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1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
B1:
a) \(\frac{x+4}{x+3}=\frac{x+9}{x+4}\)
-->(x+4)(x+4)=(x+3)(x+9)
\(x^2\)+4x+4x+16=\(x^2\)+9x+3x+27
\(x^2-x^2\)+4x+4x-9x-3x= - 16+27
- 4x=11
x=\(\frac{-4}{11}\)
b) \(\frac{x-5}{x+3}=\frac{x-4}{x+6}\)
-->(x-5)(x+6)=(x+3)(x-4)
\(x^2\)+6x-5x-30=\(x^2\)-4x+3x-12
\(x^2-x^2\)+6x-5x+4x-3x=30-12
2x=18
x=9
c)\(\frac{3x-1}{3x}=\frac{2x-1}{2x+1}\)
--> (3x-1)(2x+1)=3x.(2x-1)
\(6x^2\)+3x-2x-1=\(6x^2\)-3x
\(6x^2-6x^2\)+3x-2x+3x=1
4x=1
x=\(\frac{1}{4}\)
\(2x-\frac{1}{2}-x=\frac{3}{4}-x+\frac{2}{3}\)
\(2x-x+x=\frac{1}{2}+\frac{3}{4}+\frac{2}{3}\)
\(2x=\frac{23}{12}\)
\(x=\frac{23}{12}:2=\frac{23}{24}\)
a) \(-2\sqrt{x^2+1}=-8\)
=> \(\sqrt{x^2+1}=-8:\left(-2\right)\)
=> \(\sqrt{x^2+1}=4\)
=> \(x^2+1=16\)
=> \(x^2=16-1=15\)
=> \(\orbr{\begin{cases}x=\sqrt{15}\\x=-\sqrt{15}\end{cases}}\)
b) \(4+3\sqrt{x^2+2}=4\)
=> \(3\sqrt{x^2+2}=4-4=0\)
=> \(\sqrt{x^2+2}=0\)
=> \(x^2+2=0\)
=> \(x^2=-2\)
=> ko có giá trị x t/m
c)\(\sqrt{x+1}=3\)
=> \(x+1=9\)
=> x = 9 - 1 = 8
d) TT trên
\(\frac{1}{3}-|\frac{5}{4}-2x|=\frac{1}{4}\)
\(\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}Th1:\frac{5}{4}-2x=\frac{7}{12}\\Th2:\frac{5}{4}-2x=-\frac{7}{12}\end{cases}}\)
\(\Leftrightarrow Th1:\frac{5}{4}-2x=\frac{7}{12}\) \(\Leftrightarrow Th2:\frac{5}{4}-2x=-\frac{7}{12}\)
\(\Leftrightarrow2x=\frac{7}{12}+\frac{5}{4}\) \(\Leftrightarrow2x=-\frac{7}{12}+\frac{5}{4}\)
\(\Leftrightarrow2x=\frac{11}{6}\) \(\Leftrightarrow2x=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{11}{12}\) \(\Leftrightarrow x=\frac{1}{3}\)
P/s : Mình làm bừa ạ nếu kh đúng xin mọi người chỉ thêm ~~
\(\frac{x+1}{x-2}=\frac{3}{4}\) ( \(ĐKXĐ\) : \(x\ne2\) )
\(\Leftrightarrow\left(x+1\right).4=\left(x-2\right).3\)
\(\Leftrightarrow4x+4=3x-6\)
\(\Leftrightarrow4x-3x=-6-4\)
\(\Leftrightarrow x=-10\)
b ) \(\frac{2x-3}{x+1}=\frac{4}{7}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow7\left(2x-3\right)=4\left(x+1\right)\)
\(\Leftrightarrow14x-21=4x+4\)
\(\Leftrightarrow10x=25\)
\(\Leftrightarrow x=\frac{5}{2}\)
\(2^{2x+1}+4^{x+3}=4224\)
\(2^{2x}.2+4^x.64=4224\)
\(2^{2x}.2+2^{2x}.2^6=4224\)
\(2^{2x}.\left(2+2^6\right)=4224\)
\(2^{2x}=64\)
\(2^{2x}=2^6\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)