\(\dfrac{240.50+48.100}{42.37+21.126}\)

\(=\dfrac{24.5.10.10+48.100}{42.37+3.7.6.21}\)

\(=\dfrac{100.100+48.100}{42.37+42.63}\)

\(=\dfrac{100.\left(100+48\right)}{42.\left(37+63\right)}\)

\(=\dfrac{100.148}{42.100}\)

\(=\dfrac{148}{42}\)

\(=\dfrac{74}{21}\)

234+ 50 bàng mấy

a: Ta có: \(142-\left[50-\left(2^3\cdot10-2^3\cdot5\right)\right]\)

\(=142-\left[50-80+40\right]\)

=142-10

=132

a. 142 - [50-(2³.10-2³.5)]

= 142 - [ 50 - ( 8 . 10 - 8 . 5 )]

= 142 - [ 50 - ( 80 - 40 )]

= 142 - [ 50 - 40 ]

= 142 - 10 = 132

b. 375 : { 32[ 4+(5.3² - 42 )]} - 14

= 375 : { 32[4+(5.9 - 42 )]} - 14

= 375 : { 32[4 + ( 45 - 42 )]} - 14

= 375 : {32[4+3]} - 14

= 375 : 224 - 14

c.{210 : [ 16+3.(6+3.2^2)]} - 3

= {210 : [ 16 + 3.(6+3.4)]} - 3

= { 210 :[16+3.(6+12)]}-3

= {210 : [ 16+3.18)]} - 3

= { 210 : [ 16 + 54]} - 3

= { 210 : 70 } - 3

= 30 - 3 = 27

d.500-{5.[409-(2^3 . 3-21^2)] - 1724}

= 500-{5.[409-(8 . 63 - 21^2] - 1724}

= 500 - { 5.[409- (504 -  441) - 1724}

= 500 -{ 5.[ 409 - 63 ] - 1724}

= 500 - { 5.346 - 1724}

= 500 - { 1720 - 1724 }

= 500 - 6 

= 494

\(\left\{5.\left[42:\left(5.3+50:10\right).3:2+4\right]\right\}+\left\{35.\left[63:9.\left(5^2\right)\right]\right\}\\ =\left\{5.\left[42:\left(15+5\right).1,5+4\right]\right\}+\left\{35.\left(9.25\right)\right\}\\ =\left\{5.\left[42:20.1,5+4\right]\right\}+\left\{35.225\right\}\\ =\left\{5.\left[2,1.1,5+4\right]\right\}+7875\\ =\left\{5.\left[3,15+4\right]\right\}+7875\\ =\left\{5.7,15\right\}+7875\\ =35,75+7875\\ =7910,75\)

{5.[42:(5.3+50:10).3:2+4]}+{35.[63:9.(52)]}={5.[42:(15+5).1,5+4]}+{35.(9.25)}={5.[42:20.1,5+4]}+{35.225}={5.[2,1.1,5+4]}+7875={5.[3,15+4]}+7875={5.7,15}+7875=35,75+7875=7910,75

\(=\dfrac{12000+4800}{42\left(37+63\right)}=\dfrac{16800}{42\cdot100}=\dfrac{168}{42}=4\)

Bài 1: Tính nhanh 

A, 13 x 126 + 37 x 126 - 49 x 126

= 126 x ( 13 + 49 - 49 )

= 126 x 1

= 126

B, ( 1 + 2 + 4 + 8+.....+ 512 ) x ( 101 x 102 - 101 x 101 - 50 - 51 )

= ( 1 + 2 + 4 + 8+.....+ 512 ) x ( 101 x ( 102 - 101 - 1))

= ( 1 + 2 + 4 + 8+.....+ 512 ) x ( 101 x 0)

= ( 1 + 2 + 4 + 8+.....+ 512 ) x 0

= 0

RƯETSRYTUYUCGFXDZ

\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)

\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)

\(-18x+13=0\)

\(x=\dfrac{13}{18}\)

Vậy \(S=\left\{\dfrac{13}{18}\right\}\)

\(b.\left(x-1\right)^3-125=0\)

\(\left(x-1\right)^3=125\)

\(x-1=5\)

\(x=6\)

Vậy \(S=\left\{6\right\}\)

\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)

\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy \(S=\left\{1;-2\right\}\)

\(d.x^2-4x+4+x^2-2xy+y^2=0\)

\(\left(x-2\right)^2+\left(x-y\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Vậy \(S=\left\{2;2\right\}\)

Kết quả: 24553

96-3(x+1)=42

=>3(x+1)=96-42

=>3(x+1)=54

=>x+1=54:3

=>x+1=18

=>x=18-1

=>x=17

12x-30=3²x3³

=>12x-30=243

=>12x=243+30

=>12x=273

=>x=273:12

=>x=91/4

=>x=22,75

96-3(x+1)=42

=>3(x+1)=96-42

=>3(x+1)=54

=>x+1=54:3

=>x+1=18

=>x=18-1

=>x=17

12x-30=3²x3³

=>12x-30=243

=>12x=243+30

=>12x=273

=>x=273:12

=>x=91/4

=>x=22,75