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14 tháng 6 2016

a)\(\left(x^2-x+1\right).\left(x+1\right)-x^3+3x=15\)

\(x^3+x^2-x^2-x+x+1-x^3+3x=15\)

\(1+3x=15\)

\(3x=15-1\)

\(3x=14\)

\(x=\frac{14}{3}\)
 

 

14 tháng 6 2016

b) \(\left(x+3\right).\left(x-2\right)+3x=4\left(x+\frac{3}{4}\right)\)

\(x^2-2x+3x-6+3x=4x+3\)

\(x^2-2x+3x+3x-4x=6+3\)

\(x^2=9\)

\(x^2=3^2\) hoặc \(x^2=\left(-3\right)^2\)

 vậy x=3 hoặc x=-3

 

14 tháng 6 2016

Chúng ta sẽ sử dụng hằng đẳng thức em nhé :)

a. \(x^3+1-x^3+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)

b. \(x^2+x-6+3x=4x+3\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

c. \(x^3+2x^2-5x-10+5x=2x^2+17\Leftrightarrow x^3=27\Leftrightarrow x=3\)

14 tháng 6 2016

a,

(x2-x+1)(x+1)-x3+3x=15

x3-x2+x+x2-x+1-x3+3x=15

x3-x3-x2+x2+x-x+3x+1=15

3x+1=15

3x=15-1

3x=14

x=14/3

b,

(x+3)(x-2)+3x=\(\frac{4}{x+\frac{3}{4}}\)

x2-2x+3x-6+3x=\(\frac{4}{x+\frac{3}{4}}\)

x2-2x+3x+3x-6=\(\frac{4}{x+\frac{3}{4}}\)

Tới đây hết biết , đề có gì sai sai sao ý !

c,

(x2-5)(x+2)+5x=2x2+17

x3+2x2-5x-10+5x=2x2+17

x3+2x2-5x+5x-10=2x2+17

x3+2x2-10=2x2+17

x3-10=17

x3=17+10

x3=27

\(\Rightarrow x=3\)(Vì : 33=27)

_k_ nhé bn

14 tháng 6 2016

Nhân ra thôi bạn, có hằng đẳng thức gì đâu !

a) \(\left(x^2-x+1\right)\left(x+1\right)-x^3+3x=15\)

\(\Leftrightarrow\left(x^2-x+1\right)\cdot x+x^2-x+1-x^3+3x=15\)

\(\Leftrightarrow x^3-x^2+x+x^2-x+1-x^3+3x=15\)

\(\Leftrightarrow1+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)

b) \(\left(x+3\right)\left(x-2\right)+3x=4\cdot\left(x+\frac{3}{4}\right)\)

\(\Leftrightarrow x^2+3x-2x-6+3x=4x+3\)

\(\Leftrightarrow x^2+4x-6=4x+3\)

\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)

c) \(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)

\(\Leftrightarrow x^3-5x+2x^2-10+5x=2x^2+17\)

\(\Leftrightarrow x^3=27\Leftrightarrow x=3\)

a) Ta có: \(8x\left(2x-3\right)-4x\left(4x+3\right)=72\)

\(\Leftrightarrow16x^2-24x-16x^2-12x=72\)

\(\Leftrightarrow-36x=72\)

hay x=-2

b) Ta có: \(\left(x+2\right)\left(x+4\right)-x\left(x+2\right)=104\)

\(\Leftrightarrow x^2+6x+8-x^2-2x=104\)

\(\Leftrightarrow4x=96\)

hay x=24

c) Ta có: \(\left(x-1\right)\left(x+4\right)-x\left(x-1\right)=308\)

\(\Leftrightarrow x^2+3x-4-x^2+x=308\)

\(\Leftrightarrow4x=312\)

hay x=78

d) Ta có: \(15x\left(2x-3\right)-\left(5x+2\right)\left(6x-5\right)=-22\)

\(\Leftrightarrow30x^2-45x-30x^2+25x-12x+10=-22\)

\(\Leftrightarrow-32x=-32\)

hay x=1

a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)

=>3(3x+2)-4(3x+1)=10

=>9x+6-12x-4=10

=>-3x+2=10

=>-3x=8

=>x=-8/3

b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)

=>(x-1)(x-2)-x(x+2)=-9x+10

=>x^2-3x+2-x^2-2x=-9x+10

=>-5x+2=-9x+10

=>x=2(loại)

26 tháng 8 2023

a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)

\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)

\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)

\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)

\(x=\dfrac{-1}{3}\)

b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)

\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)

\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)

\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)

\(2x=\dfrac{-17}{24}\)

\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)

\(x=\dfrac{-17}{48}\)

c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)

26 tháng 8 2023

a, 1/4 + 3/4 : x = -2

     3/4 : x = -2 - 1/4 

     3/4 : x = -9/4

             x = 3/4 : -9/4

             x = -1/3

a: =>6x-3x^2-5=4-3x^2-2

=>6x-5=2

=>6x=7

=>x=7/6

b: =>20x+5-12x^2-3x=6x^2-10x+3x-5

=>-12x^2+17x+5-6x^2+7x+5=0

=>-18x^2+24x+10=0

=>x=5/3 hoặc x=-1/3