Tìm x:
a, 5.x - 2.3^2 = 7.2021^0
b,(3x - 1)^3 =125
c,x=ƯC (18;54)
d,84x chia hết cho 3 và chia cho 5 dư 3
Tìm x:
a)(x+2)^2-2(x+2)(x-5)=0
b)2x^2+3x-5=0
c)x+2√2x^2+2x^3=0
d)(3x-1)^2-4(x+5)^2=0
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
Tìm x:
a)(x+3)2-4x-12=0
b)x(x+5)(x-5)-(x-3)(x2+3x+9)=7
a) (x + 3)^2 - 4x - 12 = 0
<=> (x + 3)^2 - 4(x + 3) = 0
<=> (x + 3)(x - 1) = 0
<=> x = -3 hoặc x = 1
b) x(x + 5)(x- 5) - (x - 3)(x^2 + 3x + 9) = 7
<=> x^3 - 25x - x^3 + 27 = 7
<=> -25x + 27 = 7
<=> x = 4/5
a/ \(\left(x+3\right)^2-4x-12=0\)
\(\left(x+3\right)^2-4\left(x+3\right)=0\)
\(\left(x+3\right)\left(x+3-4\right)=0\)
\(\left[{}\begin{matrix}x+3=0\Rightarrow x=-3\\x+3-4=0\Rightarrow x=1\end{matrix}\right.\)
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b/ \(x\left(x+5\right)\left(x-5\right)-\left(x-3\right)\left(x^2+3x+9\right)=7\)
\(x\left(x^2-25\right)-\left(x^3-27\right)=7\)
\(x^3-25x-x^3+27=7\)
\(-25x=-20\)
\(x=\dfrac{20}{25}=\dfrac{4}{5}\)
a, <=>x2 +6x+9-4x-12=0
<=> x2 +2x -3=0
<=> x2 +3x -x-3=0
<=> x.(x+3) - (x+3) =0
<=> (x-1)(x+3)=0
<=> x=1 hoặc x=-3
b, <=> x(x2 -25) - (x-3)(x+3)2 -7=0
<=> x3 -25x + (9-x2) (x+3) -7=0
<=> x3 -25x+ 9x+27-x3 -3x2 -7=0
<=> -3x2 -16x +20=0
<=>(3x-10)(x-2) =0 (đoạn này tự phân tích nha ^ ^)
<=> x= 10/3 hoặc x=2
Chúc bạn học tốt nha!
Tìm x:
a)2x3-18x=0
b)(3x-2).(2x+1)-6x.(x+2)=11
c)(x-1)3-(x+2).(x2-2x+4)=3.(1-x2)
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
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
c. (x - 1)3 - (x + 2)(x2 - 2x + 4) = 3(1 - x2)
<=> (x3 - 3x2 + 3x - 1) - (x3 - 2x2 + 4x + 2x2 - 4x + 8) = 3 - 3x2
<=> x3 - 3x2 + 3x - 1 - x3 + 2x2 - 4x - 2x2 + 4x - 8 = 3 - 3x2
<=> x3 - x3 - 3x2 + 2x2 - 2x2 + 3x2 + 3x - 4x + 4x = 3 + 1 + 8
<=> 3x = 12
<=> x = 4
Tìm x:
a) x4-25x3=0
b) (x-5)2-(3x-2)2=0
c) x3-4x2-9x+36=0
d) (-x3+3x2-4x) : (\(-\dfrac{1}{2}\)x)=0
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.
$(x-5)^2-(3x-2)^2=0$
$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$
$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix}
-2x-3=0\\
4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=\frac{-3}{2}\\
x=\frac{7}{4}\end{matrix}\right.\)
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.
Tìm x:
a, x^2-2x+2|x-1|-7=0
b, (x^2+3x+2)(x^2+7x+12)=24
gấp ạ!!!!!!!
a. \(x^2-2x+2\left|x-1\right|-7=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2x+2\left(x-1\right)-7=0\\x^2-2x-2\left(x-1\right)-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-9=0\\x^2-4x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=9\\\left(x-5\right)\left(x+1\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm3\\x=5\\x=-1\end{matrix}\right.\)
b: Ta có: \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
\(\Leftrightarrow\left(x^2+5x\right)^2+10\cdot\left(x^2+5x\right)=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Tìm x:
a) (3x-2)(2x-1)-(6x2-3x)=0
b) x3-(x+1)(x2-x+1)=x
c) 56x4+7x=0
d) x2-5x-24=0
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: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
Tìm x:
a)(3x+5).(7-2x)+6x.(x+4)=0
b)x3-25x=0
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: \(x^3-25x=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
a. (3x + 5)(7 - 2x) + 6x(x + 4) = 0
<=> 21x - 6x2 + 35 - 10x + 6x2 + 24x = 0
<=> -6x2 + 6x2 + 21x - 10x + 24x = -35
<=> 35x = -35
<=> x = \(\dfrac{-35}{35}=-1\)
b. x3 - 25x = 0
<=> x(x2 - 52)
<=> x(x + 5)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x=0\\x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\\x=5\end{matrix}\right.\)
Tìm x:
a. 4x2 - 20x + 25 = 0
b. (x - 5)(x + 5) - (x - 3)2 = 2(x - 7)
a. `4x^2-20x+25=0`
`<=>(2x)^2-2.2x.5 +5^2=0`
`<=>(2x-5)^2=0`
`<=>2x-5=0`
`<=>x=5/2`
b. `(x-5)(x+5)-(x-3)^2=2(x-7)`
`<=>x^2-25-x^2+6x-9=2x-14`
`<=>6x-34=2x-14`
`<=>4x=20`
`<=>x=5`
\(a,4x^2-20x+25=0\Leftrightarrow\left(2x\right)^2-2.2x.5+5^2=0\)
\(\Leftrightarrow\left(2x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
b, \(\left(x-5\right)\left(x+5\right)-\left(x-3\right)^2=2\left(x-7\right)\)
\(\Leftrightarrow x^2-25-x^2+6x-9=2x-14\Leftrightarrow4x=20\Leftrightarrow x=5\)
a) Có: (2x)2 - 2.2.5.x + 52 = 0
⇒ (2x - 5)2 = 0 ⇒ 2x - 5 = 0
⇒ 2x = 5 ⇒ x = \(\dfrac{5}{2}\)
b) Có: x2 - 25 - x2 + 6x - 9 = 2x - 14
⇒ 6x - 36 = 2x - 14
⇒ 4x = 22
⇒ x = \(\dfrac{11}{2}\)
Tìm x:
a) (x-3)(x2+3x+9)-x(x2-3)=0
b) 8x4+x=0
d) x3-6x2+8x=0
a: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x^2-3\right)=0\)
\(\Leftrightarrow x^3-27-x^3+3x=0\)
\(\Leftrightarrow x=9\)
b: Ta có: \(8x^4+x=0\)
\(\Leftrightarrow x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)\left(4x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)