giả pt \(\left(x^2+2x\right)^2-6x^2+12x+9=0\)
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1 tháng 6 2019
\(\left(x^2+2x\right)^2-6x^2-12x+9=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2-6\left(x^2+2x\right)+9=0\)
\(\Leftrightarrow\left(x^2-2x-3\right)^2=0\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow x_1=3\) \(x_2=-1\)
VH
21 tháng 11 2019
a, Ta có: \(\Delta'=1-m+3=4-m\)
Phương trình có 2 nghiệm phân biệt \(\Leftrightarrow\Delta'>0\Leftrightarrow4-m>0\Leftrightarrow m< 4\)
b, ĐXXĐ: \(x\le\frac{9}{4}\)
\(pt\Leftrightarrow\sqrt{\left(9-4x\right)\left(x-3\right)^2}=\left|-2x+5\right|\sqrt{9-4x}\)
\(\Leftrightarrow\sqrt{9-4x}\left(\left|x-3\right|-\left|-2x+5\right|\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}9-4x=0\\\left|x-3\right|=\left|-2x+5\right|\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}9-4x=0\\x-3=-2x+5\\x-3=2x-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{4}\\x=\frac{8}{3}\left(l\right)\\x=2\end{matrix}\right.\)
Vậy pt đã cho có 2 nghiệm \(x=2;x=\frac{9}{4}\)
TN
12 tháng 3 2020
\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\\\Leftrightarrow\left(x+1\right)^2=4\left(x-1\right)^2\\\Leftrightarrow \left(x+1\right)^2-4\left(x-1\right)^2=0\\\Leftrightarrow \left(x+1\right)^2-\left(2x-2\right)^2=0\\\Leftrightarrow \left[\left(x+1\right)+\left(2x-2\right)\right]\left[\left(x+1\right)-\left(2x-2\right)\right] =0\\ \Leftrightarrow\left(x+1+2x-2\right)\left(x+1-2x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(3-x\right)=0\\\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{3};3\right\}\)
\(\left(2x+7\right)^2=9\left(x+2\right)^2\\ \Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\\ \Leftrightarrow\left[\left(2x+7\right)+\left(3x+6\right)\right]\left[\left(2x+7\right)-\left(3x+6\right)\right]=0\\ \Leftrightarrow\left(2x+7+3x+6\right)\left(2x+7-3x-6\right)=0\\ \Leftrightarrow\left(5x+13\right)\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+13=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-13}{5}\\x=1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-13}{5};1\right\}\)
\(4\left(2x+7\right)^2=9\left(x+3\right)^2\\\Leftrightarrow 4\left(2x+7\right)^2-9\left(x+3\right)=0\\ \Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\\\Leftrightarrow \left[\left(4x+14\right)+\left(3x+9\right)\right]\left[\left(4x+14\right)-\left(3x+9\right)\right]=0\\\Leftrightarrow \left(4x+14+3x+9\right)\left(4x+14-3x-9\right)=0\\\Leftrightarrow \left(7x+23\right)\left(x+5\right)=0\\\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-23}{7};-5\right\}\)
12 tháng 8 2019
b) \(7x\left(x-2\right)-\left(x-2\right)=0\)
<=> \(\left(7x-1\right)\left(x-2\right)=0\)
=> x=1/7 hoặc x=2
c) <=> (2x-1)3 =0
=> x=1/2
d)<=> \(\left(2x-3\right)\left(2x+3\right)-x\left(2x-3\right)=0\)
<=> \(\left(2x-3\right)\left(x+3\right)=0\)
=> x=3/2 hoặc x=-3
e) <=>\(x^2\left(x+5\right)+9\left(x+5\right)=0\)
<=> \(\left(x+5\right)\left(x^2+9\right)=0\)
=> x=-5
f) \(x^3-6x^2-x+30=0\)
<=>\(x^3+2x^2-8x^2-16x+15x+30=0\)
<=>\(x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=0\)
<=>\(\left(x+2\right)\left(x^2-5x-3x+15\right)=0\)
<=> \(\left(x+2\right)\left(x-5\right)\left(x-3\right)=0\)
=> x=-2 hoặc x=5 hoặc x=3
AT
23 tháng 8 2018
c/ đk: x khác 1; x khác -3
\(\dfrac{3x-1}{x-1}+\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
\(\Rightarrow\left(3x+1\right)\left(x+3\right)+\left(2x+5\right)\left(x-1\right)+4=x^2+2x-3\)
\(\Leftrightarrow3x^2+10x+3+2x^2+3x-5+4=x^2+2x-3\)
\(\Leftrightarrow4x^2+11x+5=0\)
\(\Leftrightarrow\left(4x^2+2\cdot2x\cdot\dfrac{11}{4}+\dfrac{121}{16}\right)-\dfrac{41}{16}=0\)
\(\Leftrightarrow\left(2x+\dfrac{11}{4}\right)^2=\dfrac{41}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{11}{4}=\dfrac{\sqrt{41}}{4}\\2x+\dfrac{11}{4}=-\dfrac{\sqrt{41}}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{41}}{8}\\x=\dfrac{-11-\sqrt{41}}{8}\end{matrix}\right.\)
Vậy.........
d/ \(\dfrac{12x+1}{6x-2}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
đk: \(x\ne\pm\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{12x+1}{2\left(3x-1\right)}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)
\(\Rightarrow2\left(12x+1\right)\left(3x+1\right)-4\left(9x-5\right)\left(3x-1\right)=108x-36x^2-9\)
\(\Leftrightarrow72x^2+24x+6x+2-108x^2+36x-60x-20-108x+36x^2+9=0\)
\(\Leftrightarrow-102x-9=0\)
\(\Leftrightarrow-102x=9\Leftrightarrow x=-\dfrac{3}{34}\)(TM)
Vậy.........
AT
23 tháng 8 2018
a/ \(\left(x+1\right)^2\left(x+2\right)+\left(x+1\right)^2\left(x-2\right)=-24\)
\(\Leftrightarrow\left(x+1\right)^2\left(x+2+x-2\right)=-24\)
\(\Leftrightarrow2x\left(x^2+2x+1\right)=-24\)
\(\Leftrightarrow2x^3+4x^2+2x+24=0\)
\(\Leftrightarrow2x^3-2x^2+8x+6x^2-6x+24=0\)
\(\Leftrightarrow x\left(2x^2-2x+8\right)+3\left(2x^2-2x+8\right)=0\)
\(\Leftrightarrow\left(2x^2-2x+8\right)\left(x+3\right)=0\)
\(\Leftrightarrow2\left(x^2-x+4\right)\left(x+3\right)=0\)
Ta thấy: \(x^2-x+4=\left(x^2-2x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\)
=> x+ 3 = 0 <=> x= -3
Vậy......
b/ \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+x^2+5x+2x^2+x+5=0\)
\(\Leftrightarrow x\left(2x^2+x+5\right)+\left(2x^2+x+5\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
Ta thấy: \(2x^2+x+5=\left(\sqrt{2}x+2\cdot\sqrt{2}x\cdot\dfrac{\sqrt{2}}{4}+\dfrac{1}{8}\right)+\dfrac{39}{8}=\left(\sqrt{2}x+\dfrac{\sqrt{2}}{4}\right)^2+\dfrac{39}{8}>0\)
=> x + 1 = 0 <=> x = -1
Vậy....
31 tháng 10 2021
a: \(=\dfrac{\left(x^4-y^4\right)^2}{x^2+y^2}=\left(x^2-y^2\right)^2\cdot\left(x^2+y^2\right)\)
b: \(=\dfrac{\left(4x+3\right)\left(16x^2-12x+9\right)}{16x^2-12x+9}=4x+3\)
\(\left(x^2+2x\right)^2-6x^2+12x+9=0\Leftrightarrow x^4+4x^3+4x^2-6x^2+12x+9=0\\ \Leftrightarrow x^4+4x^3-2x^2+12x+9=0\Leftrightarrow x^2+4x-2+\frac{12}{x}+\frac{9}{x^2}=0\\ \Leftrightarrow\left(x^2+\frac{9}{x^2}\right)+4\left(x+\frac{3}{x}\right)-2=0\)
Đặt \(k=x+\frac{3}{x}\Rightarrow x^2+\frac{9}{x^2}=k^2-6\)
Ta đc \(k^2-6+4k-2=0\Leftrightarrow k^2+4k-8=0\)
\(\left(x^2+2x\right)^2\)\(-6x^2\)\(+12x+9\)=0
⇔\(\left(x^2\right)^2\)\(+2.2x.x^2\)+\(2x^2\)-6x2+12x+9=0
⇔ x4+ 4x3+2x2-6x2+12x+9=0
⇔ x2+4x3-4x2 +12x=-9
⇔x2+ 4x(x-x+3)=-9
⇔x2+12x=-9
⇔x(x+12)=-9
⇔ {x=-9 hoặc x+12=-9}
⇔ {x=-9 hoặc x=-21}
S={-9;-21}