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Cho 2 đa thức
Ta có: A = \(\left(x+3y-5\right)^2-6xy+26\)
=> A = \(x^2+9y^2+25+6xy-10x-30y+6xy+26\)
=> A = \(\left(x^2-10x+25\right)+\left(9y^2-30y+25\right)+1\)
=> A = \(\left(x-5\right)^2+\left(3y-5\right)^2+1\)
Vì \(\left\{{}\begin{matrix}\left(x-5\right)^2\ge0\\\left(3y-5\right)^2\ge0\end{matrix}\right.\) => A \(\ge\) 1
=> Dấu bằng xảy ra <=> \(\left\{{}\begin{matrix}x-5=0\\3y-5=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=5\\y=\dfrac{5}{3}\end{matrix}\right.\)
=> GTNN của A =1 khi x = 5; y = \(\dfrac{5}{3}\)
a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) vậy \(x=2;x=-2\)
b) \(\left(2x+1\right)+x\left(2x+1\right)=0\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=-1;x=\dfrac{-1}{2}\)
c) \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\) vậy \(x=0;x=5;x=-5\)
d) \(9\left(3x-2\right)=x\left(2-3x\right)\Leftrightarrow9\left(3x-2\right)=-x\left(3x-2\right)\)
\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\Leftrightarrow\left(9+x\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}9+x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x=-9;x=\dfrac{2}{3}\)
e) \(5x\left(x-3\right)-2x+6=0\Leftrightarrow5x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\) vậy \(x=\dfrac{2}{5};x=3\)
a,(5x-2y)(x2-xy+1)=5x3-5x2+5x-2yx2+2xy2-2y
=5x3-7x2y+2xy2+5x-2y
b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x2-4.\(\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-2x+20\)
c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)
=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
d,\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
=\(x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
=\(-5x+4x-15\)
=\(-x-15\)
Chúc bạn học tốt(mỏi tay quá)
\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
a, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
*_*!!!
a, \(9x^2+6xy+y^2=9x^2+3xy+3xy+y^2\)
\(=\left(9x^2+3xy\right)+\left(3xy+y^2\right)\)
\(=3x.\left(3x+y\right)+y.\left(3x+y\right)\)
\(=\left(3x+y\right)^2\)
b, \(x^4+2x^3+x^2=x^4+x^3+x^3+x^2\)
\(=\left(x^4+x^3\right)+\left(x^3+x^2\right)\)
\(=x^3.\left(x+1\right)+x^2.\left(x+1\right)\)
\(=\left(x+1\right).\left(x^3+x^2\right)=\left(x+1\right).x^2.\left(x+1\right)\)
\(=\left(x+1\right)^2.x^2\)
Chúc bạn học tốt!!!
a)\(9x^2+6xy+y^2=\left(3x+y\right)^2\)
b)\(x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
c)\(5x^2-10xy+5xy^2-5z^2\)
\(=5\left(x^2-2xy+xy^2-z^2\right)\)
c/ \(\dfrac{x+4}{x+1}-2=\dfrac{2-x}{x}\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
a) \(5-\left(x-6\right)=4\left(3-2x\right)\\ < =>5-x+6=12-8x\\ < =>-x+8x=5+6-12\\ < =>7x=-1\\ =>x=-\dfrac{1}{7}\)
Vậy: Tập nghiệm của pt là S= {-1/7}.
A+B
=3x^2y^3-5x^3y^2-5xy+1+5x^3y^2-2x^2y^3-5xy+2
=x^2y^3-10xy+3
A+B = \(\left(3x^2y^3-5x^3y^2-5xy+1\right)+\left(5x^3y^2-2x^2y^3-5xy+2\right)\)
= \(\left(3x^2y^3-2x^2y^3\right)+\left(-5x^3y^2+5x^3y^2\right)+\left(-5xy-5xy\right)+\left(1+2\right)\)
= \(x^2y^3-10xy+3\)
Không bíc có đúng không nữa