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a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
a: \(\Leftrightarrow\dfrac{x^4+2x^2+1-x^2}{x^2}=\dfrac{x^2+x+1}{x}\)
\(\Leftrightarrow\dfrac{\left(x^2+1+x\right)\left(x^2+1-x\right)}{x^2}=\dfrac{x^2+x+1}{x}\)
\(\Leftrightarrow\dfrac{x^2-x+1}{x^2}=\dfrac{1}{x}\)
=>x^2=x(x^2-x+1)
=>x(x-x^2+x-1)=0
=>x(-x^2+2x-1)=0
=>x=0(loại) hoặc x=1(nhận)
b: =>3(x+3)^2*(x+2)^2/(x^2-4)^2+68*(x-3)^2*(x-2)^2/(x^2-4)^2-46(x^2-9)(x^2-4)=6(x^2-4)^2
=>3(x^2+5x+6)^2+68(x^2-5x+6)^2-46(x^4-13x^2+36)=6(x^4-8x^2+16)
=>\(x\simeq28,4\)
câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
bài 2: ta có : \(Q=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-\left(1-a\right)}\right)\left(\sqrt{\dfrac{1}{a^2}-1}-\dfrac{1}{a}\right).\sqrt{a^2-2a+1}\)
\(\Leftrightarrow Q=\left(\dfrac{\sqrt{1+a}\sqrt{1-a}+1-a}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\dfrac{\sqrt{1-a^2}}{a}-\dfrac{1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{1-a^2}-1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{\sqrt{1-a^2}+1}{a}\right)\left(\dfrac{\sqrt{1-a^2}-1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{1-a^2-1}{a^2}\right)\left(1-a\right)=a-1\)b) ta có : \(Q^3-Q=\left(a-1\right)\left(\left(a-1\right)^2-1\right)=a\left(a-1\right)\left(a-2\right)\)
mà ta có : \(\left\{{}\begin{matrix}a>0\\a-1< 0\\a-2< 0\end{matrix}\right.\Rightarrow a\left(a-1\right)\left(a-2\right)>0\) \(\Rightarrow Q^3-Q>0\Leftrightarrow Q^3>Q\)
vậy \(Q^3>Q\)
Nguyễn Huy TúAkai HarumaLightning FarronNguyễn Thanh Hằngsoyeon_Tiểubàng giảiMashiro ShiinaVõ Đông Anh Tuấn
Hoàng Lê Bảo NgọcTrần Việt Linh
cứu tôi với
ĐK: \(x\ge0,x\ne1\)
\(M=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(x-1\right)^2}{2}=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(x-1\right)^2}{2}=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}=\left[\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2.2}=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}=\sqrt{x}-x\)
Ta có \(M=\sqrt{x}-x=\sqrt{x}\left(1-\sqrt{x}\right)\)
Ta có \(\sqrt{x}\ge0\)
Vậy để M có GTLN thì \(1-\sqrt{x}\) có GTLN
Mà ta có \(\sqrt{x}\ge0\Leftrightarrow-\sqrt{x}\le0\Leftrightarrow1-\sqrt{x}\le1\)
\(\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)\le1\)
Dấu '=' xảy ra khi x=0
Vậy GTLN của M là 1
a/ \(\left(x+3\right)\left(3\left(x^2+1\right)^2+2\left(x+3\right)^2\right)=5\left(x^2+1\right)^3\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2+2\left(x+3\right)^3-5\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2-3\left(x^2+1\right)^3+2\left(x+3\right)^3-2\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x^2+1\right)^2\left(-x^2+x+2\right)+2\left(-x^2+x+2\right)\left(\left(x+3\right)^2+\left(x+3\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right)=0\)
\(\Leftrightarrow\left(-x^2+x+2\right)\left[3\left(x^2+1\right)^2+2\left(x+3+\dfrac{x^2+1}{2}\right)^2+\dfrac{3\left(x^2+1\right)^2}{4}\right]=0\)
\(\Leftrightarrow-x^2+x+2=0\) (phần ngoặc phía sau luôn dương)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b/ \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5\left(x^2+3x-1-\left(x^2+2x-1\right)\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}a=x^2+2x-1\\b=x^2+3x-1\end{matrix}\right.\)
\(3a^2-2b^2+5\left(b-a\right)^2=0\Leftrightarrow8a^2+3b^2-10ab=0\)
\(\Leftrightarrow\left(4a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}4a=3b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=3\left(x^2+3x-1\right)\\2\left(x^2+2x-1\right)=x^2+3x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2}{\left(x^2-1\right)^2}-\dfrac{11\left(x^4-5x^2+4\right)}{\left(x^2-1\right)^2}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)^2-6x\left(x^2+2\right)+9x^2+\left(x^2+2\right)^2+6x\left(x^2+2\right)+9x^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow2\left(x^2+2\right)^2+18x^2-11x^4+55x^2-44=0\)
\(\Leftrightarrow2\left(x^4+4x^2+4\right)-11x^4+73x^2-44=0\)
=>\(-9x^4+81x^2-36=0\)
=>9x^4-81x^2+36=0
=>x^4-9x^2+4=0
=>\(x^2=\dfrac{9\pm\sqrt{65}}{2}\)
=>\(x=\pm\sqrt{\dfrac{9\pm\sqrt{65}}{2}}\)