\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)

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

a) a^x-15 = 17

a^x=32

a^x=2^5

\(\Rightarrow\)x=5

Vậy x=5

b/       (7.x-11)^3  = 2^5.5^2+200

\(\Leftrightarrow\)(7.x-11)^3=800+200

\(\Leftrightarrow\)(7.x-11)^3= 1000

\(\Leftrightarrow\)(7x-11)^3= 10^3

\(\Leftrightarrow\) 7x-11=10

\(\Leftrightarrow\)7x=21

\(\Leftrightarrow\) x=3

Vậy x=3

a) a^x - 15 = 17

a^x = 17 + 15

a^x = 32

a^x = 2⁵

=> x = 5

Vậy x = 5

b) (7 . x - 11)³ = 2⁵ . 5² + 200

(7 . x - 11)³  = 32 . 25 + 200

(7 . x - 11)³ = 1000

(7 . x - 11)³ = 10³

=> 7 . x - 11 = 10

7 . x = 10 + 11

7 . x = 21

x = 21 : 7

x = 3

Vậy x = 3

Ủng hộ mk nha !!!  *_*

a) \(2x\left(x-3\right)+6\left(3-x\right)=0\)

\(\Leftrightarrow2\left[x\left(x-3\right)+3\left(3-x\right)\right]=0\)

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

\(\Leftrightarrow x-3=0\)

\(\Rightarrow x=3\)

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

\(\Leftrightarrow3\left[x\left(2x-5\right)-5\left(5-2x\right)\right]=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{2}\end{cases}}\)

bạn cho mình cách giải đc ko

1.4x - 5(-3+x)=7

   4x - 5(x-3)  =7

  4x - 5x + 15=7

     -1x   + 15=7

     -1x         =-8

       =>    x =8

2.5(x-3) - 2(x+6)=9

  5x - 15 -2x -12=9

 5x - 2x -15 - 12=9

             5x - 2x=9 + 12 + 15

             5x - 2x= 36

                3x   = 36

        =>     x   = 12

3.4(x-1) - 3(x-2)=15

  4x - 4 - 3x + 6=15

           4x - 3x =15 - 6 + 4

           4x - 3x = 13

              => x = 13

Nhớ mink nhoa pn

a,b: x= 1 hoặc 0

x=8

x¹⁰=1^x                        x¹⁰=x                     (2x-15)⁵=(2x-15)³

=>x=1(suy đoán)     =>x=1(suy đoán)       Hình như đề sai (suy đoán)

a)x¹⁰=1^x=>x¹⁰=1=>x=1

b)x¹⁰=x=>x¹⁰-x=0=>x(x⁹-1)=0

TH1.x=0

TH2.x⁹-1=0

=>x⁹=1=>x=1

c)(2x-15)⁵=(2x-15)³

như câu b

a) Mình chưa thấy x  ở đâu nha bạn 

b) Ta có : ( 2.x - 15) ⁵ = ( 2.x - 15 )³

<=.   ( 2.x - 15 )⁵ - ( 2.x - 15 )³ = 0

<=> ( 2.x -15 )³ . [ ( 2.x - 15 )² - 1 ] = 0

<=> \(\orbr{\begin{cases}\left(2.x-15\right)^3=0\\\left[\left(2.x-15\right)^2-1\right]=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2.x-15=0\\2.x-15=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}2.x=15\\2.x=16\end{cases}\Leftrightarrow}\orbr{\begin{cases}\frac{15}{2}\left(L\right)\\x=8\end{cases}}}\)

Vậy x = 8

câu a x có 2 cái 2 cái đều đứng trước -5

a) \(x^2+2x=\left(x-2\right).3x\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy S = {0;4}

b) \(x^3+x^2-x-1=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\mp1\end{matrix}\right.\)

Vậy: S = {-1; 1}

c) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]=40\)

\(\Leftrightarrow\left(x^2+5x+x+5\right)\left(x^2+4x+2x+8\right)=40\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)

Đặt x² + 6x + 5 = t

\(\Leftrightarrow t.\left(t+3\right)=40\)

\(\Leftrightarrow t^2+3t=40\)

\(\Leftrightarrow t^2+2.t.\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{169}{4}\)

\(\Leftrightarrow\left(t+\dfrac{3}{2}\right)^2=\dfrac{169}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}t+\dfrac{3}{2}=\dfrac{13}{2}\\t+\dfrac{3}{2}=-\dfrac{13}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{13}{2}-\dfrac{3}{2}=\dfrac{10}{2}=5\\t=-\dfrac{13}{2}-\dfrac{3}{2}=-\dfrac{16}{2}=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x+5=5\\x^2+6x+5=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\)

Mà: \(x^2+6x+13=x^2+2.x.3+9+4=\left(x+3\right)^2+4\ne0\)

=> x² + 6x = 0

<=> x. (x + 6) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy S = {0; -6}

a) Ta có: \(x^2+2x=\left(x-2\right)\cdot3x\)

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

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

\(\Leftrightarrow x\left(x+2-3x+6\right)=0\)

\(\Leftrightarrow x\left(-2x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy: S={0;4}

b) Ta có: \(x^3+x^2-x-1=0\)

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

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

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

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

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

Vậy: S={-1;1}

c) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-40=0\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+40-40=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)=0\)

\(\Leftrightarrow\left(x^2+6x\right)\left(x^2+6x+13\right)=0\)

\(\Leftrightarrow x\left(x+6\right)\left(x^2+6x+13\right)=0\)

mà \(x^2+6x+13>0\forall x\)

nên \(x\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy: S={0;-6}