Tìm x:
a)(3x+5).(7-2x)+6x.(x+4)=0
b)x3-25x=0
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C
1
C
3
7 tháng 8 2021
a) Ta có: \(36x^3-4x=0\)
\(\Leftrightarrow4x\left(9x^2-1\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)
b) Ta có: \(3x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
C
1
C
1
VD
2
27 tháng 10 2021
a: \(9x^2-30x+25=0\)
\(\Leftrightarrow3x-5=0\)
hay \(x=\dfrac{5}{3}\)
c: \(9x^2-25=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
27 tháng 10 2021
a) \(9x^2-30x+25=0\Rightarrow\left(3x-5\right)^2=0\Rightarrow x=\dfrac{5}{3}\)
b) \(25x^2-5x+\dfrac{1}{4}=0\Rightarrow\left(10x-1\right)^2=0\Rightarrow x=\dfrac{1}{10}\)
c) \(9x^2-25=0\Rightarrow\left(3x-5\right)\left(3x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
d) \(\left(2x-1\right)^2-\left(3x+2\right)^2=0\)
\(\Rightarrow\left(2x-1+3x+2\right)\left(2x-1-3x-2\right)=0\)
\(\Rightarrow-\left(5x+1\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
C
1
7 tháng 9 2021
a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)
\(\Leftrightarrow x^3-x^3-1=x\)
hay x=-1
c: Ta có: \(56x^4+7x=0\)
\(\Leftrightarrow7x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: Ta có: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
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.
S
1
25 tháng 12 2021
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
TP
3
4 tháng 9 2021
a: Ta có: \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)
\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)
\(\Leftrightarrow-13x=13\)
hay x=-1
c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)
\(\Leftrightarrow3x=12\)
hay x=4
4 tháng 9 2021
a) 2x3-18x=0
⇔ 2x(x2-9)=0
⇔ 2x(x-3)(x+3)=0
⇔ \(\left\{{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b)(3x-1)(2x+1)-6x(x+2)=11
⇔ 6x2+x-1-6x2-12x=11
⇔ -11x=12
\(\Leftrightarrow x=-\dfrac{12}{11}\)
c) (x-1)3-(x+2).(x2-2x+4)=3.(1-x2)
⇔ x3-3x2+3x-1-x3-8-3+3x2=0
⇔ 3x=12
⇔ x=4
a) \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)
\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)
\(\Leftrightarrow35x=-35\Leftrightarrow x=-1\)
b) \(x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
a: Ta có: \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)
\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)
\(\Leftrightarrow x=1\)
b: Ta có: \(x^3-25x=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)