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Ta có (\(^{x^{2^{ }}^{ }+3x}\)) (\(^{x^{2^{ }}+3x+4}\))
Đặt \(x^{2^{ }^{ }}+3x\) là a ta có
a.(a+4)=-4
4a+\(a^2\) -4=0
\(^{ }\left(a-2\right)^2\)=0
Suy ra a=2
hay \(x^{2^{ }^{ }^{ }}+3x=2\)
\(x^2+3x-2=0\)
𝑥=−3±17√/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(x\notin\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)
Ta có: \(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
\(\Leftrightarrow\dfrac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}\)
Suy ra: \(9x^2-6x+1-9x^2-6x-1=12\)
\(\Leftrightarrow-12x=12\)
hay x=-1(thỏa ĐK)
Vậy: S={-1}
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x-1\right)^2=3x\left(2x-1\right)\)
\(\left(2x-1\right)^2-3x\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(2x-1-3x\right)=0\)
\(\left(2x-1\right)\left(-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(a,\left(2x-1\right)^2=3x\left(2x-1\right)\\ \Leftrightarrow4x^2-4x+1=6x^2-3x\\ \Leftrightarrow6x^2-3x-4x^2+4x-1=0\\ \Leftrightarrow2x^2+x-1=0\\ \Leftrightarrow2x^2+2x-x-1=0\\ \Leftrightarrow2x\left(x+1\right)-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(b,ĐKXĐ:x\ne0,2\\ \dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\\ \Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2+2x-x+2-2}{x\left(x-2\right)}=0\\ \Rightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(4x-16=3x\left(x-4\right)\)
\(4\left(x-4\right)=3x\left(x-4\right)\)
\(3x\left(x-4\right)-4\left(x-4\right)=0\)
\(\left(x-4\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\left(đk:x\ne0,2\right)\)
\(\dfrac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
\(x^2+2x-x+2=2\)
\(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)}{6\left(x-2\right)}-\dfrac{3\left(x-2\right)}{6\left(x-2\right)}=\dfrac{3\left(2x-3\right)}{6\left(x-2\right)}\)
Suy ra: \(2x+5-3x+6=6x-9\)
\(\Leftrightarrow-x+11-6x+9=0\)
\(\Leftrightarrow20-7x=0\)
\(\Leftrightarrow7x=20\)
hay \(x=\dfrac{20}{7}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{20}{7}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(3x+1\right)^2-\left(2x-5\right)^2=0\\ \Leftrightarrow\left(3x+1+2x-5\right)\left(3x+1-2x+5\right)=0\\ \Leftrightarrow\left(5x-4\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-6\end{matrix}\right.\\ b,\left(x+3\right)\left(4-3x\right)=x^2+6x+9\\ \Leftrightarrow\left(x+3\right)\left(4-3x\right)-\left(x+3\right)^2=0\\ \Leftrightarrow\left(x+3\right)\left(4-3x-x-3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(1-4x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow3x+2\left(x+2\right)=5\left(x-1\right)\)
=>3x+2x+4=5x-5
=>4=-5(vô lý)
b: \(\Leftrightarrow\dfrac{2}{x\left(x+4\right)}-\dfrac{3x}{x+4}=-3\)
\(\Leftrightarrow2-3x^2=-3x\left(x+4\right)\)
\(\Leftrightarrow2-3x^2+3x^2+12x=0\)
=>12x+2=0
hay x=-1/6
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (=) x2-2x+4=4 b) (=) 3x-2=0 hoặc 4x+5=0 (=) x2-2x=0 (=) 3x=2 hoặc 4x=5 (=) x(x-2)=0 (=) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{5}{4}\) (=) x=0 hoặc x-2=0 (=) x=0 hoặc x=2
a) |x – 2| = |3x| ⇔ x – 2 = 3x hoặc x – 2 = –3x
⇔ 2x = –2 hoặc 4x = 2 ⇔ x = –1 hoặc x = 1/2
Tập nghiệm: S = {-1;1/2}
b) Điều kiện: 3x ≥ 0 ⇔ x ≥ 0. Khi đó:
|x – 2| = 3x
⇔ x – 2 = 3x hoặc x – 2 = –3x
⇔ 2x = –2 hoặc 4x = 2
⇔ x = –1 hoặc x = 1/2
Vì x ≥ 0, nên ta lấy x = 1/2.
Tập nghiệm: S = 1/2.