Tìm x biết :
\(\left|3x+4\right|=2\left|2x-9\right|\)
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9 tháng 11 2017
Ta có: \(\left(8x^2-2x+7\right)\left(4x-6x^2-3\right)=\left(6x^2+3x+4\right)\left(9x-8x^2-6\right)\)
\(\Rightarrow\left(8x^2-2x+7\right)\left(4x-6x^2-3\right)-\left(6x^2+3x+4\right)\left(9x-8x^2-6\right)=0\)
\(\Rightarrow14x^3-33x^2+16x+3=0\) (Rút gọn vế đầu)
\(\Rightarrow14x^2\left(x-1\right)-19x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Rightarrow\left(14x^2-19x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\left[7x\left(2x-3\right)+\left(2x-3\right)\right]\left(x-1\right)=0\)
\(\Rightarrow\left(7x+1\right)\left(2x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{7}\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\).
Vậy \(x\in\left\{-\dfrac{1}{7};1;\dfrac{3}{2}\right\}.\)
26 tháng 6 2022
\(D=\dfrac{9x^8y^6\cdot\dfrac{1}{6}x^2y+\left(-16\right)}{15x^2y^2\cdot0.4\cdot ax^2y^2z^2}=\dfrac{\dfrac{3}{2}x^{10}y^7-16}{6ax^4y^4z^2}\)
KK
5 tháng 9 2021
a. (x - 22) - 1 = 0
<=> x - 4 - 1 = 0
<=> x = 5
b. 4 - (x - 2)2 = 0
<=> 22 - (x - 2)2 = 0
<=> (2 - x + 2)(2 + x - 2) = 0
<=> x(4 - x) = 0
<=> \(\left[{}\begin{matrix}x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
KK
5 tháng 9 2021
d. (3x - 2)2 - (2x + 3)2 = 5(x + 4)(x - 4)
<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x2 - 16)
<=> (x - 5)(5x + 1) = 5x2 - 80
<=> 5x2 + x - 25x - 5 = 5x2 - 80
<=> 5x2 - 5x2 + x - 25x = -80 + 5
<=> -24x = -75
<=> x = \(\dfrac{25}{8}\)
15 tháng 7 2021
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)