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a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
3x^3-5x^2+9x-15 3x-5 x^2+3 3x^3-5x^2 9x-15 9x-15 0
Vậy \(3x^2-5x^2+9x-15=\left(3x-5\right)\left(x^2+3\right)\)
b
\(\left(x+1\right)\left(x-2\right)-x\left(x-3\right)=0\)
\(\Leftrightarrow x^2-2x+x-2-x^2+3x=0\)
\(\Leftrightarrow2x-2=0\)
\(\Leftrightarrow x=1\)
b
\(x^2+4x+3=0\)
\(\Leftrightarrow\left(x^2+4x+4\right)-1=0\)
\(\Leftrightarrow\left(x+2\right)^2-1=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=-1;x=-3\)
a) \(4x^4-101x^2+25=0\)
\(\Leftrightarrow4x^4-100x^2-x^2+25=0\)
\(\Leftrightarrow4x^2\left(x^2-25\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x-5\right)\left(x+5\right)=0\)
Từ đây cậu suy ra đc tập nghiệm của ptr là : \(S=\left\{\frac{1}{2};-\frac{1}{2};5;-5\right\}\)
b) Tớ chịu :>
c) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left[\left(x^2-1\right)^2-\left(x^4+x^2+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4-2x^2+1-x^4-x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(-3x^2\right)=0\)
Từ đây thấy rằng tập nghiệm ptr là : \(S=\left\{1;-1;0\right\}\)
Chúc cậu học tốt !
a) (3x+2)(2x+9) - (x+2)(6x+1) = (x+1) - (x-6)
<=> 6x2 + 27x + 4x + 18 - 6x2 - x - 12x - 2 = x+1 - x+6
<=> 18x + 16 = 7
<=> 18x = -9
<=> x = \(-\dfrac{1}{2}\)
b) 3(2x-1)(3x-1) - (2x-3)(9x-1) = 0
<=> 3.(6x2-2x-3x+1) - (18x2-2x-27x+3) = 0
<=> 3.(6x2-5x+1) - 18x2+29x-3 = 0
<=> 18x2-15x+3 - 18x2+29x - 3 = 0
<=> 14x = 0
<=> x = 0
a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0
TH2 3x-4=0 <=>x=4/3
KL:.....
b) (x+3)(x−1)+2x(x+3)=0.
<=> (x+3)(x-1+2x)=0
TH1: x+3=0 <=> x=-3
TH2 x-1=0 <=> x=1
KL:.....
c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 \)
KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)
KL:....
a) \(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
d) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
câu 1
a)\(ĐKXĐ:x^3-8\ne0=>x\ne2\)
b)\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2-2x+4\right)}{\left(x-2\right)\left(x^2-2x+4\right)}=\frac{3}{x-2}\left(#\right)\)
Thay \(x=\frac{4001}{2000}\)zô \(\left(#\right)\)ta được
\(\frac{3}{\frac{4001}{2000}-2}=\frac{3}{\frac{4001}{2000}-\frac{4000}{2000}}=\frac{3}{\frac{1}{2000}}=6000\)
a) 3�(�−1)−1+�=03x(x−1)−1+x=0
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b) �2−9�=0x2
−9x=0
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a) 3�(�−1)−1+�=03x(x−1)−1+x=0
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b) �2−9�=0x2
−9x=0
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a) 3�(�−1)−1+�=03x(x−1)−1+x=0
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b) �2−9�=0x2
−9x=0
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a) 3�(�−1)−1+�=03x(x−1)−1+x=0
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b) �2−9�=0x2
−9x=0
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a) 3�(�−1)−1+�=03x(x−1)−1+x=0
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b) �2−9�=0x2
−9x=0
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a) 3�(�−1)−1+�=03x(x−1)−1+x=0
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b) �2−9�=0x2
−9x=0
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a) 3�(�−1)−1+�=03x(x−1)−1+x=0
=(3x+1)(x-1)
⇒TH1:3x+1=0⇒x=-\(\dfrac{1}{3}\)
⇒TH2:x-1=0⇒x=1
Vậy xϵ{-\(\dfrac{1}{3}\);1}
b) �2−9�=0x2
−9x=0
x(x-9)=0
⇒TH1:x=0
⇒TH2:x-9=0⇒x=9
Vậy xϵ{0;9}
a) 3�(�−1)−1+�=03x(x−1)−1+x=0
3�(�−1)+(�−1)=03x(x−1)+(x−1)=0
(3�+1)(�−1)=0(3x+1)(x−1)=0
Suy ra 3�+1=03x+1=0 hoặc �−1=0x−1=0
Vậy �=−13x=−31 hoặc �=1x=1
b) �2−9�=0x2−9x=0
�(�−9)=0x(x−9)=0
Suy ra �=0x=0 hoặc �=9x=9.
a) 3�(�−1)−1+�=03x(x−1)−1+x=0
3�(�−1)+(�−1)=03x(x−1)+(x−1)=0
(3�+1)(�−1)=0(3x+1)(x−1)=0
Suy ra 3�+1=03x+1=0 hoặc �−1=0x−1=0
Vậy �=−13x=−31 hoặc �=1x=1
b) �2−9�=0x2−9x=0
�(�−9)=0x(x−9)=0
Suy ra �=0x=0 hoặc �=9x=9.
a) 3�(�−1)−1+�=03x(x−1)−1+x=0
3�(�−1)+(�−1)=03x(x−1)+(x−1)=0
(3�+1)(�−1)=0(3x+1)(x−1)=0
Suy ra 3�+1=03x+1=0 hoặc �−1=0x−1=0
Vậy x =1/3 hoặc x =1
b)x^2−9x=0
�(�−9)=0x(x−9)=0
Suy ra �=0x=0 hoặc �=9x=9.
a)3x(x-1)-1+x=0
(3x+1)(x−1)=0
TH1:3x+1=0
x=-1/3
TH2:x-1=0
x=1
Vậy x thuộc {-1/3;1}
b)x^2-9x=0
x(x-9)=0
TH1:x=0
TH2:x-9=0
x=9
Vậy x thuộc {0;9}
a) 3x(x−1)−1+x=0 = -0,3
b) �2−9�=0x\(^2\) −9x=0 = 0
a) 3x(x - 1) - (1+x) = 0
⇔3x(x - 1) - (1+x) = 0
⇔ (x -1)(3x - 1) = 0⇔
⇔⇔(x – 1)(3x + 1) = 0
⇔[x−1=03x+1=0⇔[x=1x=\(\dfrac{-1}{3}\)
b)x2-9x=0
⇔x(x-9)=0
⇔x=0 hoặc x-9=0