\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)

\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)

\(\text{⇔}8x+5=-11\) 

\(\text{⇔}8x=-16\)

\(\text{⇔}x=-2\)

Vậy: \(x=-2\)

==========

\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)

\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)

\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)

\(\text{⇔}-26x=-\dfrac{4}{3}\)

\(\text{⇔}x=\dfrac{2}{39}\)

\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)

a) \(\Leftrightarrow x^3-8-x^3+9x=1\)

\(\Leftrightarrow9x=9\Leftrightarrow x=1\)

b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)

\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

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

\(\Leftrightarrow2x^2-4x+x-2-2x^2=0\)

\(\Leftrightarrow\left(2x^2-2x^2\right)-\left(4x-x\right)-2=0\)

\(\Leftrightarrow-3x-2=0\)

\(\Leftrightarrow-3x=2\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

b) \(\left(x+3\right)\left(2x-1\right)+x^2=9\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)+x^2-9=0\)

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

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

\(\Leftrightarrow\left(x+3\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\3x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{4}{3}\end{matrix}\right.\)

`#3107.101107`

a)

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

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

`<=> -3x - 2 = 0`

`<=> -3x = 2`

`<=> x = -2/3`

Vậy, `x=-2/3`

b)

`(x + 3)(2x - 1) + x^2 = 9`

`<=> 2x^2 - 5x - 3 + x^2 = 9`

`<=> 3x^2 - 5x - 3 = 9`

`<=> 3x^2 - 3x - 12 = 0`

`<=> 3x^2 + 4x - 9x - 12 = 0`

`<=> (3x^2 - 9x) + (4x - 12) = 0`

`<=> 3x(x - 3) + 4(x - 3) = 0`

`<=> (3x + 4)(x - 3) = 0`

`<=>` TH1: `3x + 4 = 0`

`<=> 3x = -4`

`<=> x = -4/3`

TH2: `x - 3 = 0`

`<=> x = 3`

Vậy,` x \in {-4/3; 3}.`

a: ta có: \(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)

\(\Leftrightarrow2x^2+4x-5x-10-2x^2+2x=15\)

\(\Leftrightarrow x=25\)

b: Ta có: \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow4x^2-25+\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+3\right)=0\)

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

c: Ta có: \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Leftrightarrow-9x=1\)

hay \(x=-\dfrac{1}{9}\)

a: ta có: $\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15$

$\iff 2{}^{}+4x-5x-10-2{}^{}+2x=15$

$\iff x=25$

b: Ta có: $\left(5-2x\right)\left(2x+7\right)=4{}^{}-25$

$\iff 4{}^{}-25+\left(2x-5\right)\left(2x+7\right)=0$

$\iff \left(2x-5\right)(2x+5+2x+$

a,(2x+1)(y-3)=12

2x+1 và y-3 Ư(12)=

=>x=0,y=15

c) Ta có: \(36^{25}=\left(6^2\right)^{25}=6^{50}\)

\(25^{36}=\left(5^2\right)^{36}=5^{72}\)

Ta có: \(6^{50}=\left(6^5\right)^{10}=7776^{10}\)

mà \(5^{70}=\left(5^7\right)^{10}=78125^{10}\)

nên \(6^{50}< 5^{70}\)

mà \(5^{70}< 5^{72}\)

nên \(6^{50}< 5^{72}\)

hay \(36^{25}< 25^{36}\)

a/

Với $x,y$ là số tự nhiên $2x+1, y-3$ là số nguyên. Mà $(2x+1)(y-3)=12$ nên $2x+1$ là ước của 12. 

$2x+1>0, 2x+1$ lẻ nên $2x+1\in \left\{1;3\right\}$

Nếu $2x+1=1\Rightarrow y-3=12$

$\Rightarrow x=0; y=15$

Nếu $2x+1=3\Rightarrow y-3=4$

$\Rightarrow x=1; y=7$ 

Vậy...........

b/

$2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-8$

$2^x(1+2+2^2+2^3+...+2^{2015})=2^{2019}-8(1)$
$2^x(2+2^2+2^3+2^4+...+2^{2016})=2^{2020}-16(2)$ (nhân 2 vế với 2)

Lấy (2) trừ (1) theo vế thì:

$2^x(2^{2016}-1)=2^{2020}-2^{2019}-8$

$2^x(2^{2016}-1)=2^{2019}(2-1)-8=2^{2019}-8$

$2^x(2^{2016}-1)=2^3(2^{2016}-1)$

$\Rightarrow 2^x=2^3$

$\Rightarrow x=3$

c/

$25^{36}=(5^2)^{36}=5^{72}$

$36^{25}=(6^2)^{25}=6^{50}=(6^5)^{10}< (5^7)^{10}=5^{70}< 5^{72}$

$\Rightarrow 25^{36}> 36^{25}$

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

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

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

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

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

=>2x=-4

hay x=-2