: Tìm x, biết: |75x+23|=|3x−1||75x+23|=|3x−1| ?
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Ta có 3x3 - 75x=0
x.3.x2-75.x=0
x(3x2-75)=0
=>hoặc x=0
hoặc 3x2 - 75=0 => 3x2 = 75 => x2 = 25 => hoặc x = 25
hoặc x = -25
Vậy x thuộc {0;25;-25}
Tick nha
![](https://rs.olm.vn/images/avt/0.png?1311)
*vn:vô nghiệm.
a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).
b. \(16x^2-8x+5=0\)
\(\Leftrightarrow16x^2-8x+1+4=0\)
\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)
-Vậy S=∅.
c. \(2x^3-x^2-8x+4=0\)
\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)
-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).
d. \(3x^3+6x^2-75x-150=0\)
\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)
-Vậy \(S=\left\{-2;\pm5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(3x+2\right)^2+4x-3x^2+2\left(5x-2\right)\left(5x+2\right)-75x^2\)
\(=9x^2+12x+4+4x-3x^2+50x^2-8-75x^2\)
\(=-19x^2+16x-4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
<=> 75x+23 = 3x-1 or 75x+23 = -(3x-1)
<=> x = -1/3 or x = -11/39