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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 \(x=-2\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a: \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\le\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)
\(\Leftrightarrow2x-3+5x\left(x-2\right)\le5x^2-7\left(2x-3\right)\)
\(\Leftrightarrow2x-3+5x^2-10x< =5x^2-14x+21\)
=>-8x-3<=-14x+21
=>6x<=24
hay x<=4
b: \(\dfrac{6x+1}{18}+\dfrac{x+3}{12}>=\dfrac{5x+3}{6}+\dfrac{12-5x}{9}\)
=>2(6x+1)+3(x+3)>=6(5x+3)+4(12-5x)
=>12x+2+3x+9>=30x+18+48-20x
=>15x+11>=10x+66
=>5x>=55
hay x>=11
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
\(A=2x-x^2=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\)
Vậy GTLN của A là 1 khi x = 1
\(B=-x^2-4x-y^2+2=-\left(x^2+4x+4\right)-y^2+6=-\left(x+2\right)^2-y^2+6\le6\)
Vậy GTLN của B là 6 khi x = -2; y = 0
\(C=19-6x-9x^2=-\left(9x^2+6x+1\right)+20=-\left(3x+1\right)^2+20\le20\)
Vậy GTLN của C là 20 khi x = \(-\dfrac{1}{3}\)
\(D=-4x^2-6x-4=-\left(4x^2+6x+\dfrac{9}{4}\right)-\dfrac{7}{4}=-\left(2x+\dfrac{3}{2}\right)^2-\dfrac{7}{4}\le-\dfrac{7}{4}\)
Vậy GTLN của D là \(-\dfrac{7}{4}\) khi x = \(-\dfrac{3}{4}\)
\(E=-\dfrac{1}{3}x^2+2x-5=-\dfrac{1}{3}\left(x^2-6x+9\right)-2=-\dfrac{1}{3}\left(x-3\right)^2-2\le-2\)\
Vậy GTLN của E là -2 khi x = 3
a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)
=>-12x-4=2x-10
=>-14x=-6
hay x=3/7
b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)
=>15x-9-10x+2=-4
=>5x-7=-4
=>5x=3
hay x=3/5(loại)
c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\)
=>4x=2
hay x=1/2(nhận)
\(C=\dfrac{4}{x^2-x+1}=\dfrac{4}{x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{4}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)
\(D=-5x^2-6x+2=-\left(5x^2+6x-2\right)\\ =-5\left(x^2+\dfrac{6}{5}x\right)+2\\ =-5\left(x^2+2.x.\dfrac{3}{5}+\dfrac{9}{25}\right)+\dfrac{19}{5}\\ =-5\left(x+\dfrac{3}{5}\right)^2+\dfrac{19}{5}\le\dfrac{19}{5}\)