Tìm X biết : x3+1=2\(\sqrt[3]{2x-1}\)
Tìm x, biết:
a) 2-x = 2 ( x - 2 ) 3 ; b) 8 x 3 - 72x = 0;
c) ( x - 1 , 5 ) 6 + 2 ( 1 , 5 - x ) 2 = 0; d) 2 x 3 +3 x 2 +3 + 2x = 0;
e) x 3 - 4x- 14x(x - 2) = 0; g) x 2 (x + 1)- x(x + 1) + x(x - 1) = 0.
tìm x biết a)x(2x-1)-(x-1)2-x2=0
b)(x+2)3-x3-6x2=4
a,x(2x-1)-(x-1)^2-x^2=0
<=>x(2x-1-x)-(x-1)^2=0
<=>x(x-1)-(x-1)^2=0
<=>(x-x+1)(x-1)=0
<=>x-1=0
<=>x=1
b,(x+2)^3-x^3-6x^2=4
<=>x^3+6x^2+12x+8-x^3-6x^2=4
<=>12x+8=4
<=>x=-1/3
tick mik nha
`a)x(2x-1)-(x-1)^2-x^2=0`
`<=>2x^2-x-x^2+2x-1-x^2=0`
`<=>x-1=0`
`<=>x=1`
Vậy `x=1.`
`b)(x+2)^3-x^3-6x^2=4`
`<=>x^3+6x^2+12x+8-x^3-6x^2=4`
`<=>12x+8=4`
`<=>12x=-4`
`<=>x=-1/3`
Vậy `x=-1/3.`
a: Ta có: \(x\left(2x-1\right)-\left(x-1\right)^2-x^2=0\)
\(\Leftrightarrow2x^2-x-x^2+2x-1-x^2=0\)
\(\Leftrightarrow x=1\)
b: Ta có: \(\left(x+2\right)^3-x^3-6x^2=4\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=4\)
\(\Leftrightarrow12x=-4\)
hay \(x=-\dfrac{1}{3}\)
tìm x biết a) ( x + 3 )2 - ( 2x + 1 ).( x+3 ) = 0 ; b) x3 - 12x2 + 36x = 0
\(a,\Leftrightarrow\left(x+3\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-12x+36\right)=0\\ \Leftrightarrow x\left(x-6\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
a, (x+3)2 - ( 2x + 1 ).( x+3)=0 b, x3-12x2+36x =0
=> (x+3).(x+3-2x-1) => x(x2-12x+36) = 0
=>(x+3).(-x+2) => x(x-6)2 = 0
=> x+3=0 <=> x=-3 => x=0 <=> x=0
-x+2=0 <=> x=-2 x-6= 0 <=> x=6
Tìm x, biết:
a) 2(5x-8)-3(4x-5) = 4(3x-4) + 11;
b) 2 x ( 6 x - 2 x 2 ) + 3 x 2 ( x - 4 ) = 8;
c) 2 ( x 3 - 1 ) - 2 x 2 ( x + 2 x 4 ) + ( 4 x 5 + 4 ) x = 6;
d)(2x)2(4x-2)-(x3 -8x2) = 15.
a) x = 2 7 b) x = 2.
c) x = 2 d) x = 1.
Tìm các giới hạn
1. lim ( x đến +--∞) (x3 +3x2+2)
2. lim (x đến -∞) (\(\sqrt{4x^2-x+5}\))
3. lim ( x đến +- ∞) (\(\sqrt{x^2-2x-1}-\sqrt{x^2-7x+3}\))
1.
\(\lim\limits_{x\to +\infty}(x^3+3x^2+2)=+\infty\)
2.
\(\lim\limits_{x\to -\infty}\sqrt{4x^2-x+5}=\lim\limits_{x\to -\infty}-x.\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}=+\infty\) do $-x\to +\infty$ và $\lim\limits_{x\to -\infty}\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}=4>0$
3.
\(\lim\limits_{x\to +\infty}(\sqrt{x^2-2x-1}-\sqrt{x^2-7x+3})=\lim\limits_{x\to +\infty}\frac{x^2-2x-1-(x^2-7x+3)}{\sqrt{x^2-2x-1}+\sqrt{x^2-7x+3}}\)
\(=\lim\limits_{x\to +\infty}\frac{5x-4}{\sqrt{x^2-2x-1}+\sqrt{x^2-7x+3}}=\lim\limits_{x\to +\infty}\frac{5-\frac{4}{x}}{\sqrt{1-\frac{2}{x}-\frac{1}{x^2}}+\sqrt{1-\frac{7}{x}+\frac{3}{x^2}}}\)
\(=\frac{5}{1+1}=\frac{5}{2}\)
Tìm x, biết:
a) 3x(x - 1) + x - 1 = 0;
b) (x - 2)( x 2 + 2x + 7) + 2( x 2 - 4) - 5(x - 2) = 0;
c) ( 2 x - 1 ) 2 - 25 = 0;
d) x 3 + 27 + (x + 3)(x - 9) = 0.
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
Bài 8. Tìm x ϵ N biết:
a) x3=27
b) (2x-1)3=8
c) (x-2)2=16
d) (2x – 3)2=9
e) 2x+5=34:32
f) (3x-24).73=2.74
Bài 8. Tìm x ϵ N biết:
a) x3=27
b) (2x-1)3=8
c) (x-2)2=16
d) (2x – 3)2=9
e) 2x+5=34:32
f) (3x-24).73=2.74
Cho đa thức P ( x ) = x 3 - 4 x 2 + 3 - 2 x 3 + x 2 + 10 x - 1
Tìm đa thức Q(x) biết P ( x ) + Q ( x ) = x 3 + x 2 + 2 x - 1
A. - 4 x 2 - 8 x - 3
B. 2 x 3 - 4 x 2 + 8 x - 3
C. 2 x 3 + 4 x 2 - 8 x - 3
D. 4 x 2 - 8 x - 3
Chọn C
Ta có: P(x) + Q(x) = x3+ x2+ 2x-1
⇒ Q(x) = (x3 + x2 + 2x-1) - P(x)
= 2x3 + 4x2 - 8x - 3.
tìm x biết:
a)x2 + 3x = 0 b) x3 – 4x = 0
c) 5x(x-1) = x-1 d) 2(x+5) - x2-5x = 0
e) 2x(x-5)-x(3+2x)=26 f) 5x.(x – 2012) – x + 2012 = 0
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)