\sqrt{2}}\right)^2\\ y^2=11+6\sqrt{2}\)
\(\Rightarrow x^2=y^2=11+6\sqrt{2}\)
1. ta có : \(4\sqrt{7}=\sqrt{112}\)
\(3\sqrt{3}=\sqrt{27}\)
ta thấy : \(\sqrt{112}>\sqrt{27}\) hay \(4\sqrt{7}>3\sqrt{3}\)
2. \(\dfrac{1}{4}\sqrt{82}=\sqrt{\dfrac{41}{8}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
ta thấy :\(\sqrt{\dfrac{41}{8}}< \sqrt{\dfrac{36}{7}}\) hay \(\dfrac{1}{4}\sqrt{82}< 6\sqrt{\dfrac{1}{7}}\)
3. \(x^2=\left(3+\sqrt{2}\right)^2\)
\(y^2=11+6\sqrt{2}\)=\(\left(3+\sqrt{2}\right)^2\)
ta thấy : \(x^2=y^2\Rightarrow x=y\)
Ta có:
\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Áp dụng BĐT Cosi ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\)
Cmtt:
\(\dfrac{y^3}{y\sqrt{1-y^2}}\ge2y^3\)
\(\dfrac{z^3}{z\sqrt{1-z^2}}\ge2z^3\)
\(\Rightarrow\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}+\dfrac{y^3}{y\sqrt{1-y^2}}+\dfrac{z^3}{z\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\) (ĐPCM)
Bài 1 :
\(\dfrac{x+4}{x^2-9}-\dfrac{2}{x+3}=\dfrac{4x}{3x-x^2}\) ( ĐK : \(\left\{{}\begin{matrix}x\ne0\\x\ne-3\\x\ne3\end{matrix}\right.\) )
\(\Leftrightarrow\dfrac{x\left(x+4\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{2x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{-4x\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow x\left(x+4\right)-2x\left(x-3\right)=-4x\left(x+3\right)\)
\(\Leftrightarrow x^2+4x-2x^2+6x+4x^2+12x=0\)
\(\Leftrightarrow3x^2+22x=0\)
\(\Leftrightarrow x\left(3x+22\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+22=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=-\dfrac{22}{3}\left(N\right)\end{matrix}\right.\)
Vậy \(x=-\dfrac{22}{3}\)
Bài 2 : \(x\left(x+1\right)\left(x^2+x+1\right)=42\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)
Đặt \(x^2+x=t\) . Phương trình trở thành :
\(t\left(t+1\right)=42\)
\(\Leftrightarrow t^2+t-42=0\)
\(\Leftrightarrow\left(t-6\right)\left(t+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-6=0\\t+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=6\\t=-7\end{matrix}\right.\)
Với \(t=6\)
\(\Leftrightarrow x^2+x=6\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Với \(t=-7\)
\(\Leftrightarrow x^2+x=-7\)
\(\Leftrightarrow x^2+x+7=0\)
---> Phương trình vô nghiệm !
Vậy \(x=-3;x=2\)
a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
Biến đổi vế trái ta có:
\(a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)+c^3\)
\(=\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b\right)\)
\(=\left(a+b+c\right)^3-3\left(a+b\right)\left(ac+bc+c^2+ab\right)\)
\(=\left(a+b+c\right)^3-3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)*
Vì \(a+b+c=0\)\(\Rightarrow\)*\(=-3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
cũng có \(\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\) Thay vào biểu thức trên ta được
\(-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=3abc\)
\(VT=VP\)=> đpcm
vì \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)
ta có \(B=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\)
mà \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\Rightarrow B=xyz.\dfrac{3}{xyz}=3\)