\(a,\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\\ \Leftrightarrow4x^2-4x^2+4x-16x+1-16-9=0\\ \Leftrightarrow-12x=24\\ \Leftrightarrow x=\dfrac{24}{-12}=-2\\ b,\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Leftrightarrow x^2-x^2+6x-4x=1-9-32\\ \Leftrightarrow2x=-40\\ \Leftrightarrow x=-20\\ c,3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\\ \Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\\ \Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\\ \Leftrightarrow3x^2+4x^2-7x^2+12x-4x=36-12-1-63\\ \Leftrightarrow8x=-40\\ \Leftrightarrow x=\dfrac{-40}{8}=-5\)

a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Rightarrow\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\\ \Rightarrow\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\\ \Rightarrow-3\left(4x+5\right)=9\\ \Rightarrow4x+5=-3\\ \Rightarrow4x=-8\\ \Rightarrow x=-2\)

b c d nữa bạn

\(b) (x+3)^2-(x-4)(x+8)=1 <=>x^2+6x=9-(x^2+8x-4x-32)=0 \)

\(.<=> X^2+6x+9-x^2-8x+4x+32-1=0\)

\(<=>2x=-40<=>x=-20\)

=> ptrình có tập nghiêm S={-20}

c) 3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36

\(<=>3(x^2+4x+4)+(4x^2-4x+1)-7(x^2-9)=36\)

\(<=>3x^2+12x+12+4x^2-4x+1-7x^2+49=0\)

\(<=>8x=-62<=>x=7,75\)

=> ptrình có tập nghiệm S={7,75}

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

\(<=> x^3+3x^2+9x-3x^2-9x-27-x(x^2-4)=1\)

\(<=>x^3+3x^2+9x-3x^2-9x-27-x^3+4x-1=0\)

\(<=> 4x=28<=> x=7\)

=> ptrình có tập nghiệm S={7}

1) 5.(3-x)+2.(x-7)=-14

    15-5x+2x-14=-14

    1-3x=-14

    3x=15

    X=5

2) 30.(x+2)-6.(x-5)-24x=100

    30x+60-6x+30-24x=100

    0X+90=100

    0X=10 vô lí

=> ko có giá trị x thỏa mãn điều kiện

3) (3x-9)^2=36

    3x-9=6    

    3x-9=-6

TH1:3x-9=6                      TH2:3x-9=-6

        3x=15                       3x=3

        X=5                 x=1

Vậy….

4) (1-2x)^3=-27

    (1-2x)³=(-3)³

     1-2x=-3

      2x=4

    X=2

Vậy…

5) (x-3).(x-2)<0

=>x-3 và x-2 cùng dấu

TH1:x-3>0              TH2:x-3<0                   

        x-2<0                    x-2>0

    =>X>3                      =>x<3

        X<2                          x>2

=>x>3                           =>x<2

Vậy 3<x<2

câu 6 chịuuuu

câu 5 hơi khó ko bt có đúng hay ko đâu :)))

3) \(\left(3x-9\right)^2=36\Leftrightarrow\orbr{\begin{cases}3x-9=6\\3x-9=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=15\\3x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=1\end{cases}}}\)

4) \(\left(1-2x\right)^3=-27\)

<=> 1-2x=-3

<=> 3x=4

<=> \(x=\frac{4}{3}\)

5) (x-3)(x-2)<0

=> x-3 và x-2 trái dấu nhau

thấy x-3<x-2 => \(\hept{\begin{cases}x-3< 0\\x-2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 3\\x>2\end{cases}\Leftrightarrow}2< x< 3}\)

6) làm tương tự

a, Xét : x-4 = 0 => x= 4

            2x+1 = 0 => x= \(\frac{1}{2}\)

            x+3 = 0 => x = -3

            x + 9 = 0 => x = -9

Khi đó ta có bảng xét dấu : 

=> có 5 trường hợp:

TH1 : \(x\le-9\)

TH2 : \(-9\le x< -3\)

TH3 : \(-3\le x< \frac{1}{2}\)

TH4 : \(\frac{1}{2}\le x< 4\)

Do đó :

TH1 : \(x\le-9\)

Ta có :  /x-4/ = -(x-4) = 4 - x

            /2x+1/ = -(2x+1) = -2x -1

           /x+3/   = -(x + 3 ) = -x - 3

          /x-9/ = -(x-9) = -x + 9                  Thay vào đề bài ta có:

                                               3.(4-x) + 2x-1 +5(-x - 3) -x-9 = 5

                                    => 12 - 3x + 2x - 1 + -5x - 15 - x - 9 = 5

                                    =>(12 - 1 - 15 -9 ) +(-3x +2x -5x -x) = 5

                                   => -13 - 7x                                        = 5

                                             7x                                =     -13 - 5

                                                 7x =      -18

                                              x = \(\frac{-18}{7}\)( Ko TM)

Tương tự với 4 trường hợp còn lại.

1: =>3^x=81

=>x=4

2: =>2^x=8

=>x=3

3: =>x^3=2^3

=>x=2

4: =>x^20-x=0

=>x(x^19-1)=0

=>x=0 hoặc x=1

5: =>2^x=32

=>x=5

6: =>(2x+1)^3=9^3

=>2x+1=9

=>2x=8

=>x=4

7: =>x^3=115

=>\(x=\sqrt[3]{115}\)

8: =>(2x-15)^5-(2x-15)^3=0

=>(2x-15)^3*[(2x-15)^2-1]=0

=>2x-15=0 hoặc (2x-15)^2-1=0

=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1

=>x=15/2 hoặc x=8 hoặc x=7

1. Tìm số tự nhiên x biết:

1) \(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

_____

2) \(7.2^x=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

_____

3) \(x^3=8\)

\(x^3=2^3\)

\(\Rightarrow x=3\)

_____

4) \(x^{20}=x\)

\(x^{20}-x=0\)

\(x\left(x^{19}-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x=1\)

5) \(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

_____

6) \(\left(2x+1\right)^3=9.81\)

\(\left(2x+1\right)^3=729=9^3\)

\(\rightarrow2x+1=9\)

\(2x=9-1\)

\(2x=8\)

\(x=8:2\)

\(\Rightarrow x=4\)

_____

7) \(x^6:x^3=125\)

\(x^3=125\)

\(x^3=5^3\)

\(\Rightarrow x=5\)

_____

8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)

_____

9) \(3^{x+2}-5.3^x=36\)

\(3^x.\left(3^2-5\right)=36\)

\(3^x.\left(9-5\right)=36\)

\(3^x.4=36\)

\(3^x=36:4\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

_____

10) \(7.4^{x-1}+4^{x+1}=23\)

\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)

\(4^{x-1}.\left(7+4^2\right)=23\)

\(4^{x-1}.\left(7+16\right)=23\)

\(4^{x-1}.23=23\)

\(4^{x-1}=23:23\)

\(4^{x-1}=1\)

\(4^{x-1}=4^1\)

\(\rightarrow x-1=0\)

\(x=0+1\)

\(\Rightarrow x=1\)

Chúc bạn học tốt

- Gửi lẻ câu hỏi ra nha bạn 2 3 câu 1 lần thôi .

a) (x-3)²-4=0

⇒ (x-3)²=4

⇒ hoặc x-3=2⇒x=5

hoặc x-3=-2⇒x=1

c) (x+4)²-(x+1)(x-1)=16

⇒ x²+8x+16-x²+1=16

⇒ 8x+17=16

⇒ 8x=-1

⇒ x=-1/8

a) \(\left(x-3\right)^2-4=0\)

\(\left(x-7\right)\left(x+1\right)=0\)

\(\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)

b) \(x^2-2x=24\)

\(x^2-2x-24=0\)

\(\left(x-6\right)\left(x+4\right)=0\)

\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(5x^2+10x+10-5x^2+245=0\)

\(10x+255=0\)

\(x=-25.5\)

A) \(\left(x-3\right)^2-4=0\)

\(\left(x-3\right)^2=4\Rightarrow\left(x-3\right)^2=\left(-2\right)^2;2^2\)

th1\(\left(x-3\right)^2=2^2\)

\(\Rightarrow x-3=2\)

\(\Rightarrow x=2+3\)

\(\Rightarrow x=5\)

th2: \(\left(x-3\right)^2=\left(-2\right)^2\)

\(\Rightarrow x-3=-2\)

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

\(\Rightarrow x=1\)

\(\Leftrightarrow x\in\left\{1;5\right\}\)

a) \(\left(x-3\right)^2-4=0\)

Ta xét 2TH:

TH1: \(x-3=2\)

\(x-3+3=2+3\)

\(x=5\)

TH2: \(x-3=-2\)

\(x-3+3=-2+3\)

\(x=1\)

=> \(x\in\left\{1;5\right\}\)

(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)

 20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1

 10x2−19x−33=010x2−19x−33=0

 (10x+11)(x−3)=0

chỉ bt lm con b thoy

..army,,,,,,,,,,

a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+20+2\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\frac{22}{5}\)

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2\)

\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)

\(\Leftrightarrow20x^2-16x-33=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

a, ( 2x + 3 )(x - 4) + ( x - 5 )( x - 2) =( 3x - 5 ) ( x - 4 )
<-> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20
<-> 3x² - 10x - 2 = 3x² - 15x + 20
<-> 3x² - 3x² - 10x + 15x = 20 + 2
<-> 5x = 22
<-> x = 22/5