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

5 + ( x + 27 ) = 64

a) \(\Rightarrow x+27=59\Rightarrow x=32\)

b) \(\Rightarrow x-2=39\Rightarrow x=41\)

c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)

d) \(\Rightarrow5x=35\Rightarrow x=7\)

e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)

f) \(\Rightarrow15+x=37\Rightarrow x=22\)

g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)

h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)

a) \(4x^3-9x=0\)

\(\Leftrightarrow x\left(4x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2=9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)

b) \(3x\left(x-2\right)-5x+10=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)

c) \(4x\left(x+3\right)-x^2+9=0\)

\(\Leftrightarrow4x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)

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

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

\(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

e) \(16x^2-25=\left(4x-5\right)\left(2x+1\right)\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-2\end{cases}}\)

f) \(\left(x+\frac{1}{5}\right)^2=\frac{64}{9}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{8}{3}\\x+\frac{1}{5}=-\frac{8}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{37}{15}\\x=-\frac{43}{15}\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x+6=x+3\\3x+6=-x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{9}{4}\end{cases}}\)

a) 4x³ - 9x = 0

<=> x( 4x² - 9 ) = 0

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

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

<=> x = 0 hoặc x = ±3/2

b) 3x( x - 2 ) - 5x + 10 = 0

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

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

<=> x - 2 = 0 hoặc 3x - 5 = 0

<=> x = 2 hoặc x = 5/3

c) 4x( x + 3 ) - x² + 9 = 0

<=> 4x( x + 3 ) - ( x² - 9 ) = 0

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

<=> ( x + 3 )[ 4x - ( x - 3 ) ] = 0

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

<=> ( x + 3 )( 3x + 3 ) = 0

<=> x + 3 = 0 hoặc 3x + 3 = 0

<=> x = -3 hoặc x= -1

d) ( 2x + 5 )( x - 4 ) = ( x - 4 )( 5 - x )

<=> ( 2x + 5 )( x - 4 ) - ( x - 4 )( 5 - x ) = 0

<=> ( x - 4 )[ ( 2x + 5 ) - ( 5 - x ) ] = 0

<=> ( x - 4 )( 2x + 5 - 5 + x ) = 0

<=> ( x - 4 ).3x = 0

<=> x - 4 = 0 hoặc 3x = 0

<=> x = 4 hoặc x = 0

e) 16x² - 25 = ( 4x - 5 )( 2x + 1 )

<=> ( 4x - 5 )( 4x + 5 ) - ( 4x - 5 )( 2x + 1 ) = 0

<=> ( 4x - 5 )[ ( 4x + 5 ) - ( 2x + 1 ) ] = 0

<=> ( 4x - 5 )( 4x + 5 - 2x - 1 ) = 0

<=> ( 4x - 5 )( 2x + 4 ) = 0

<=> 4x - 5 = 0 hoặc 2x + 4 = 0

<=> x = 5/4 hoặc x = -2

f) ( x + 1/5 )² = 64/9

<=> ( x + 1/5 )² = ( ±8/3 )²

<=> x + 1/5 = 8/3 hoặc x + 1/5 = -8/3

<=> x = 37/15 hoặc x = -43/15

g) 9( x + 2 )² = ( x + 3 )²

<=> 3²( x + 2 )² - ( x + 3 )² = 0

<=> [ 3( x + 2 ) ]² - ( x + 3 )² = 0

<=> ( 3x + 6 )² - ( x + 3 )² = 0

<=> [ ( 3x + 6 ) - ( x + 3 ) ][ ( 3x + 6 ) + ( x + 3 ) ] = 0

<=> ( 3x + 6 - x - 3 )( 3x + 6 + x + 3 ) = 0

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

<=> 2x + 3 = 0 hoặc 4x + 9 = 0

<=> x = -3/2 hoặc x = -9/4

a) x = 8.

b) x = -5.

c) x = 5.

d) x = -3.

a,\((3x)^3\)=-64(=)3x=-4(=)x=\(-\frac{4}{3}\)

ý b thiếu đề bài

a)\(27x^3+64=0\)

\(\Rightarrow27x^3=-64\)

\(\Rightarrow x^3=-\frac{64}{27}\)

\(\Rightarrow x^3=\left(-\frac{4}{3}\right)^3\)

\(\Rightarrow x=-\frac{4}{3}\)

a) x= -4/3

b) chiu

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