= x³ + 3³ -x(x² -1) -27 =0 ( tổng các lập phuong)

x =0 

CX100%

bạn chỉ cần phá hết hằng đẳng thức ra thôi 

câu 1

(x+3)^2+(x-2)(x+2)-2(x-1)^2=7

<=> x^2+6x+9+x^2-4-2x^2-4x-2-7=0

<=> x^2+x^2-2x^2+6x-4x+9-4-2-7=0

<=> 2x-4=0

<=> x=2

câu 2

(x+3)(x^2-3x+9)-x(x-1)(x+1)-27=0

<=> x^3+27-x(x^2-1)-27=0

<=> x^3-x^3+x+27-27=0

<=>x=0

a/ => 4x² - 4x + 1 + 4x² + 4x + 1 = 16

=> 8x² = 14

=> x² = 14/8

=> x = \(\frac{\sqrt{7}}{2}\) hoặc x = \(-\frac{\sqrt{7}}{2}\)

b/ => 6x² - (6x² - 11x - 10) = 17

=> 6x² - 6x² + 11x + 10 = 17

=> 11x = 7

=> x = 7/11

c/ => 2x(x + 5) - x² - 5x = 0

=> 2x(x + 5) - x(x + 5) = 0 

=> (x + 5)(2x - x) = 0

=> x(x + 5) = 0 

=> x = 0

hoặc x + 5 = 0 => x = -5

Vậy x = 0 ; x = -5

d/ \(x^2+\frac{1}{x^2}+2x+\frac{2}{x}=-3\)

đề là như vầy hả

smile, giúp mình lần này đi! đi nah! miao~~~~~

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\(a;\frac{2x+1}{6-x}=-\frac{2x+1}{x-6}=-\frac{2x-12+13}{x-6}=-2-\frac{13}{x-6}\)

\(\Rightarrow13⋮\left(x-6\right)\)

\(\Rightarrow x-6\in\left(1;-1;13;-13\right)\)

\(\Rightarrow x\in\left(7;5;19;-7\right)\)

\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2+x}+2x-1\right)\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{4x^2+x-\left(2x-1\right)^2}{\sqrt{4x^2+x}-2x+1}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{4x^2+x-4x^2+4x-1}{\sqrt{4x^2+x}-2x+1}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{5x-1}{-x\cdot\sqrt{4+\dfrac{1}{x}}-2x+1}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{5-\dfrac{1}{x}}{-\sqrt{4+\dfrac{1}{x}}-2+\dfrac{1}{x}}\)

\(=\dfrac{5-0}{-\sqrt{4+0}-2+0}=\dfrac{5}{-4}=-\dfrac{5}{4}\)

\(2^x+3\cdot2^x=9\cdot2^9\)

\(\Rightarrow2^x\cdot\left(1+3\right)=9\cdot2^9\)

\(\Rightarrow2^x\cdot4=9\cdot2^9\)

\(\Rightarrow2^x=\dfrac{9\cdot2^9}{4}\)

\(\Rightarrow2^x=9\cdot2^7\)

Xem lại đề ! 

Ta có:

\(\frac{2}{3}-\frac{1}{3}.\left(x-\frac{3}{2}\right)-\frac{1}{2}.\left(2x+1\right)=5\)

\(\Rightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{3}.\frac{3}{2}-x-\frac{1}{2}=5\)

\(\Rightarrow\frac{2}{3}-\left(\frac{1}{3}x+x\right)+\frac{1}{2}-\frac{1}{2}=5\)

\(\Rightarrow\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)

\(\Rightarrow x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{3}.\frac{3}{4}=-\frac{13}{4}\)

Vậy x=-13/4

a: \(\left[\left(10-x\right)\cdot2+51\right]:3-2=3\)

=>\(\left[2\left(10-x\right)+51\right]:3=5\)

=>\(\left[2\left(10-x\right)+51\right]=15\)

=>\(2\left(10-x\right)=15-51=-36\)

=>10-x=-36/2=-18

=>\(x=10-\left(-18\right)=10+18=28\)

b: \(\left(x-12\right)-15=20-\left(17+x\right)\)

=>\(x-12-15=20-17-x\)

=>\(x-27=3-x\)

=>\(2x=30\)

=>\(x=\dfrac{30}{2}=15\)

c: \(720-\left[41-\left(2x-5\right)\right]=2^3\cdot5\)

=>\(720-\left[41-2x+5\right]=8\cdot5=40\)

=>\(\left[41-2x+5\right]=720-40=680\)

=>-2x+46=680

=>-2x=680-46=634

=>\(x=\dfrac{634}{-2}=-317\)

\(x+\left(2x+1\right).2=137\)

         \(x+4x+2=137\)

                 \(5x+2=137\)

                         \(5x=137-2\)

                         \(5x=135\)

                            \(x=27\)

\(x+\left(2x+1\right)\cdot2=137\)

\(x+4x+2=137\)

\(5x+2=137\)

\(5x=135\)

\(x=135:5\)

\(x=27\)

VẬY \(x=27\)

\(x+\left(2x+1\right)\cdot2=137\)

\(x+4x+2=137\)

\(5x+2=137\)

\(5x=135\)

\(x=135:5\)

\(x=27\)

VẬY \(x=27\)

2(x-1)=6(10-x)

<=>2x-2=60-6x

<=>2x+6x=60+2

<=>8x=62

<=>x=62:8

<=>x=7,75

=> 2x - 2 = 60 - 6x

=> 2x + 6x = 2 + 60

=> 8x = 62

=> x = 7,75