a) 

(2018 - 1002.2) + 96 . 100 

= (2018 - 2004) + 9600

= 14 + 9600

= 9614

b)

50 . [ 720 : (10 + 70 . 5 ] - 36

= 50 . [ 720 : ( 10 + 350] - 36

= 50 . [ 720 : 360] - 36

= 50 . 2 - 36

= 100 - 36

= 643

tìm x

[ 3 . (2x + 8) - 15] . 4 = 84

[3 . (2x + 8) - 15]       = 84 : 4 

[3 . (2x + 8) - 15]       = 21

[3 . (2x + 8) ]              = 21 + 15 

[3 . (2x + 8) ]              = 36

(2x + 8)                     = 36 : 3 

(2x + 8)                     = 12

(2x)                           = 12 - 8 

(2x)                           = 4

   x                            = 4 : 2 

   x                            = 2

1\

a)(2018 - 1002 . 2) + 96 .100

=(2018 - 2004) + 9600

=14 + 9600

=9614

b 50.[720:(10 +70.5)]-36

=50.[720:(10+350)]-36

=50.(720:360)-36

=50.2-36

=100-36

=64

2\Tìm x

[3.(2x + 8)-15].4=84

=>3.(2x+8)-15=84:4

=>3.(2x+8)-15=21

=>3.(2x+8)=21+15

=>3.(2x +8)=36

=>2x+8=36:3

=>2x+8=12

=>2x=12-8

=>2x=4

=>X=2

Vậy X=2

CHÚC BẠN HỌC GIỎI

\(\left(2x-1\right)^3-2x\left(2x+1\right)^2=5x-20x^2\)

\(\Rightarrow8x^3-12x^2+6x-1-2x\left(4x^2+4x+1\right)-5x+20x^2=0\)

\(\Rightarrow8x^3-12x^2+6x-1-8x^3-8x^2-2x-5x+20x^2=0\)

\(\Rightarrow-x-1=0\)

\(\Rightarrow x=-1\)

a)16.4^x=4⁸

^{\(4^2.4^x=4^{8^{ }}\)}

^{\(4^x=4^{8^{ }}:4^2\)}

\(4^x=4^{6^{ }}\)

\(x=6\)
b)(X-2)(X-5)=0

\(\Rightarrow x-2=0\rightarrow x=2\)

\(x-5=0\rightarrow x=5\)

Vậy x∈ {2;5}
c)2^x+2^x+1=96

\(2^x.1+2^{x^{ }}.2\)

\(2^x.\left(1+2\right)=96\)

\(2^x.3=96\)

\(2^x=96:3\)

\(2^x=32\)

\(2^x=2^5\)

\(x=5\)

\(Zzz\) 😪

(2x + 1) + (2x + 2) + ... + (2x + 2015) = 0

=> 2015.2x + (1 + 2 + 3 + ... + 2015) = 0

=> 4030x + (2015 + 1).2015 : 2 = 0

=> 4030x = -2031120

=> x = -504

x=-504

ủng hộ mk nha các bạn

\(a,\Rightarrow 2x^2-10x-3x-2x^2=26\\ \Rightarrow -13x=26\\ \Rightarrow x=-2\\ b, \Rightarrow -2x^2+3x+3-3x-3+2x^2-x=18\\ \Rightarrow -x=18\Rightarrow x=-18\)

Câu 1 :

\(\text{ a) }12-2x-x^2=0\\ \Leftrightarrow2\left(6-x-x^2\right)=0\\ \Leftrightarrow6-x-x^2=0\\ \Leftrightarrow6-3x+2x-x^2=0\\ \Leftrightarrow\left(6-3x\right)+\left(2x-x^2\right)=0\\ \Leftrightarrow3\left(2-x\right)+x\left(2-x\right)=0\\ \Leftrightarrow\left(3+x\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3+x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy \(x=-3\) hoặc \(x=2\)

\(\text{b) }\left(x^2-\dfrac{1}{2}x\right):2x-\left(3x-1\right):\left(3x-1\right)=0\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{4}-1=0\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{5}{4}=0\\ \Leftrightarrow\dfrac{1}{2}x=\dfrac{5}{4}\\ \Leftrightarrow x=\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}\)

Câu 2:

\(N=x^2+5y^2+2xy-2y+2005\\ N=x^2+4y^2+y^2+2xy-2y+1+2004\\ N=\left(x^2+2xy+y^2\right)+\left(4y^2-2y+1\right)+2004\\ N=\left(x+y\right)^2+\left(2y-1\right)^2+2004\\ \text{Do }\left(x+y\right)^2\ge0\forall x;y\\ \left(2y-1\right)^2\ge0\forall y\\ \Rightarrow\left(x+y\right)^2+\left(2y-1\right)^2\ge0\forall x;y\\ \Rightarrow N=\left(x+y\right)^2+\left(2y-1\right)^2+2004\ge0\forall x;y\\ \text{Dấu "=" xảy ra khi : }\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(2y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\2y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(N_{\left(Min\right)}=2004\) khi \(x=-\dfrac{1}{2};y=\dfrac{1}{2}\)