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NM
1
26 tháng 12 2021
\(a,892^2+216.892+108^2=892^2+2.108.892+108^2=\left(892+108\right)^2=1000^2=1000000\)
b, \(9x^2+42x+49=9.1^2+42.1+49=9+42+49=100\)
c,\(\left(x^2-3x\right)+x-3=x\left(x-3\right)+\left(x-3\right)=\left(x+1\right)\left(x-3\right)=\left(6+1\right)\left(6-3\right)=7. 3=21\)
NT
1
31 tháng 7 2022
a: \(A=\dfrac{\left(37+12\right)\left(37^2+12^2-37\cdot12\right)}{49-37\cdot12}\)
\(=\dfrac{49\cdot1069}{49-37\cdot12}\simeq-132.61\)
b: \(=\dfrac{\left(52-48\right)\left(52^2+48^2+52\cdot48\right)}{4+52\cdot48}\)
\(=\dfrac{4\cdot7504}{4+52\cdot48}=\dfrac{7504}{625}\)
18 tháng 5 2022
\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)
\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)
\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)
\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)
C
3
H
3 tháng 7 2021
a) \(153^2-53^2=\left(153-53\right)\left(153+53\right)=100.206=20600\)
b)
\(\left(2020^2-2019^2\right)+\left(2018^2-2017^2\right)+...+\left(2^2-1^2\right)\\ =\left(2020+2019\right)\left(2020-2019\right)+\left(2018+2017\right)\left(2018-2017\right)+...+\left(2+1\right)\left(2-1\right)\\ =2020+2019+2018+2017+...+2+1\\ =\dfrac{\left(2020+1\right)2020}{2}=2041210\)
AH
Akai Haruma
Giáo viên
3 tháng 7 2021
Lời giải:
a. $153^2-53^2=(153-53)(153+53)=100.206=20600$
b.
$2020^2-2019^2+2018^2-2017^2+...+2^2-1^2$
$=(2020^2-2019^2)+(2018^2-2017^2)+...+(2^2-1^2)$
$=(2020-2019)(2020+2019)+(2018-2017)(2018+2017)+...+(2-1)(2+1)$
$=2020+2019+2018+2017+...+2+1$
$=\frac{2020.2021}{2}=2041210$
NH
3
13 tháng 8 2021
a) 2,83.5,68-2,83.4,68+1,17.5,68-1,17.4,68
= 2,83.(5,86-4,86)+1,17.(5,86-4,86)=2,83.1+1,17.1=4
b) 1112-1372-482+96.137
= 1112-(1372-2.48.137+482)=1112-(137-48)2=1112-892=(111-89)(111+89)=22.200=4400
D
19 tháng 1 2022
a/ (x-1)2-(4x+3)(2-x)=x2-2x+1-(8x-4x2+6-3x)
=x2-2x+1-8x+4x2-6+3x=5x2-7x-6
b/ (15x3y2 - 6x2y3) : 3x2y2 = 5x - 2y
c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)
21 tháng 7 2022
a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
a) Ta có: \(A=\dfrac{37^3+12^3}{49}-37\cdot12\)
\(=\dfrac{\left(37+12\right)\left(37^2-37\cdot12+12^2\right)}{49}-37\cdot12\)
\(=37^2-2\cdot37\cdot12+12^2\)
\(=\left(37-12\right)^2\)
\(=25^2=625\)