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1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
1/\(4x^4+12x^3-47x^2+12x+4=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x^3+20x^2-7x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(2x^2+11x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\\x=\frac{-11\pm\sqrt{105}}{4}\end{matrix}\right.\)
Vậy ....
a,\(\left|9+x\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}9+x=2x\\9x+x=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}9=x\\9=-3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
Vậy...
Trường hợp 2 chưa chắc chắn lắm!!!
a) \(\left|9+x\right|=2x\)
Xét trường hợp 1:
\(9+x=2x\)
\(\Leftrightarrow9+x-2x=0\)
\(\Leftrightarrow9-x=0\)
\(\Leftrightarrow x=9\)
Xét trường hợp 2:
\(9+x=-2x\)
\(\Leftrightarrow9+x-\left(-2x\right)=0\)
\(\Leftrightarrow9+x+2x=0\)
\(\Leftrightarrow9+3x=0\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-9:3\)
\(\Leftrightarrow x=-3\)
Vậy x=9 hoặc x=-3
b) \(\left|x+6\right|-9=2x\)
\(\Leftrightarrow\left|x+6\right|=2x+9\)
Xét trường hợp 1:
\(x+6=2x+9\)
\(\Leftrightarrow x+6-\left(2x+9\right)=0\)
\(\Leftrightarrow x+6-2x-9=0\)
\(\Leftrightarrow-3-x=0\)
\(\Leftrightarrow x=-3\)
Xét trường hợp 2:
\(x+6=-\left(2x+9\right)\)
\(\Leftrightarrow x+6-\left[-\left(2x+9\right)\right]=0\)
\(\Leftrightarrow x+6+\left(2x+9\right)=0\)
\(\Leftrightarrow x+6+2x+9=0\)
\(\Leftrightarrow3x+15=0\)
\(\Leftrightarrow3x=-15\)
\(\Leftrightarrow x=-15:3\)
\(\Leftrightarrow x=-5\)
Vậy x=-3 hoặc x=-5
a/ ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow\sqrt{x+1}-1+\sqrt{x+4}-2>0\)
\(\Leftrightarrow\frac{x}{\sqrt{x+1}+1}+\frac{x}{\sqrt{x+4}+2}>0\)
\(\Leftrightarrow x>0\)
b/
Chắc bạn ghi nhầm đề, thấy đề hơi kì lạ
c/ ĐKXĐ: \(\left[{}\begin{matrix}-\frac{3}{2}\le x\le\frac{3-\sqrt{57}}{8}\\x\ge\frac{3+\sqrt{57}}{8}\end{matrix}\right.\)
\(\Leftrightarrow2x+3>4x^2-3x-3\)
\(\Leftrightarrow4x^2-5x-6< 0\) \(\Rightarrow-\frac{3}{4}< x< 2\)
Kết hợp ĐKXĐ ta được nghiệm của BPT: \(\left[{}\begin{matrix}-\frac{3}{4}< x\le\frac{3-\sqrt{57}}{8}\\\frac{3+\sqrt{57}}{8}\le x< 2\end{matrix}\right.\)
d/
\(\Leftrightarrow x^2+5x+28-5\sqrt{x^2+5x+28}-24< 0\)
Đặt \(\sqrt{x^2+5x+28}=t>0\)
\(\Leftrightarrow t^2-5t-24< 0\) \(\Rightarrow-3< t< 8\)
\(\Rightarrow t< 8\Rightarrow\sqrt{x^2+5x+28}< 8\)
\(\Leftrightarrow x^2+5x-36< 0\Rightarrow-9< x< 4\)
Lời giải:
Ta có:
\(x^4-4x^3-2x^2+12x+5=0\)
\(\Leftrightarrow (x^4-4x^3+4x^2)-6x^2+12x+5=0\)
\(\Leftrightarrow (x^2-2x)^2-6(x^2-2x)+5=0\)
Đặt $x^2-2x=a$. Khi đó:
\(a^2-6a+5=0\)
\(\Leftrightarrow (a-1)(a-5)=0\Rightarrow \left[\begin{matrix} a=1\\ a=5\end{matrix}\right.\)
Nếu $a=1$ thì $x^2-2x=1$
\(\Leftrightarrow x^2-2x-1=0\Rightarrow x=1\pm \sqrt{2}\)
Nếu $a=5$ thì $x^2-2x=5$
\(\Leftrightarrow x^2-2x-5=0\Rightarrow x=1\pm \sqrt{6}\)
a) \(\left[\left(4x+28\right).3+55\right]:5=35\)
\(\Leftrightarrow\left(4x+28\right).3+55=35.5\)
\(\Leftrightarrow\left(4x+28\right).3+55=175\)
\(\Leftrightarrow\left(4x+28\right).3=175-55\)
\(\Leftrightarrow\left(4x+28\right).3=120\)
\(\Leftrightarrow4x+28=120:3\)
\(\Leftrightarrow4x+28=40\)
\(\Leftrightarrow4x=40-28\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=12:4\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
b) \(\left(12x-4^3\right).8^3=4.8^4\)
\(\Leftrightarrow12x-4^3=4.8^4:8^3\)
\(\Leftrightarrow12x-4^3=4.8^{4-3}\)
\(\Leftrightarrow12x-4^3=4.8\)
\(\Leftrightarrow12x-4^3=32\)
\(\Leftrightarrow12x-64=32\)
\(\Leftrightarrow12x=32+64\)
\(\Leftrightarrow12x=96\)
\(\Leftrightarrow x=96:12\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
c) \(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=8.5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x-5\right)=720:40\)
\(\Leftrightarrow41-\left(2x-5\right)=18\)
\(\Leftrightarrow2x-5=41-18\)
\(\Leftrightarrow2x-5=23\)
\(\Leftrightarrow2x=23+5\)
\(\Leftrightarrow2x=28\)
\(\Leftrightarrow x=28:2\)
\(\Leftrightarrow x=14\)
Vậy \(x=14\)
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