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\(4x^4+9=0\)
\(4x^4=-9\)
mà \(4x^4\ge0\) với mọi x
=> Vô nghiệm
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1) \(\left|\dfrac{1}{2}x-\dfrac{1}{6}\right|=\dfrac{1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{6}=\dfrac{1}{3}\\\dfrac{1}{2}x-\dfrac{1}{6}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{1}{2}\\\dfrac{1}{2}x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(---\)
2) \(\left|\dfrac{1}{2}x+\dfrac{3}{5}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+\dfrac{3}{5}=\dfrac{1}{2}\\\dfrac{1}{2}x+\dfrac{3}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{1}{10}\\\dfrac{1}{2}x=-\dfrac{11}{10}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{11}{5}\end{matrix}\right.\)
\(---\)
3) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\left|\dfrac{-3}{4}\right|\)
\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{4}\\\dfrac{3}{4}x-\dfrac{3}{4}=-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{3}{2}\\\dfrac{3}{4}x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
\(---\)
4) \(14-\left|\dfrac{3x}{2}-1\right|=9\)
\(\Rightarrow\left|\dfrac{3x}{2}-1\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x}{2}-1=5\\\dfrac{3x}{2}-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=6\\\dfrac{3x}{2}=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\3x=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(---\)
5) \(17-\left|\dfrac{2}{3}-4x\right|=9\)
\(\Rightarrow\left|\dfrac{2}{3}-4x\right|=8\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-4x=8\\\dfrac{2}{3}-4x=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{22}{3}\\4x=\dfrac{26}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{6}\\x=\dfrac{13}{6}\end{matrix}\right.\)
\(---\)
6) \(5-\left|2x-3\right|=\dfrac{1}{2}\)
\(\Rightarrow\left|2x-3\right|=\dfrac{9}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
#\(Toru\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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a \(f\left(x\right)-h\left(x\right)=g\left(x\right)\)
\(h\left(x\right)=f\left(x\right)-g\left(x\right)\)
\(h\left(x\right)=\left(2x^4+5x^3-x+8\right)-\left(x^4-x^2-3x+9\right)\)
\(h\left(x\right)=2x^4+5x^3-x+8-x^4+x^2+3x-9\)
\(h\left(x\right)=3x^4+5x^3+x^2+2x-1\)
b \(h\left(x\right)-g\left(x\right)=f\left(x\right)\)
\(h\left(x\right)=f\left(x\right)+g\left(x\right)\)
\(h\left(x\right)=2x^4+5x^3-x+8+x^4-x^2-3x+9\)
\(h\left(x\right)=3x^4+5x^3-x^2-4x+17\)
![](https://rs.olm.vn/images/avt/0.png?1311)
theo t/c dãy tỉ số = nhau;
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}=\frac{\left(x+5\right)-\left(x+3\right)}{\left(y+7\right)-\left(y+5\right)}=\frac{x+5-x-3}{y+7-y-5}=\frac{2}{2}=1\)
=>x+3=y+5
=>x-y=5-3
=>x-y=2
vậy x-y=2
thắng à chép cốc cốc toán à em
\(\Leftrightarrow\frac{7}{x-1}=\frac{x}{9}+\frac{1}{9}\)
\(\Rightarrow-\frac{x}{9}+\frac{7}{x-1}-\frac{1}{9}=0\)
\(\Leftrightarrow-\frac{x^2-64}{9\left(x-1\right)}=0\)
=>x2-64=0
=>x2=64
=>x=±8