Tìm x:
a) x(x-1)+x=4
b) 3x(x-5)-2x+10=0
c) 5x2-3x-2=0
d) x4-11x2+18=0
K
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Những câu hỏi liên quan
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14 tháng 8 2021
a:Ta có: \(x\left(x-1\right)+x=4\)
\(\Leftrightarrow x^2-x+x=4\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b: Ta có: \(3x\left(x-5\right)-2x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c: Ta có: \(5x^2-3x-2=0\)
\(\Leftrightarrow5x^2-5x+2x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: Ta có: \(x^4-11x^2+18=0\)
\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
14 tháng 8 2021
a) x(x-1)+x=4
⇔x2=4⇔\(x=\pm2\)
b)3x(x-5)-2x+10=0
⇔3x(x-5)-2(x-5)=0
⇔(x-5)(3x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
c)5x2-3x-2=0
⇔ 5x(x-1)+2(x-1)=0
⇔ (x-1)(5x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d)x4-11x2+18=0
⇔ x2(x2-2)-9(x2-2)=0
⇔ (x2-2)(x2-9)=0
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
a.
$x^4-25x^3=0$
$\Leftrightarrow x^3(x-25)=0$
\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)
b.
$(x-5)^2-(3x-2)^2=0$
$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$
$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix}
-2x-3=0\\
4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=\frac{-3}{2}\\
x=\frac{7}{4}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
c.
$x^3-4x^2-9x+36=0$
$\Leftrightarrow x^2(x-4)-9(x-4)=0$
$\Leftrightarrow (x-4)(x^2-9)=0$
$\Leftrightarrow (x-4)(x-3)(x+3)=0$
\(\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.\)
d. ĐK: $x\neq 0$
$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$
$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$
$\Leftrightarrow -2(-x^2+3x-4)=0$
$\Leftrightarrow x^2-3x+4=0$
$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)
Vậy pt vô nghiệm.
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25 tháng 12 2021
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
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3
31 tháng 7 2021
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
31 tháng 7 2021
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
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21 tháng 8 2021
a: Ta có: \(\left(2x-3\right)^2+6\left(2x-1\right)=7\)
\(\Leftrightarrow\left(2x-3\right)^2+6\left(2x-1\right)-7=0\)
\(\Leftrightarrow4x^2-12x+9+12x-6-7=0\)
\(\Leftrightarrow4x^2=4\)
\(\Leftrightarrow x^2=1\)
hay \(x\in\left\{1;-1\right\}\)
b: Ta có: \(x^2-7x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
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21 tháng 8 2021
a) \(\left(2x-3\right)^2+6\left(2x-1\right)=7\\ \Rightarrow4x^2-12x+9+12x-6-7=0\\ \Rightarrow4x^2-4=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
b) \(x^2-7x+10=0\\ \Rightarrow\left(x^2-2x\right)-\left(5x-10\right)=0\\ \Rightarrow\left(x-2\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
c) \(-6x^2+13x-5=0\\ \Rightarrow-\left(6x^2-13x+5\right)=0\\ \Rightarrow-\left[\left(6x^2-10x\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left[2x\left(3x-5\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left(2x-1\right)\left(3x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\left(2x-1\right)=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\)
a) \(x^2-x+x=4\)
\(x^2=4\)
\(x=\pm2\)
b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)
\(\left(x-5\right)\left(3x-2\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(a+b+c=5-3-2=0\)
\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)
d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :
\(t^2-11t+18=0\)
\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)
\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)
\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)