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a: \(N=\left(\dfrac{1}{y-1}+\dfrac{1}{\left(y-1\right)\left(y^2+y+1\right)}\cdot\dfrac{y^2+y+1}{y+1}\right)\cdot\left(y^2-1\right)\)
\(=\dfrac{y+1+1}{\left(y-1\right)\left(y+1\right)}\cdot\left(y^2-1\right)=y+2\)
b: Thay y=1/2 vào N, ta được:
N=1/2+2=5/2
c: Để N>0 thì y+2>0
hay y>-2
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}y>-2\\y\notin\left\{-1;1\right\}\end{matrix}\right.\)
Lời giải:
a. ĐKXĐ: $y\neq \pm 1$
\(N=\left(\frac{1}{y-1}-\frac{1}{(1-y)(1+y+y^2)}.\frac{y^2+y+1}{y+1}\right).(y^2-1)\)
\(=(\frac{1}{y-1}-\frac{1}{(1-y)(y+1)})(y-1)(y+1)\)
\(=\frac{1}{y-1}(y-1)(y+1)-\frac{1}{-(y-1)(y+1)}.(y-1)(y+1)=y+1-(-1)=y+2\)
b. Khi $y=\frac{1}{2}$ thì:
$N=\frac{1}{2}+2=\frac{5}{2}$
c. Để $N>0\Leftrightarrow y+2>0\Leftrightarrow y>-2$
Kết hợp đkxđ suy ra $y>-2$ và $y\neq \pm 1$ thì $N$ dương.
ĐK: \(x\ne-\dfrac{2}{3};x\ne3\)
\(\dfrac{6x-1}{3x+2}=\dfrac{2x+5}{x-3}\Rightarrow\left(6x-1\right)\left(x-3\right)=\left(2x+5\right)\left(3x+2\right)\)
\(\Leftrightarrow6x^2-19x+3=6x^2+19x+10\Leftrightarrow38x=-7\Leftrightarrow x=-\dfrac{7}{38}\).
ĐKXĐ : x ≠ -2/3 ; x ≠ 3
\(\dfrac{6x-1}{3x+2}=\dfrac{2x+5}{x-3}\Rightarrow\left(6x-1\right)\left(x-3\right)=\left(3x+2\right)\left(2x+5\right)\)
\(\Leftrightarrow6x^2-19x+3=6x^2+19x+10\)
\(\Leftrightarrow-38x=7\Leftrightarrow x=-\dfrac{7}{38}\)(tm)
Vậy ...
theo đầu bài ta có\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)=>\(3x^2+3y^2=10xy\)
A=\(\dfrac{x-y}{x+y}\)
=>\(A^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}=\dfrac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\dfrac{10xy-6xy}{10xy+6xy}=\dfrac{4xy}{16xy}=\dfrac{1}{4}\)
=>A=\(\sqrt{\dfrac{1}{4}}=\dfrac{-1}{2}hoặc\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\) (cộng trừ căn 1/4 nhé)
vì y>x>0=> A=-1/2
4
ta có : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{-1}{z}\)
Ta có: \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{1}{x^3}+3\times\dfrac{1}{x^2}\times\dfrac{1}{y}+3\times\dfrac{1}{x}\times\dfrac{1}{y^2}+\dfrac{1}{y^3}-3\times\dfrac{1}{x^2}\times\dfrac{1}{y}-3\times\dfrac{1}{x}\times\dfrac{1}{y^2}+\dfrac{1}{z^3}\) \(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3-3\times\dfrac{1}{xy}\times\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\left(\dfrac{-1}{z}\right)^3-3\times\dfrac{1}{xy}\times\left(\dfrac{-1}{z}\right)+\dfrac{1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-\dfrac{1}{z^3}+3\times\dfrac{1}{xyz}+\dfrac{1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\Leftrightarrow xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=3\)(ĐPCM)
ĐK: \(3x\ne\pm y;x\ne0\)
A = \(\dfrac{3x}{3x+y}-\dfrac{x}{3x-y}+\dfrac{2x}{\left(3x-y\right)\left(3x+y\right)}\)
= \(\dfrac{3x\left(3x-y\right)-x\left(3x+y\right)+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{6x^2-4xy+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{2x\left(3x-2y+1\right)}{\left(3x-y\right)\left(3x+y\right)}\)
Thay x = 1; y=2, ta có:
A = \(\dfrac{2.1\left(3.1-2.2+1\right)}{\left(3.1-2\right)\left(3.1+2\right)}=0\)
\(A+1=1+\dfrac{3y^2-4y}{y^2+1}=\dfrac{4y^2-4y+1}{y^2+1}=\dfrac{\left(2y-1\right)^2}{y^2+1}\ge0\)
\(A+1\ge0\Rightarrow A\ge-1\)
ĐÂY NÀY:
( x +y) ^2 = a^2 => x^2 + 2xy + y^2 = a^2
=> 2xy = a^2 - ( x^2 + y^2) = a^2 -b
=> xy = a^2-b/2
Ta có E = x^3 + y^3 = ( x+ y)( x^2 - xy + y^2)
E = a ( b - a^2-b/2)
Giải:
Ta có:
\(P=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\)
\(\Leftrightarrow P=\dfrac{1}{2}\left[\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)+\left(\dfrac{yz}{x}+\dfrac{zx}{y}\right)+\left(\dfrac{zx}{y}+\dfrac{xy}{z}\right)\right]\)
Áp dụng BĐT AM-GM, có:
\(P=\dfrac{1}{2}\left[\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)+\left(\dfrac{yz}{x}+\dfrac{zx}{y}\right)+\left(\dfrac{zx}{y}+\dfrac{xy}{z}\right)\right]\ge\dfrac{1}{2}.\left(2\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}+2\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+2\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)\)
\(\Leftrightarrow P\ge\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\)
\(\Leftrightarrow P\ge x+y+z\)
\(\Leftrightarrow P\ge2019\)
\(\Leftrightarrow P_{Min}=2019\)
\("="\Leftrightarrow x=y=z=\dfrac{2019}{3}\)
Vậy ...
ĐK: \(y\ne1;y\ne3\).
Ta có \(\dfrac{y+5}{y-1}-\dfrac{y+1}{y-3}=\dfrac{-8}{\left(y-1\right)\left(y-3\right)}\)
\(\Leftrightarrow\dfrac{\left(y+5\right)\left(y-3\right)-\left(y+1\right)\left(y-1\right)}{\left(y-1\right)\left(y-3\right)}=\dfrac{-8}{\left(y-1\right)\left(y-3\right)}\)
\(\Rightarrow\left(y+5\right)\left(y-3\right)-\left(y+1\right)\left(y-1\right)=-8\Leftrightarrow\left(y^2+2y-15\right)-\left(y^2-1\right)=-8\Leftrightarrow2y-14=-8\Leftrightarrow y=3\). (loại)
Vậy không tồn tại y thỏa mãn