5.
\(2x-\dfrac{2}{3}+\dfrac{1}{2}x=-1\)
\(\left(2+\dfrac{1}{2}\right)x=-1+\dfrac{2}{3}\)
\(\dfrac{5}{2}x=-\dfrac{1}{3}\)
\(x=-\dfrac{1}{3}:\dfrac{5}{2}\)
\(x=-\dfrac{2}{15}\)
6.
\(\dfrac{31}{36}-\left(\dfrac{1}{3}-x\right)^2=\dfrac{5}{6}\)
\(\left(\dfrac{1}{3}-x\right)^2=\dfrac{31}{36}-\dfrac{5}{6}\)
\(\left(\dfrac{1}{3}-x\right)^2=\dfrac{1}{36}\)
\(\left(\dfrac{1}{3}-x\right)^2=\left(\dfrac{1}{6}\right)^2\)
\(\dfrac{1}{3}-x=\dfrac{1}{6}\) hoặc \(\dfrac{1}{3}-x=-\dfrac{1}{6}\)
\(x=\dfrac{1}{3}-\dfrac{1}{6}\) hoặc \(x=\dfrac{1}{3}+\dfrac{1}{6}\)
\(x=\dfrac{1}{6}\) hoặc \(x=\dfrac{1}{2}\)
1.
\(\dfrac{2}{3}:x+\dfrac{1}{3}=-7\)
\(\dfrac{2}{3}:x=-7-\dfrac{1}{3}\)
\(\dfrac{2}{3}:x=-\dfrac{22}{3}\)
\(x=\dfrac{2}{3}:\left(-\dfrac{22}{3}\right)\)
\(x=-\dfrac{1}{11}\)
2.
\(-\dfrac{2}{3}-3x=-2\)
\(3x=-\dfrac{2}{3}-\left(-2\right)\)
\(3x=\dfrac{4}{3}\)
\(x=\dfrac{4}{3}:3\)
\(x=\dfrac{4}{9}\)
3.
\(-\dfrac{2}{3}+2\left(x+\dfrac{1}{2}\right)=1\)
\(2\left(x+\dfrac{1}{2}\right)=1+\dfrac{2}{3}\)
\(2\left(x+\dfrac{1}{2}\right)=\dfrac{5}{3}\)
\(x+\dfrac{1}{2}=\dfrac{5}{3}:2\)
\(x+\dfrac{1}{2}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}-\dfrac{1}{2}\)
\(x=\dfrac{1}{3}\)
4.
\(\dfrac{1}{2}x-2=2x+1\)
\(\dfrac{1}{2}x-2x=1+2\)
\(\left(\dfrac{1}{2}-2\right)x=3\)
\(-\dfrac{3}{2}x=3\)
\(x=3:\left(-\dfrac{3}{2}\right)\)
\(x=-2\)
7.
\(\dfrac{2}{3}\left|2x+\dfrac{1}{2}\right|=5\)
\(\left|2x+\dfrac{1}{2}\right|=5:\dfrac{2}{3}\)
\(\left|2x+\dfrac{1}{2}\right|=\dfrac{15}{2}\)
\(2x+\dfrac{1}{2}=\dfrac{15}{2}\) hoặc \(2x+\dfrac{1}{2}=-\dfrac{15}{2}\)
\(2x=\dfrac{15}{2}-\dfrac{1}{2}\) hoặc \(2x=-\dfrac{15}{2}-\dfrac{1}{2}\)
\(2x=7\) hoặc \(2x=-8\)
\(x=\dfrac{7}{2}\) hoặc \(x=-4\)
8.
\(\dfrac{3}{4}-\left|x-\dfrac{3}{2}\right|=\dfrac{1}{6}\)
\(\left|x-\dfrac{3}{2}\right|=\dfrac{3}{4}-\dfrac{1}{6}\)
\(\left|x-\dfrac{3}{2}\right|=\dfrac{7}{12}\)
\(x-\dfrac{3}{2}=\dfrac{7}{12}\) hoặc \(x-\dfrac{3}{2}=-\dfrac{7}{12}\)
\(x=\dfrac{7}{12}+\dfrac{3}{2}\) hoặc \(x=-\dfrac{7}{12}+\dfrac{3}{2}\)
\(x=\dfrac{25}{12}\) hoặc \(x=\dfrac{11}{12}\)
9.
\(\left(x^2+4\right)\left(9x^2-1\right)=0\)
\(\left(x^2+4\right)\left(3x-1\right)\left(3x+1\right)=0\)
\(3x-1=0\) hoặc \(3x+1=0\) hoặc \(x^2+4=0\) (vô nghiệm)
\(x=\dfrac{1}{3}\) hoặc \(x=-\dfrac{1}{3}\)
10.
\(\left(\dfrac{1}{2}x-x+\dfrac{2}{3}\right)\left(2x^2-4\right)=0\)
\(\left(-\dfrac{1}{2}x+\dfrac{2}{3}\right)\left(2x^2-4\right)=0\)
\(-\dfrac{1}{2}x+\dfrac{2}{3}=0\) hoặc \(2x^2-4=0\)
\(-\dfrac{1}{2}x=-\dfrac{2}{3}\) hoặc \(x^2=2\)
\(x=\dfrac{4}{3}\) hoặc \(x=\sqrt{2}\) hoặc \(x=-\sqrt{2}\)
11.
\(\left(\left|x\right|-\dfrac{2}{3}\right)\left(2x^2-\dfrac{9}{2}\right)=0\)
\(\left|x\right|-\dfrac{2}{3}=0\) hoặc \(2x^2-\dfrac{9}{2}=0\)
\(\left|x\right|=\dfrac{2}{3}\) hoặc \(x^2=\dfrac{9}{4}\)
\(x=-\dfrac{2}{3}\) hoặc \(x=\dfrac{2}{3}\) hoặc \(x=\dfrac{3}{2}\) hoặc \(x=-\dfrac{3}{2}\)
12.
\(x-2\sqrt{x}=0\)
\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\sqrt{x}=0\) hoặc \(\sqrt{x}-2=0\)
\(x=0\) hoặc \(\sqrt{x}=2\)
\(x=0\) hoặc \(x=4\)
13.
\(\left(\sqrt{x}+3\right)\left(x^2-3\right)=0\)
\(\sqrt{x}+3=0\) hoặc \(x^2-3=0\)
\(\sqrt{x}=-3\) (vô nghiệm) hoặc \(x^2=3\)
\(x=\sqrt{3}\) hoặc \(x=-\sqrt{3}\) (loại)
Vậy \(x=\sqrt{3}\)
14.
\(\dfrac{3x-2}{5}=\dfrac{x+4}{2}\)
\(2\left(3x-2\right)=5\left(x+4\right)\)
\(6x-4=5x+20\)
\(6x-5x=20+4\)
\(x=24\)
15.
\(2^x+2^{x+3}=144\)
\(2^x+2^x.2^3=144\)
\(2^x\left(1+2^3\right)=144\)
\(2^x.9=144\)
\(2^x=144:9\)
\(2^x=16\)
\(2^x=2^4\)
\(x=4\)
16.
\(\left(\left|x-1\right|-2\right)\left(-\dfrac{1}{2}x-2\right)=0\)
\(\left|x-1\right|-2=0\) hoặc \(-\dfrac{1}{2}x-2=0\)
\(\left|x-1\right|=2\) hoặc \(-\dfrac{1}{2}x=2\)
\(x-1=2\) hoặc \(x-1=-2\) hoặc \(x=2:\left(-\dfrac{1}{2}\right)\)
\(x=3\) hoặc \(x=-1\) hoặc \(x=-4\)
17.
\(\left(\left|x\right|-1\right)\left(x^2+2\right)=0\)
\(\left|x\right|-1=0\) hoặc \(x^2+2=0\)
\(\left|x\right|=1\) hoặc \(x^2=-2\) (vô nghiệm)
\(x=1\) hoặc \(x=-1\)
18.
\(\left(8x^3+1\right)\left(2x^2-6\right)=0\)
\(8x^3+1=0\) hoặc \(2x^2-6=0\)
\(8x^3=-1\) hoặc \(2x^2=6\)
\(x^3=-\dfrac{1}{8}=\left(-\dfrac{1}{2}\right)^3\) hoặc \(x^2=3\)
\(x=-\dfrac{1}{2}\) hoặc \(x=\sqrt{3}\) hoặc \(x=-\sqrt{3}\)
19.
\(\left(3x+3\right)^2+\left(4x^2-4\right)^2=0\)
Do \(\left\{{}\begin{matrix}\left(3x+3\right)^2\ge0\\\left(4x^2-4\right)^2\ge0\end{matrix}\right.\) ;\(\forall x\)
nên \(\left(3x+3\right)^2+\left(4x^2-4\right)^2\ge0\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}3x+3=0\\4x^2-4=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\x^2=1\end{matrix}\right.\)
\(\Rightarrow x=-1\)
20.
\(x\left(x-3\right)+9=x^2+5x\)
\(x^2-3x+9=x^2+5x\)
\(x^2-3x-x^2-5x=-9\)
\(-8x=-9\)
\(x=\dfrac{9}{8}\)
