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\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)
\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)
\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)
\(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\)
\(=\frac{2x+1}{x-3}\)
b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)
thay \(x=-\frac{3}{2}\) vào P tâ đc: \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)
c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)
\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)
\(\Leftrightarrow4x+2=x^2-3x\)
\(\Leftrightarrow x^2-7x-2=0\)
\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)
bạn tự giải nốt nhé!!
d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)
\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
bạn tự làm nốt nhé
a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)
\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\)
b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)
a)\(\left(4x^3-xy^2+y^3\right)\left(x^2y+2xy^2-2y^3\right)\)
\(=x^2y\left(4x^3-xy^2+y^3\right)+2xy^2\left(4x^3-xy^2+y^3\right)\)
\(-2y^3\left(4x^3-xy^2+y^3\right)\)
\(=4x^5y-x^3y^3+x^2y^4+8x^4y^2-2x^2y^4+2xy^5\)
\(-8x^3y^3+2xy^5-2y^6\)
\(=-2y^6+4x^5y+\left(2xy^5+2xy^5\right)+8x^4y^2+\left(x^2y^4-2x^2y^4\right)\)
\(-\left(x^3y^3+8x^3y^3\right)\)
\(=-2y^6+4x^5y+4xy^5+8x^4y^2-x^2y^4-9x^3y^3\)
b)
(!) \(2\left(x+y\right)^2-7\left(x+y\right)+5\)
\(=2\left(x+y\right)^2-2\left(x+y\right)-5\left(x+y\right)+5\)
\(=2\left(x+y\right)\left(x+y-1\right)-5\left(x+y-1\right)\)
\(=\left(2x+2y-5\right)\left(x+y-1\right)\)
(!!) \(\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-x^2-y^2-z^2\)
\(=2\left(xy+yz+zx\right)\)
a/ giá trị nhỏ nhất của A là 2
b/ giá trị lớn nhất của B là 51
tớ chỉ có bài tham khảo trên mạng thôi bạn thông cảm
Ta có: x + y = 1
<=> (x + y)3 = 1
<=> x3 + y3 + 3xy(x + y) = 1
<=> x3 + y3 + 3xy = 1 (do x + y = 1)
<=> x3 + y3 = 1 - 3xy
Áp dụng BĐT Cô - si, ta có:
xy >= (x+y)24=14(x+y)24=14
<=> -3xy≥−34≥−34
Ta có x3 + y3 = 1 - 3xy ≥1−34=14≥1−34=14
Dấu "=" xảy ra khi x = y = 1212
Vậy GTNN của x3 + y3 là 1414khi x = y = 12
\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Trường hợp x=y=z thì không phải bàn,ns cái trường hợp x+y+z=0
\(\frac{1}{x^2+y^2-z^2}=\frac{1}{\left(x+y\right)^2-2xy-z^2}=\frac{1}{\left(-z\right)^2-z^2-2xy}=\frac{1}{-2xy}\)
Tương tự rồi cộng lại thì \(BT=0\) thì phải
Condition\(\hept{\begin{cases}x\ne0\\y\ne0\\z\ne0\end{cases}}\)
Put \(P=\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}\)
\(=\frac{1}{x^2+\left(y-z\right)\left(y+z\right)}+\frac{1}{y^2+\left(z-x\right)\left(z+x\right)}+\frac{1}{z^2+\left(x-y\right)\left(x+y\right)}\left(4\right)\)
Because \(x^2+y^2+z^2=3xyz\)
\(\Leftrightarrow x^2+y^2+z^2-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=0\)ư\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\end{cases}}\)
The first case: If \(x+y+z=0\left(1\right)\)
\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\z+x=-y\end{cases}\left(2\right)}\)
From \(\left(1\right)\Rightarrow\hept{\begin{cases}x-y=-2y-z\\y-z=-2z-x\\z-x=-2x-y\end{cases}\left(3\right)}\)
\(\left(2\right)\)and \(\left(3\right)\)into \(\left(4\right)\)we have
\(P=\frac{1}{x^2-x\left(-2z-x\right)}+\frac{1}{y^2-y\left(-2x-y\right)}+\frac{1}{z^2-z\left(-2y-z\right)}\)
\(=\frac{1}{2x^2+2xz}+\frac{1}{2y^2+2xy}+\frac{1}{2z^2+2yz}\)
\(=\frac{1}{2x\left(x+z\right)}+\frac{1}{2y\left(x+y\right)}+\frac{1}{2z\left(z+y\right)}\)
\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)
\(\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}\)
\(=\frac{z+x+y}{-2xyz}=0\)( Because x+y+z=0)
The second case:\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\left(5\right)\)
We have \(\hept{\begin{cases}\left(x-y\right)^2\ge0;\forall x,y,z\\\left(y-z\right)^2\ge0;\forall x,y,z\\\left(z-x\right)^2\ge0;\forall x,y,z\end{cases}}\)\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0;\forall x,y,z\left(6\right)\)
From \(\left(5\right),\left(6\right)\)\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow x=y=z}\)
Because \(x=y=z\Rightarrow x^2=y^2=z^2=xy=yz=zx\)
So \(P=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)
\(=\frac{z+x+y}{xyz}=0\)
So...
a) Rút gọn C = 3 2 ( x − 2 ) 2 , thay x = 3 tính được C = 3 2 .
b) Rút gọn D = - ( x – y + z ) 2 , thay x = 17; y = 16; z = 1 tính được D = -4.