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mng giúp em với tối em nộp bài rồi a

cức + điên= lan ngọc cức điên

lớp 8 sao giống lớp 6 bọn em vậy

Đề bài bn ghi thek thì ai làm nổi cho bn :V ?

ko có biết

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² - 2² = 0

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

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

<=> x = 5 hoặc x = 1

b( 2x + 3 )² - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x² + 12x + 9 - ( 4x² - 1 ) = 22

<=> 4x² + 12x + 9 - 4x² + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )² = 16

<=> 16x² - 9 - ( 16x² - 40x + 25 ) = 16

<=> 16x² - 9 - 16x² + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> x = 50/40 = 5/4

d) x³ - 9x² + 27x - 27 = -8

<=> ( x - 3 )³ = -8

<=> ( x - 3 )³ = (-2)³

<=> x - 3 = -2

<=> x = 1 

e) ( x + 1 )³ - x²( x + 3 ) = 2

<=> x³ + 3x² + 3x + 1 - x³ - 3x² = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

f) ( x - 2 )³ - x( x - 1 )( x + 1 ) + 6x² = 5

<=> x³ - 6x² + 12x - 8 - x( x² - 1 ) + 6x² = 5

<=> x³ + 12x - 8 - x³ + x = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

a) \(\left(x-3\right)^2-4=0\)

=> \(\left(x-3\right)^2-2^2=0\)

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

=> \(\left(x-5\right)\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)

=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)

=> \(4x^2+12x+9-4x^2+1=22\)

=> \(12x+9+1=22\)

=> \(12x+10=22\)

=> 12x = 12

=> x = 1

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)

=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)

=> \(16x^2-9-16x^2+40x-25=16\)

=> \(-9+40x-25=16\)

=> \(40x=16+25-\left(-9\right)=16+25+9=50\)

=> x = 50/40 = 5/4

d) \(x^3-9x^2+27x-27=-8\)

=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)

=> \(\left(x-3\right)^3=-8\)

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

=> x - 3  = -2 => x = 1

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

=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)

=> \(3x+1=2\)

=> \(3x=1\)=> x = 1/3

f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)

=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)

=> \(\left(12x+x\right)-8=5\)

=> 13x  = 13

=> x = 1

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

=>(x+3)^2 = 4

=>(x+3)^2 = 2^2 = (-2)^2

=>x+3 = 2 hoặc -2

=> x= -1 hoặc -5

125 : 5^2 . (x-3) = 3

125 : 25 .(x-3) = 3

5 (x-3) =3

x-3 = 3 : 5

Đề bài sai hay sao ý?

2x -5 = 3

2x = 3+5 =8

x =4

 (19 .x+2.5^2 ): 14 = (13-8)^2 - 4^2

19x + 50) : 14 = 25 - 16

(19x+50) : 14 = 9

19x+50 = 3*14 = 126

 19x = 126-50 = 76

x = 76 / 19 = 4

\(a\)) \(125:5^2.\left(x-3\right)=3\)

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

\(5.\left(x-3\right)=3\)

\(5x-15=3\)

\(5x=3+15\)

\(5x=18\)

\(x=\frac{18}{5}\)\(.\)Vậy \(x=\frac{18}{5}\)

\(b\)) \(\left(2x-5\right)^3=27\)

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

\(\Rightarrow2x-5=3\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=4\)\(.\)Vậy \(x=4\)

\(c\))\(\left(19.x+2.5^2\right):14=\left(13-8\right)^2-4^2\)

\(\left(19.x+2.25\right):14=5^2-16\)

\(\left(19.x+2.25\right):14=25-16\)

\(\left(19.x+50\right):14=9\)

\(19.x+50=126\)

\(19.x=126-50\)

\(19.x=76\)

\(x=76:19\)

\(x=4\)\(.\)Vậy \(x=4\)

ko biết mình mới học lớp 4 thôi

Con " Nguyễn Huyền Trang " đéo biết thì trả lời làm cái l*n gì

thôi ko cần giúp nữa chiều nay mik học xong rồi

a)  2x-3=-x+6

2x - 3 + x - 6 =0

3x -9 = 0

3x = 9

x = 9 : 3

x= 3

c/ \(\left(-12x-4^3\right).8^3=4.8^4\)

\(\left(-12x-64\right).512=16384\)

\(-12x-64=\dfrac{16384}{512}=32\)

\(-12x=32+64=96\)

\(x=\dfrac{96}{\left(-12\right)}=-8\)

a,   2x-3=-x+6

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

\(\Leftrightarrow\)3x-9=0

\(\Leftrightarrow\)3x=9

\(\Leftrightarrow\)x=3

a) Ta có: \(2x-3=-x+6\)

\(\Leftrightarrow2x+x=6+3\)

\(\Leftrightarrow3x=9\)

hay x=3

Vậy: x=3

b) Ta có: \(\left|2x-7\right|+2=13\)

\(\Leftrightarrow\left|2x-7\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=11\\2x-7=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=18\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{9;-2\right\}\)

c) Ta có: \(\left(-12x-4^3\right)\cdot8^3=4\cdot8^4\)

\(\Leftrightarrow\left(-12x-64\right)=\dfrac{4\cdot8^4}{8^3}=32\)

\(\Leftrightarrow-12x=96\)

hay x=-8

Vậy: x=-8