Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
2)
a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)
\(=\dfrac{6x}{xy}\)
\(=\dfrac{6}{y}\)
b) \(\dfrac{2x^2}{y}.3xy^2\)
\(=\dfrac{2x^2.3xy^2}{y}\)
\(=\dfrac{6x^3y^2}{y}\)
\(=6x^3y\)
c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)
\(=\dfrac{15x.2y^2}{7y^3.x^2}\)
\(=\dfrac{30xy^2}{7x^2y^3}\)
\(=\dfrac{30}{7xy}\)
d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
\(A=\dfrac{1}{-x^2+2x-2}\)
A min \(\Leftrightarrow\dfrac{1}{A}\)max
ta có \(\dfrac{1}{A}=-x^2+2x-2=-\left(x^2-2x+2\right)=-\left(x-1\right)^2-1\le-1\)
\(\dfrac{1}{A}\)max= -1 tại x=1
=> A min = -1 tại x=1
\(B=\dfrac{2}{-4x^2+8x-5}\) ( phải là -4x2 nha bn)
B min \(\Leftrightarrow\dfrac{1}{B}\) max
ta có \(\dfrac{1}{B}=\dfrac{-4x^2+8x-5}{2}=\dfrac{-\left(4x^2-8x+5\right)}{2}=\dfrac{-\left(2x-4\right)^2+11}{2}=\dfrac{\left(-2x-4\right)^2}{2}+\dfrac{11}{2}\le\dfrac{11}{2}\)
\(\dfrac{1}{B}\)max=\(\dfrac{11}{2}\) tại x=2
\(\Rightarrow B\) min = \(\dfrac{1}{\dfrac{11}{2}}=\dfrac{2}{11}\) tại x=2
\(A=\dfrac{3}{2x^2+2x+3}=\dfrac{3}{2\left(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{5}{2}}=\dfrac{3}{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}\)
A max \(\Leftrightarrow\dfrac{1}{A}\) min
\(\Leftrightarrow\dfrac{2\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{\dfrac{5}{2}}{3}=\dfrac{2\left(x+\dfrac{1}{2}\right)^2}{3}+\dfrac{5}{6}\ge\dfrac{5}{6}\)
\(\dfrac{1}{A}\) min = \(\dfrac{5}{6}\)tại x= \(-\dfrac{1}{2}\)
\(\Rightarrow A\)max = \(\dfrac{6}{5}\) tại x= \(-\dfrac{1}{2}\)
B\(=\dfrac{5}{3x^2+4x+15}=\dfrac{5}{3.\left(x^2+\dfrac{4}{3}x+5\right)}=\dfrac{5}{3\left(x^2+2.x.\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{41}{9}\right)}=\dfrac{5}{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}\)
B max \(\Leftrightarrow\dfrac{1}{B}\) min
\(\Leftrightarrow\dfrac{3\left(x+\dfrac{2}{3}\right)^2+\dfrac{41}{3}}{5}=\dfrac{3\left(x+\dfrac{2}{3}\right)^2}{5}+\dfrac{41}{15}\ge\dfrac{41}{15}\)
\(\dfrac{1}{B}\) min = \(\dfrac{41}{15}\) tại x=\(-\dfrac{2}{3}\)
=> \(B\) max = \(\dfrac{15}{41}\) tại x=\(-\dfrac{2}{3}\)
Đây chỉ là gợi ý !! bn pải tự lí luận nha
tik 
1) a) Đặt biểu thức là A
\(A=2x^2+4y^2-4xy-4x-4y+2017\)
\(A=\left(x-2y\right)^2+x^2-4x-4y+2017\)
\(A=\left(x-2y\right)^2+2\left(x-2y\right)+x^2-6x+2017\)
\(A=\left(x-2y-1\right)^2+\left(x+3\right)^2+2008\)
Vậy: MinA=2008 khi x=-3; y=-2
3) a) \(A=\dfrac{1}{x^2+x+1}\)
\(B=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(\Rightarrow B\ge\dfrac{3}{4}\Rightarrow A\ge\dfrac{4}{3}\)
Vậy MinA là \(\dfrac{4}{3}\) khi x=-0,5
\(a,\)
\(A=\dfrac{x^2-2x}{x^2+x-6}.\dfrac{x^2-9}{x^2-3x}\)
\(A=\dfrac{x\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}.\dfrac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}\)
\(A=\dfrac{x}{\left(x+3\right)}.\dfrac{\left(x+3\right)}{x}=1\)
\(b,\)
\(B=\dfrac{2x^2+5x-3}{x^2-9}.\dfrac{4x^2-1}{x^2-6x+9}\)
\(B=\dfrac{\left(x+3\right)\left(2x-1\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(x-3\right)^2}\)
\(B=\dfrac{\left(2x-1\right)}{\left(x-3\right)}.\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(x-3\right)^2}\)
\(B=\dfrac{\left(2x-1\right)^2.\left(2x+1\right)}{\left(x-3\right)^3}\)
\(C=\dfrac{4}{x^2-x+1}=\dfrac{4}{x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{4}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)
\(D=-5x^2-6x+2=-\left(5x^2+6x-2\right)\\ =-5\left(x^2+\dfrac{6}{5}x\right)+2\\ =-5\left(x^2+2.x.\dfrac{3}{5}+\dfrac{9}{25}\right)+\dfrac{19}{5}\\ =-5\left(x+\dfrac{3}{5}\right)^2+\dfrac{19}{5}\le\dfrac{19}{5}\)
1.
a.
\(\dfrac{5x+10}{4x-8}\cdot\dfrac{2x-4}{x+2}=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{2\left(x-2\right)}{x+2}=\dfrac{5\cdot2}{4}=\dfrac{5}{2}\)
b.
\(\dfrac{1-4x^2}{x+4x}:\dfrac{2-4x}{3x}=\dfrac{\left(1-2x\right)\left(2x+1\right)}{5x}:\dfrac{2\left(1-2x\right)}{3x}=\dfrac{\left(1-2x\right)\left(2x+1\right)\cdot3x}{5x\cdot2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{10}\)
2.
\(-x^2+6x-11=-\left(x^2-2\cdot x\cdot3+9\right)-2=-\left(x-3\right)^2-2\le-2\)
Max = -2 khi x = 3
.