a, \(x\left(x^2+x+1\right)-x^2\left(x+1\right)=2x+5\)

\(\Rightarrow x^3+x^2+x-x^3-x^2-2x=5\)

\(\Rightarrow-x=5\Rightarrow x=-5\)

b, \(\left(x-3\right)\left(x-2\right)-\left(x+1\right)\left(x-5\right)=0\)

\(\Rightarrow x^2-2x-3x+6-\left(x^2-5x+x-5\right)=0\)

\(\Rightarrow x^2-5x+6-x^2+4x+5=0\)

\(\Rightarrow-x=-5-6\Rightarrow x=11\)

c, \(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Rightarrow x\left(2x^2+10x-x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\)

\(\Rightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Rightarrow-6x=3,5+4,5\Rightarrow-6x=8\Rightarrow x=-\dfrac{4}{3}\)

Chúc bạn học tốt!!!

a)\(\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow6x=36\Leftrightarrow x=6\)

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

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`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

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`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

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`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

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`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

a/ 2x - 10 - [3x - 14 - (4 - 5x) - 2x] = 2

=> 2x - 10 - (3x - 14 - 4 + 5x - 2x) = 2

=> 2x - 10 - 3x + 14 + 4 - 5x + 2x = 2

=> -4x + 6 = 0

=> -4x = -6

=> x = 3/2

b/ \(\left(\frac{1}{4}x-1\right)+\left(\frac{5}{6}x-2\right)-\left(\frac{3}{8}x+1\right)=4,5\)

\(\Rightarrow\frac{1}{4}x-1+\frac{5}{6}x-2-\frac{3}{8}x-1-\frac{9}{2}=0\)

\(\Rightarrow\frac{17}{24}x-\frac{17}{2}=0\)

\(\Rightarrow\frac{17}{24}x=\frac{17}{2}\)

\(\Rightarrow x=12\)

a) \(x-2=\left(x-2\right)^2\)

\(\left(x-2\right)^2-\left(x-2\right)=0\)

\(\left(x-2\right)\left(x-2-1\right)=0\)

\(\left(x-2\right)\left(x-3\right)=0\)

\(\Rightarrow x-2=0\) hoặc \(x-3=0\)

*) \(x-2=0\)

\(x=2\)

*) \(x-3=0\)

\(x=3\)

Vậy \(x=2;x=3\)

b) \(x+5=2\left(x+5\right)^2\)

\(2\left(x+5\right)^2-\left(x+5\right)=0\)

\(\left(x+5\right)\left[2\left(x+5\right)-1\right]=0\)

\(\left(x+5\right)\left(2x+10-1\right)=0\)

\(\left(x+5\right)\left(2x+9\right)=0\)

\(\Rightarrow x+5=0\) hoặc \(2x+9=0\)

*) \(x+5=0\)

\(x=-5\)

*) \(2x+9=0\)

\(2x=-9\)

\(x=-\dfrac{9}{2}\)

Vậy \(x=-5;x=-\dfrac{9}{2}\)

c) \(\left(x^2+1\right)\left(2x-1\right)+2x=1\)

\(\left(x^2+1\right)\left(2x-1\right)+2x-1=0\)

\(\left(x^2+1\right)\left(2x-1\right)+\left(2x-1\right)=0\)

\(\left(2x-1\right)\left(x^2+1+1\right)=0\)

\(\left(2x-1\right)\left(x^2+2\right)=0\)

\(\Rightarrow2x-1=0\) hoặc \(x^2+2=0\)

*) \(2x-1=0\)

\(2x=1\)

\(x=\dfrac{1}{2}\)

*) \(x^2+2=0\) 

\(x^2=-2\) (vô lí)

Vậy \(x=\dfrac{1}{2}\)

d) Sửa đề:

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

\(\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\)

\(\left(x+1\right)\left(x^2+3+1\right)=0\)

\(\left(x+1\right)\left(x^2+4\right)=0\)

\(\Rightarrow x+1=0\) hoặc \(x^2+4=0\)

*) \(x+1=0\)

\(x=-1\)

*) \(x^2+4=0\)

\(x^2=-4\) (vô lí)

Vậy \(x=-1\)

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

len google di ban

mk chua hoc bai nay

a)\(2\left|2x-3\right|=\frac{1}{2}\)

\(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{8}\\x=\frac{11}{8}\end{matrix}\right.\)

Vậy....

b)\(7,5-3\left|5-2x\right|=-4,5\)

\(\Leftrightarrow\left|5-2x\right|=4\)

\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{9}{2}\end{matrix}\right.\)

VẬy...

c)\(\left|3x-4\right|+\left|5-2x\right|=0\)

Có: \(\left|3x-4\right|\ge0với\forall x\\ \left|5-2x\right|\ge0với\forall x\)

\(\Rightarrow\left[{}\begin{matrix}3x-4=0\\5-2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=\frac{5}{2}\end{matrix}\right.\)

\(\Rightarrow x\in\varnothing\)