cứu

\(\left(x+42\right)^2\cdot115=4\cdot115\)

=>\(\left(x+42\right)^2=4\)

=>\(\left[{}\begin{matrix}x+42=2\\x+42=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=-44\end{matrix}\right.\)

2: \(\Leftrightarrow x-34=97\)

hay x=133

a) 1 267 + 99 + 501 = 1 267 + (99 + 501)

                                = 1 267 + 600 = 1 867

     25 x 14 x 4 = (25 x 4) x 14

                       = 100 x 14 = 1 400

b) 3 905 x (50 – 1) = 3 905 x 50 – 3 905 x 1

                             = 195 250 – 3 905 = 191 345

     270 : ( 27 x 2 ) = 270 : 27 : 2

                             = 10 : 2 = 5

c) 115 x 58 + 115 x 42 = 115 x (58 + 42)

                                   = 115 x 100 = 11 500

136 x 67 – 136 x 66 = 136 x (67 – 66)

                              = 136 x 1 = 136

ngu vê ka lờ

= 115 x ( 9-7+8)

= 115 x 10

= 1150 

a)115x9-115x7+8x115

=115x(9-7+8)

=115x10

=1150

b)4667+3378:(4555:5-905)

=4667+3378:(911-905)

=4667+3378:6

=4667+563

=5230

p/s tham khảo nhé

Trả lời:

a, \(x+5x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}\)

Vậy x = 0; x = - 1/5 là nghiệm của pt.

b, \(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

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

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy x = 5 là nghiệm của pt.

Trả lời:

a) x² + 4y² + 4xy = x² + 2.x.2y + (2y)² = ( x + 2y )²

b) \(\frac{1}{64}-27x^3=\left(\frac{1}{4}\right)^3-\left(3x\right)^3=\left(\frac{1}{4}-3x\right)\left(\frac{1}{16}+\frac{3}{4}x+9x^2\right)\)

c) x³ - 6x² + 12x - 8 = x³ - 3.x².2 + 3.x.2² - 2³ = ( x - 2 )³ 

d) x² -  x - y² - y = ( x² - y² ) - ( x + y ) = ( x - y )( x + y ) - ( x + y ) = ( x + y )( x - y - 1 )

e) 5x - 5y + ax  - ay = ( 5x - 5y ) + ( ax - ay ) = 5 ( x - y ) + a ( x - y ) = ( x - y )( 5 + a )

Đáp án:

\(\dfrac{115}{x}+\dfrac{37}{x}=8\)

\(\dfrac{115+37}{x}=8\)

\(\dfrac{152}{x}=8\)

\(x=152:8\)

\(x=19\)

x = 19