Tìm x biết x3-16x=0
Tìm giá trị của x, biết:
a. x3 - 16x = 0 b. (2x + 1)2 - (x - 1)2 = 0
a) \(x^3-16x=0\)
⇔\(x\left(x^2-16\right)=0\)
⇒\(x=0\) hoặc \(x^2-16=0\)
\(TH_1:x=0\)
\(TH_2:x^2-16=0\) ⇔ \(x^2=16\) ⇔ \(x=\pm4\)
Vậy \(x\in\left\{0;\pm4\right\}\)
b) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)
⇒ \(2x+1=x-1\)
⇒ \(2x+2=x\)
⇒ \(2\left(x+1\right)=x\) ⇒ x = -2
Vậy x = -2
Tìm x:
a) x2+9x=0
b) (x+4)2-16=0
c) x3-16x=0
d) x2-10x+25=0
\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
a) \(\Leftrightarrow x\left(x+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)
b) \(\Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
c) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) \(\Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)
Tìm x
a, 3/4x*(x2-9)=0
b, x3-16x=0
c, (x-1)(x+2)-x-2=0
d, 3x3-27x=0
e, x2(x+1)+2x(x+1)=0
f, x(2x-3)-2(3-2x)=0
c: =>(x-1)(x+1)=0
hay \(x\in\left\{1;-1\right\}\)
a,
\(=\dfrac{3}{4x}.\left(x-3\right)\left(x+3\right)\)=0
\(\left\{{}\begin{matrix}\dfrac{3}{4x}=0\\x-3=0\\x+3=0\end{matrix}\right.\)
=>\(x=\left\{3,-3\right\}\)
b,
\(x^3-16x=0\\x\left(x^2-16\right)\\ x\left(x-4\right)\left(x+4\right)\)
\(\left\{{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
=>\(x=\left\{-4,0,4\right\}\)
d,
\(3x^3-27x=0\\ 3x\left(x^2-9\right)=0\\ 3x\left(x-3\right)\left(x+3\right)=0\)
\(\left\{{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
=>\(x=\left\{-3,0,3\right\}\)
e,
\(x^2+\left(x+1\right)+2x\left(x+1\right)=0\\ x\left(x+1\right)\left(x+2\right)=0\)
\(\left\{{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)
=>\(x=\left\{-2,-1,0\right\}\)
f,
\(x\left(2x-3\right)-2\left(3-2x\right)=0\\ \left(2x-3\right)\left(x+2\right)=0\)
\(\left\{{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
Bài 1: Phân tích đa thức thành nhân tử: a) 4y3 + 16y2 + 16y b) 8x2-48x+6xy-36y c) 8x2-48x-6xy+36y d) a2 –2ab+b2 –4 e) 4–x2 –4xy–4y2 f) 8a2 –16a+8ax–16x g) 16–4x2 +8xy–4y2 h) –4x2 –16xy–16y2 Bài 2: Tìm x, biết: a) x3 – 6x2 + 9x = 0 b) 5x(x–6)+3x–18=0 c) 5x(x – 6) – 18 + 3x = 0 d) 5x(x – 6) – 3x + 18 = 0 e) (2x – 3)2 = (5 – x)2 f) (2x + 1)2 = (3x – 2)2 g) 16(2x–3)=-25x2 (3–2x)
b: \(8x^2-48x+6xy-36y\)
\(=8x\left(x-6\right)+6y\left(x-6\right)\)
\(=2\left(x-6\right)\left(4x+3y\right)\)
d: \(a^2-2ab+b^2-4\)
\(=\left(a-b\right)^2-4\)
\(=\left(a-b-2\right)\left(a-b+2\right)\)
1.Tìm x:
a)2x(x+1)-2x2=4
b)x3-16x=0
c)(3x+1)2-8x2+2x=-6
2.Tìm m để đa thức f(x)=x3+6x2+12x+m chia hết cho đa thức h(x)=x+2
Bài 2:
x^3+6x^2+12x+m chia hết cho x+2
=>x^3+2x^2+4x^2+8x+4x+8+m-8 chia hết cho x+2
=>m-8=0
=>m=8
Tìm x biết: 3-16x^2=0
`3-16x^2=0`
`<=>(\sqrt3)^2-(4x)^2=0`
`<=>(\sqrt3+4x)(\sqrt3-4x)=0`
`<=> [(\sqrt3=-4x),(\sqrt3=4x):}`
`<=> [(x=-\sqrt3/4),(x=\sqrt3/4):}`
Vậy `S={\pm \sqrt3/4}`.
Ta có: \(3-16x^2=0\)
\(\Leftrightarrow16x^2=3\)
\(\Leftrightarrow x^2=\dfrac{3}{16}\)
hay \(x\in\left\{\dfrac{\sqrt{3}}{4};-\dfrac{\sqrt{3}}{4}\right\}\)
Tìm giá trị lớn nhất của hàm số f (x) = x 3 - 8 x 2 + 16 x - 9 trên đoạn [1;3]
A. m a x [ 1 ; 3 ] f ( x ) = - 6
B. m a x [ 1 ; 3 ] f ( x ) = 13 27
C. m a x [ 1 ; 3 ] f ( x ) = 0
D. m a x [ 1 ; 3 ] f ( x ) = 5
phân tích đa thức thành nhân tử , tìm x
1, x^3-x=0
2,x3-16x=0
1) \(x^3-x=0\)
\(\Leftrightarrow x.\left(x^2-1\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy :.....
2) \(x^3-16x=0\)
\(\Leftrightarrow x.\left(x^2-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-16=0\Rightarrow x^2=16\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\\x=-8\end{matrix}\right.\)
Vậy :....
Tìm x biết:
a. x3 – 25x = 0 b. 3x(x- 2) – x + 2 = 0
c. x2 – 4x - 5 = 0 d.x3 – x2 + 3x – 3 = 0
e. x3 + 27 + ( x + 3)( x – 9) = 0
a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
Tìm x biết
a)16x<1284
b)5x.5x+1.5x+2< 100….0:218
18 số 0
a: =>2^4x<2^28
=>4x<28
=>x<7
b: =>5^3x+3<5
=>3x+3<1
=>3x<-2
=>x<-2/3
a) \(16^x< 128^4\)
= (24)x < (27)4
= 24x < 228
= 4x < 28
= x < 7
Vậy \(x=\left\{0;1;2;3;4;5;6;\right\}\)
\(#Tuyết\)