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CD
0
PN
17 tháng 5 2017
3x.|x+1|−2x|x+2|=12
Với x < -2 ta có: 3x.(-x-1)-2x(-x-2)-12=0
<=> -3x2 - 3x + 2x2 + 4x -12 =0
<=> -x2 - x - 12=0
$\Leftrightarrow $ -(x2 +x+12)=0 ( vô lý)
Làm tương tự với 2 trường hợp còn lại:
begin{align}
\begin{cases}
-2 bé hơn hoặc bằng x bé hơn -1 \\
x lớn hơn hoặc bằng -1 \\
\end{cases}
\end{align}
MM
1
26 tháng 11 2022
a: \(=\dfrac{2x^2-16x+44+6}{x^2-8x+22}=2+\dfrac{6}{x^2-8x+22}\)
\(=2+\dfrac{6}{\left(x-4\right)^2+6}\)
(x-4)^2+6>=6
=>6/(x-4)^2+6<=1
=>A<=3
Dấu = xảy ra khi x=4
b: \(B=\dfrac{5x^2+4x-1}{x^2}=\dfrac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\dfrac{\left(2x-1\right)^2}{x^2}< =9\)
Dấu = xảy ra khi x=1/2
3 tháng 3 2018
\(\Leftrightarrow1+\dfrac{2}{x+2}+1+\dfrac{8}{x+8}=1+\dfrac{4}{x+4}+1+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x+2}+\dfrac{4}{x+8}=\dfrac{2}{x+4}+\dfrac{3}{x+6}\)
\(\Leftrightarrow\dfrac{4}{x+8}-\dfrac{3}{x+6}=\dfrac{2}{x+4}-\dfrac{1}{x+2}\)
\(\Leftrightarrow\dfrac{4x+24-3\left(x+8\right)}{\left(x+8\right)\left(x+6\right)}=\dfrac{2x+4-\left(x+4\right)}{\left(x+4\right)\left(x+2\right)}\)
\(\dfrac{x}{\left(x+8\right)\left(x+6\right)}=\dfrac{x}{\left(x+4\right)\left(x+2\right)}\)
x=0 là nghiệm
x khác 0
\(\left\{{}\begin{matrix}x\ne\left\{-8;-6;-4;-2\right\}\\\left(x+4\right)\left(x+2\right)=\left(x+8\right)\left(x+6\right)\end{matrix}\right.\)<=>x^2 +6x+8 =x^2 +14x+48
-40 =8x=> x =-5 nhận
x={-5;0}
15 tháng 4 2018
\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\left(đkxđ:x\ne-2;-8;-4;-6\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}-1+\dfrac{8}{x+8}-1=\dfrac{4}{x+4}-1+\dfrac{6}{x+6}-1\)
\(\Leftrightarrow\dfrac{-x}{x+2}+\dfrac{-x}{x+8}=\dfrac{-x}{x+4}+\dfrac{-x}{x+6}\)
\(\Leftrightarrow\left(-x\right)\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
27 tháng 6 2017
a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....
b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........
c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......
d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......
TK
30 tháng 3 2017
ĐKXĐ: x\(\ne\) -2; x\(\ne\) -4; x\(\ne\) -6; x\(\ne\) -8;
\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\) \(\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow\left(x+2+\dfrac{2}{x+2}\right)+\left(x+8+\dfrac{8}{x+8}\right)=\)
\(\left(x+4+\dfrac{4}{x+4}\right)+\left(x+6+\dfrac{6}{x+6}\right)\)
\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
=> 2.(x+4)(x+8)(x+6) + 8(x+2)(x+4)(x+6)=4(x+2)(x+6)(x+8)
+ 6(x+2)(x+4)(x+8)
<=>(2x+8)(x2 + 14x+64) + (8x+48)(x2+6x+8) - (4x+8)(x2 + 14x+64)
-(6x+48)(x2+6x+8)
<=> (x2 + 14x+64)(2x+8 -4x -8) + (x2+6x+8)(8x+48+6x-48)=0
<=> -2x(x2 + 14x+64)+ 2x(x2+6x+8) = 0
<=> -2x3 -28x2 -128x+ 2x3 +12x2 +16x = 0
<=> -16x2 - 112x = 0
<=> -x(16x+112) = 0
<=>\(\left[{}\begin{matrix}x=0\\16x+112=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=7\left(tmđk\right)\end{matrix}\right.\)
vậy S={0;7}
TK
30 tháng 3 2017
sửa bài:
<=>﴾2x+8﴿﴾x2 + 14x+48﴿ + ﴾8x+48﴿﴾x2 +6x+8﴿ ‐ ﴾4x+8﴿﴾x2 + 14x+48﴿
‐﴾6x+48﴿﴾x2 +6x+8﴿
<=> ﴾x2 + 14x+48﴿﴾2x+8 ‐4x ‐8﴿ + ﴾x2 +6x+8﴿﴾8x+48+6x‐48﴿=0
<=> ‐2x﴾x2 + 14x+48﴿+ 2x﴾x2 +6x+8﴿ = 0
<=> ‐2x3 ‐28x2 ‐96x+ 2x3 +12x2 +16x = 0
<=> ‐16x2 ‐ 80x = 0
<=> ‐x﴾16x+80﴿ = 0
<=>\(\left[{}\begin{matrix}x=0\\16x+80=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
vậy : S={0;-5}
\(\dfrac{8x^2+16x^2+8x}{4x^2+4x}\)
= \(\dfrac{24x^2+8x}{4x^2+4x}\)
= \(\dfrac{4x(6x+2)}{4x(x+1)}\)
= \(\dfrac{6x+2}{x+1}\)