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1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)
\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)
\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)
\(\Leftrightarrow5x-6=0\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy: x=-2
3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)
\(\Leftrightarrow15x-30=0\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
Vậy: x=2
4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)
\(\Leftrightarrow83x-83=0\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy: x=1
\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
=> \(2x^2+3\left(x^2-1\right)=5x^2+5x\)
=> \(2x^2+3x^2-3-5x^2-5x=0\)
=> \(-3-5x=0\)
=> \(5x=-3\Rightarrow x=-\frac{3}{5}\)
\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+\frac{9}{2}\right)=\frac{7}{2}\)
=> \(x\left[2x\left(x+5\right)-1\left(x+5\right)\right]-2x^2\left(x+\frac{9}{2}\right)-1\left(x+\frac{9}{2}\right)=\frac{7}{2}\)
=> \(x\left(2x^2+10x-x-5\right)-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)
=> \(2x^3+10x^2-x^2-5x-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)
=> \(\left(2x^3-2x^3\right)+\left(10x^2-x^2-9x^2\right)+\left(-5x-x\right)-\frac{9}{2}=\frac{7}{2}\)
=> \(-6x-\frac{9}{2}=\frac{7}{2}\)
=> \(-6x=8\Rightarrow x=-\frac{8}{6}=-\frac{4}{3}\)
\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
=> 12x(4x - 1) - 5(4x - 1) + 3x(1 - 16x) - 7(1 - 16x) = 81
=> 48x2 - 12x - 20x + 5 + 3x - 48x2 - 7 + 112x = 81
=> -12x - 20x + 3x + 112x + 5 - 7 = 81
=> 83x + 5 - 7 = 81
=> 83x = 81 + 7 - 5
=> 83x = 83
=> x = 1
1) \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow5x=-3\)
\(\Rightarrow x=-\frac{3}{5}\)
2) \(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+\frac{9}{2}\right)=\frac{7}{2}\)
\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)
\(\Leftrightarrow-6x=8\)
\(\Rightarrow x=-\frac{4}{3}\)
3) \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-32x+5-48x^2+115x-7=81\)
\(\Leftrightarrow83x=83\)
\(\Rightarrow x=1\)
a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0
=>-5x-4=0
=>x=-4/5
b: =>6x^2-9x+2x-3-6x^2-12x=16
=>-19x=19
=>x=-1
c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81
=>83x=83
=>x=1
a/ \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
<=> \(48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
<=> \(83x-2=81\)
<=> \(83x=83\)
<=> \(x=1\)
b/ \(\left(2x-3\right)\left(2x+3\right)-\left(4x+1\right)x=1\)
<=> \(4x^2-9-4x^2-x=1\)
<=> \(-\left(9+x\right)=1\)
<=> \(9+x=-1\)
<=> \(x=-10\)
c/ \(3x^2-\left(x+2\right)\left(3x-1\right)=-7\)
<=> \(3x^2-\left(3x^2-x+6x-2\right)=-7\)
<=> \(3x^2-3x^2+x-6x+2=-7\)
<=> \(-5x+2=-7\)
<=> \(-5x=-9\)
<=> \(x=\frac{9}{5}\)
a, <=> x2 -2x +1 + 5x -x2 =8
<=> 3x +1 =8
<=> 3x = 7
<=> x= 7/3
b, thiếu đề
c, <=> 2x3 -1 + 2x(4 -x2) = 7
<=> 2x3 + 8x -23 = 8
<=> 8x =8
<=> x=1
1.
\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)
\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)
\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max
2.
\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)
\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)
\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)
\(E_{min}=-1\) khi \(x=0\)
\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)
\(G_{min}=-2\) khi \(x=2\)
b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)
\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)
\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)
\(\Leftrightarrow3x\left(x+4\right)=0\)
=>x=0(nhận) hoặc x=-4(loại)