a) `(2x+5)^3-(2x-5)^3-(120x^2+49)`

`=(2x+5-2x+5)[(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)`

`=10(12x^2+25)-(120x^2+49)`

`=120x^2+250-120x^2-49`

`=201`

b) `(4-5x)^2-(3+5x)^2=(4-5x+3+5x)(4-5x-3-5x)=7.(-10x+1)=-70x+7`

Lời giải:
a. 

$(2x+5)^3-(2x-5)^3-(120x^2+49)$

$=[(2x+5)-(2x-5)][(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)$

$=10(4x^2+20x+25+4x^2-25+4x^2-20x+25)-(120x^2+49)$

$=10(12x^2+25)-(120x^2+49)=250-49=201$

b.

$(4-5x)^2-(3+5x)^2=[(4-5x)+(3+5x)][(4-5x)-(3+5x)]$

$=7(1-10x)$

a) Ta có: \(\left(2x+5\right)^3-\left(2x-5\right)^3-\left(120x^2+49\right)\)

\(=8x^3+60x^2+150x+125-8x^3+60x^2-150x+125-120x^2-49\)

\(=201\)

b) Ta có: \(\left(4-5x\right)^2-\left(3+5x\right)^2\)

\(=\left(4-5x-3-5x\right)\left(4-5x+3+5x\right)\)

\(=7\left(-10x+1\right)\)

\(=-70x+7\)

\(\left(2x+5\right)^2-2\left(2x+5\right)\left(x-1\right)+\left(x+1\right)^2=49\)

<=>\(\left\{\left(2x-5\right)-\left(x-1\right)\right\}^2=49\)

<=> \(\left(2x-5-x+1\right)^2=49\)

<=> \(\left(x-4\right)^2=49\)

<=> \(\hept{\begin{cases}x-4=7\\x-4=-7\end{cases}}\) 

<=> \(\hept{\begin{cases}x=11\\x=-3\end{cases}}\)

học tốt

\((2x-5)^2=49\\\Rightarrow \left[\begin{array}{} 2x-5=7\\ 2x-5=-7 \end{array} \right. \\\Rightarrow \left[\begin{array}{} 2x=12\\ 2x=-2 \end{array} \right. \\\Rightarrow \left[\begin{array}{} x=6\\ x=-1 \end{array} \right.\)

\(Vậy:x\in\left\{-1;6\right\}\)

(2x-5)²=7² 

^2x-5=7

^2x=7+5

^2x=12

^x=6

x=6

a) \(\left(2x-5\right)^2=49\)    

     \(\left(2x-5\right)^2=\left(\pm7\right)^2\)   

      \(=>2x-5=7\)  hoặc \(2x-5=-7\)  

      \(\cdot2x-5=7\)                                        \(\cdot2x-5=-7\)   

       \(2x=5+7\)                                          \(2x=-7+5\)    

       \(2x=12\)                                                \(2x=-2\)   

         \(x=12:2\)                                             \(x=-2:2\)  

         \(x=6\)                                                     \(x=-1\)  

   Vậy x=6 hoặc x=-1

b/ \(\left(2x+5\right)^2-\left(1-2x\right)^2=10\)   

     \(4x^2+20x+25-\left(1-4x+4x^2\right)=10\)   

     \(4x^2+20x+25-1+4x-4x^2=10\)   

                  \(24x+24=10\)   

                   \(24x=10-24\)   

                    \(24x=-14\)    

                         \(x=\frac{-14}{24}\)

                          \(x=\frac{-7}{12}\)   

c/ \(\left(9-2x\right)^3=27\)   

     \(\left(9-2x\right)^3=3^3\)

      \(9-2x=3\)   

       \(2x=9-3\)

        \(2x=6\)

         \(x=6:2\)

         \(x=3\)

a/ (2x-5)^2=49

TH1: 2x-5=7

x=6

TH2: 2x-5=-7

x=-1

2x - 49 = 5 . 3²

2x - 49 = 5 . 9

2x - 49 = 45

2x = 45 + 49

2x = 94

--> x = 94 : 2 = 47

Vậy x = 47

a) \(\Rightarrow\left(2x-3\right)^2=49\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)

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

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

a, ⇒ (2x - 3)² = 49

    ⇒  (2x - 3)² = \(\left(\pm7\right)^2\)

    ⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0

    ⇒ (x - 5).(2x + 7)  = 0

    ⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)

c, ⇒ x² - 5x + 2x - 10 = 0

    ⇒ (x² - 5x) + (2x - 10) = 0

    ⇒ x.(x - 5) +2.(x - 5)    = 0

    ⇒ (x - 5).(x + 2)=0

    \(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

a: \(\left(2x-3\right)^2-49=0\)

\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)

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

a. (2x - 3)² - 49 = 0

<=> (2x - 3)² - 7² = 0

<=> (2x - 3 + 7)(2x - 3 - 7) = 0

<=> (2x + 4)(2x - 10) = 0

<=> \(\left[{}\begin{matrix}2x+4=0\\2x-10=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

b. 2x(x - 5) - 7(5 - x) = 0

<=> 2x(x - 5) + 7(x - 5) = 0

<=> (2x + 7)(x - 5) = 0

<=> \(\left[{}\begin{matrix}2x+7=0\\x-5=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)

c. x² - 3x - 10 = 0

<=> x² - 5x + 2x - 10 = 0

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

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

<=> \(\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

a, (2x - 3)² - 49 = 0

(2x - 3)² - 7² = 0

(2x - 3 + 7)( 2x - 3 - 7) = 0

(2x + 4)( 2x - 10) = 0

=> 2x + 4 = 0                => 2x - 10 = 0

     2x       = - 4                   2x         = 10

       x       = - 2                     x         = 5

3.|x+1|-2=4

3.|x+1|=4+2

3.|x+1|=6                                    

|x+1|=6:3

|x+1|=2

Trường hợp 1       x+1=2

x=2-1

x=1

trường hợp 2

x+1=-2

x=(-2)-1

x=-3

==> x thuộc {1; -3}

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