Giúp tớ với tớ cần gấp

a: =>(x-5)(x+5)=0

=>x=5 hoặc x=-5

b: \(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow x\in\left\{1;-3\right\}\)

c: =>x-3=4

hay x=7

d: =>x+1=7

hay x=6

e: =>x=0 và y-2=0

hay x=0 và y=2

9x² - 4 - ( 3x - 2 )( x + 5 ) = 0

<=> ( 3x - 2 )( 3x + 2 ) - ( 3x - 2 )( x + 5 ) = 0

<=> ( 3x - 2 )( 3x + 2 - x - 5 ) = 0

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

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

x³ + 64 + ( x + 4 )( 2x - 3 ) = 0

<=> ( x + 4 )( x² - 4x + 16 ) + ( x + 4 )( 2x - 3 ) = 0

<=> ( x + 4 )( x² - 4x + 16 + 2x - 3 ) = 0

<=> ( x + 4 )( x² - 2x + 13 ) = 0

<=> \(\orbr{\begin{cases}x+4=0\\x^2-2x+13=0\end{cases}}\Leftrightarrow x=-4\)( vì x² - 2x + 13 = ( x² - 2x + 1 ) + 12 = ( x - 1 )² + 12 ≥ 12 > 0 ∀ x )

( x - 3 )( x² + 4x + 9 ) + 2( x² - 9 ) - 10( x - 3 ) = 0

<=> ( x - 3 )( x² + 4x + 9 ) + 2( x - 3 )( x + 3 ) - 10( x - 3 ) = 0

<=> ( x - 3 )( x² + 4x + 9 + 2x + 6 - 10 ) = 0

<=> ( x - 3 )( x² + 6x + 5 ) = 0

<=> ( x - 3 )( x + 1 )( x + 5 ) = 0

<=> x = 3 hoặc x = -1 hoặc x = -5

<=> ( x - 3 )( 

a. x mũ 2 - 2x + 1 = 25 

= x^2 + 2.x.1 + 1^2

= ( x + 1 ) ^2

ko bt có đúng ko nữa, mấy câu kia tui ko bt lm

Sos

hahaha

a/ => x³ = 64 => x³ = 4³ => x = 4

b/ => 4x² - 12x + 9 - x² - 10x - 25 = 0 

=> 3x² - 22x - 16 = 0

=> (x - 8)(3x + 2) = 0

=> x - 8 = 0 => x = 8

hoặc 3x + 2 = 0 => 3x = -2 => x = -2/3

Vậy x = 8 ; x = -2/3

c/ => x³ - x² - 4x² + 8x - 4 = 0 

=> x³ - 5x² + 8x - 4 = 0 

=> (x - 2)² (x - 1) = 0

=> (x - 2)² = 0 => x - 2 = 0 => x = 2

hoặc x - 1 = 0 => x = 1 

Vậy x = 2 ; x = 1

1: \(2^x=64\)

=>\(x=log_264=6\)

2: \(2^x\cdot3^x\cdot5^x=7\)

=>\(\left(2\cdot3\cdot5\right)^x=7\)

=>\(30^x=7\)

=>\(x=log_{30}7\)

3: \(4^x+2\cdot2^x-3=0\)

=>\(\left(2^x\right)^2+2\cdot2^x-3=0\)

=>\(\left(2^x\right)^2+3\cdot2^x-2^x-3=0\)

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

=>\(2^x-1=0\)

=>\(2^x=1\)

=>x=0

4: \(9^x-4\cdot3^x+3=0\)

=>\(\left(3^x\right)^2-4\cdot3^x+3=0\)

Đặt \(a=3^x\left(a>0\right)\)

Phương trình sẽ trở thành:

\(a^2-4a+3=0\)

=>(a-1)(a-3)=0

=>\(\left[{}\begin{matrix}a-1=0\\a-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\left(nhận\right)\\a=3\left(nhận\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3^x=1\\3^x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

5: \(3^{2\left(x+1\right)}+3^{x+1}=6\)

=>\(\left[3^{x+1}\right]^2+3^{x+1}-6=0\)

=>\(\left(3^{x+1}\right)^2+3\cdot3^{x+1}-2\cdot3^{x+1}-6=0\)

=>\(3^{x+1}\left(3^{x+1}+3\right)-2\left(3^{x+1}+3\right)=0\)

=>\(\left(3^{x+1}+3\right)\left(3^{x+1}-2\right)=0\)

=>\(3^{x+1}-2=0\)

=>\(3^{x+1}=2\)

=>\(x+1=log_32\)

=>\(x=-1+log_32\)

6: \(\left(2-\sqrt{3}\right)^x+\left(2+\sqrt{3}\right)^x=2\)
=>\(\left(\dfrac{1}{2+\sqrt{3}}\right)^x+\left(2+\sqrt{3}\right)^x=2\) 

=>\(\dfrac{1}{\left(2+\sqrt{3}\right)^x}+\left(2+\sqrt{3}\right)^x=2\)

Đặt \(b=\left(2+\sqrt{3}\right)^x\left(b>0\right)\)

Phương trình sẽ trở thành:

\(\dfrac{1}{b}+b=2\)

=>\(b^2+1=2b\)

=>\(b^2-2b+1=0\)

=>(b-1)²=0

=>b-1=0

=>b=1

=>\(\left(2+\sqrt{3}\right)^x=1\)

=>x=0

7: ĐKXĐ: \(x^2+3x>0\)

=>x(x+3)>0

=>\(\left[{}\begin{matrix}x>0\\x< -3\end{matrix}\right.\)
\(log_4\left(x^2+3x\right)=1\)

=>\(x^2+3x=4^1=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

Tìm x à nếu thế thì mk làm cùi lắm

c)(x+4)\(^2\)-64=0

=>(x+4)\(^2\)=64

=>(x+4)\(^2\)=8\(^2\)

=>x+4=8

=>x=4

d)(x+13)\(^3\)+27=0

(x+13)\(^3\)=-27

(x+13)\(^3\)=(-3)\(^3\)

a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)

=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)

=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)

b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)

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

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

=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)

d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)

=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)

=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)

e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)

=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)

=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)

f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)

=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)

=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)

=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)