Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
\(x^2-x+\dfrac{1}{2}=x^2-2\cdot\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{2}\\ =\left(x^2-2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)-\dfrac{1}{4}+\dfrac{1}{2}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
ta có: \(\left(x-\dfrac{1}{2}^{ }\right)^2\ge0\forall x\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}>0\forall x\left(vì\dfrac{1}{4}>0\right)\)
hay \(x^2-x+\dfrac{1}{2}>0\forall x\)
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)
a, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
Bạn ơi! Bạn xem lại đề chứ mình làm thử kết quả ra một số thập phân tương đối dài đó bạn
Bạn xem lại đề xem thử có sai gì không nha :))
\(\left(x+3\right)\left(2x-4\right)-\left(x-3\right)=0\)
\(\Leftrightarrow2x^2+2x-12-x+3=0\)
\(\Leftrightarrow2x^2+x-9=0\)
\(\Delta=1^2-4\cdot2\cdot\left(-9\right)=73\)
\(\Rightarrow x_{1,2}=\dfrac{-1\pm\sqrt{73}}{4}\)
\(\dfrac{x+1}{39}+\dfrac{x+2}{38}+\dfrac{x+3}{37}=0\)
\(\Leftrightarrow\dfrac{x+1}{39}+1+\dfrac{x+2}{38}+1+\dfrac{x+3}{37}+1-3=0\)
\(\Leftrightarrow\dfrac{x+40}{39}+\dfrac{x+40}{38}+\dfrac{x+40}{37}=3\)
\(\Leftrightarrow\left(x+40\right)\left(\dfrac{1}{39}+\dfrac{1}{38}+\dfrac{1}{37}\right)=3\)
\(\Leftrightarrow\left(x+40\right).\dfrac{4331}{54834}=3\)
\(\Leftrightarrow x+40=\dfrac{164502}{4331}\)
\(\Leftrightarrow x=\dfrac{-8738}{4331}\)
-Vậy \(S=\left\{\dfrac{-8738}{4331}\right\}\)