\(a,5\cdot\left(2x-3\right)=55\)

\(\left(2x-3\right)=55:5\)

\(2x-3=11\)

\(2x=11+3\)

\(2x=14\)

\(x=14:2\)

\(x=7\)

\(b,3x+17=2\)

\(3x=2-17\)

\(3x=-15\)

\(x=\left(-15\right):3\)

\(x=-5\)

1) 40 + 15 + (-10) + (-15)                     2) -13 + (-750) + (-17) + 750                    3) (35 - 17) + (17 + 120 - 35)

= 40 + 15 - 10 - 15                               = -13 - 750 - 17 + 750                               = 35 - 17 + 17 + 120 - 35

= (40 - 10) + (15 - 15)                          = (-13 - 17) + (-750 + 750)                        = (35 - 35) + (-17 + 17) + 120

= 30                                                     = -30                                                          = 120

4) (55 + 45 + 15) - (15 - 55 + 45)               5) -(12 + 21 - 23) - (23 - 21 + 10)             6) (2020 - 79 + 15) - (-79 + 15)

= 55 + 45 + 15 - 15 + 55 - 45                     = -12 -21 + 23 - 23 + 21 - 10                    = 2020 - 79 + 15 + 79 - 15

= (45 - 45) + (15 - 15) + (55 + 55)              = (-12 - 10) + (-21 + 21) + (23 - 23)          = 2020 + (-79 + 79) + (15 - 15)

= 110                                                          = -22                                                         = 2020

7) -(515 - 80 + 91) - (2010 + 80 - 91)                     8) 25 - (-17) + 24 - 12               9) 235 - (34 + 135) - 100        

= -515 + 80 - 91 - 2010 - 80 + 91                           = 25 + 17 + 24 - 12                   = 235 - 34 - 135 - 100

= (-515 -2010) + (80 - 80) + (-91 + 91)                   = 54                                           = -34

= -2525       

10) (13 + 49) - (13 - 135 + 49)

= 13 + 49 - 13 + 135 - 49

= (13 - 13) + (49 - 49) +135

= 135                          

tự tính 

~HT~

a/ 60x45-60x15-45x60+45x15

=(60x45-60x45)-(60x15-45x15)=0-15x(60-45)=-15x15=-225

b/68x12+68x32-32x68-12=68x12-12=12x67=804

c/-17x33+17x5+33x17-33x5

=5x(17-33)=5x(-16)=-80

aq)

(x+3)(y+2)=1

=> x+3 và y+2 thuộc Ư(1)={-1,1}

Ta có bảng :

Vậy ...

b) (2x-5)(y-6)=17

=> 2x-5 và y-6 thuộc Ư(17)={-1,-17,1,17}

Ta có bảng :

Vậy...

c) (2x+1)(3y-2)=-55

=> 2x+1 ; 3y-2 thuộc Ư(55)={-1,-5,-11,-55,1,5,11,55}

Ta có bảng :

Vậy ....

a) (x + 3)(y + 2) = 1

Vì (x + 3)(y + 2) = 1

=> Ta có 1 = 1.1 = (-1) . (-1)

Lập bảng xét 2 trường hợp :

Vậy các cặp (x,y) thỏa mãn là : (-2 ; - 1) ; (- 1 ; - 3)

b) (2x - 5)(y - 6) = 17

Vì (2x - 5)(y - 6) = 17

=> Ta có 17 = 1.17 = (-1).(-17)

Lập bảng xét 4 trường hợp:

Vậy các cặp (x;y) thỏa mãn là : (3;11) ; (11;7) ; (2; - 11) ; (-6 ; 5)

c) (2x + 1)(3y - 2) = 55

Vì (2x + 1)(3y - 2) = 55

=> Ta có : 55 = 11.5 = (-11) . (- 5) = 1.55 = (- 1) . (- 55) 

Lập bảng xét 8 trường hợp: 

Vậy các cặp (x,y) thỏa mãn là :(-6; - 1) ; (-3; - 3) ; (0;19); (27;1)

a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)

b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)

c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)

d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)

e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)

f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)

g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)

h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)

i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)

k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)

Cho x^2_60x+900=0

Suy ra:x^2_2.x.30+30²=0

(x-30)²=0

suy ra x-30=0

vậy x=30

ai nhanh nhất mình k cho

Bài 1:

\(A=\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+\frac{1}{\left(1+2+3+4\right)}\)\(+\frac{1}{\left(1+2+3+4+5\right)}+...+\)\(\frac{1}{\left(1+2+3+...+10\right)}\)

\(A=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)

\(A=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)

\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(A=\frac{9}{11}\)

Bài 2 :

2)  Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15

= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15

Mẫu số =  11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19  + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17 lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17  

=  (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x  17 = (1+27+125+729) x 13 x 15 x 17 

\(\Rightarrow A< \frac{\left(1+27+125+729\right)\times11\times13\times15}{\left(1+27+125+729\right)\times13\times15\times17}\)

\(=\frac{11}{17}\)

\(=\frac{1111}{1717}=B\)

Vậy \(A=B\)

mình đâu biết

Answer:

Bài 1: 

\(\left(135-35\right).\left(-37\right)+37.\left(-42-58\right)\)

\(=100.\left(-37\right)+37.\left(-100\right)\)

\(=\left(-3700\right)+\left(-3700\right)\)

\(=-7400\)

\(-65.\left(87-17\right)-87.\left(17-65\right)\)

\(=\left(-65\right).87-\left(-65\right).17-87.17+87.65\)

\(=-\left(-65\right).17-87.17\)

\(=17.\left(65-87\right)\)

\(=-374\)

\([3.\left(-2\right)-\left(-8\right)].\left(-7\right)-\left(-2\right).\left(-5\right)\)

\(=[\left(-6\right)+8].\left(-7\right)-\left(-2\right).\left(-5\right)\)

\(=2.\left(-7\right)-10\)

\(=-24\)

Bài 2:

\(\left(-55\right).\left(-25\right).\left(-8\right)\)

\(=\left(-55\right).[\left(-25\right).\left(-8\right)]\)

\(=\left(-55\right).200\)

\(=-11000\)

\(\left(-1\right).\left(-2\right).\left(-3\right).\left(-4\right).\left(-5\right).10\)

\(=[\left(-1\right).10].[\left(-2\right).\left(-5\right)].[\left(-3\right).\left(-4\right)]\)

\(=\left(-10\right).10.12\)

\(=-1200\)

a.

\(\Leftrightarrow\sqrt[3]{3x-5}=\left(2x-3\right)^3+2x-3-\left(3x-5\right)\)

Đặt \(\left\{{}\begin{matrix}2x-3=a\\\sqrt[3]{3x-5}=b\end{matrix}\right.\)

\(\Rightarrow b=a^3+a-b^3\)

\(\Leftrightarrow a^3-b^3+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt[3]{3x-5}=2x-3\)

\(\Leftrightarrow3x-5=\left(2x-3\right)^3\)

\(\Leftrightarrow8x^3-36x^2+51x-22=0\)

\(\Leftrightarrow\left(x-2\right)\left(8x^2-20x+11\right)=0\)

\(\Leftrightarrow...\)

b.

\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x+3x-2-\sqrt[3]{81x-8}=0\)

\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x+\dfrac{\left(3x-2\right)^3-\left(81x-8\right)}{\left(3x-2\right)^2+\left(3x-2\right)\sqrt[3]{81x-8}+\sqrt[3]{\left(81x-8\right)^2}}=0\)

\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x+\dfrac{27\left(x^3-2x^2-\dfrac{5}{3}x\right)}{\left(3x-2\right)^2+\left(3x-2\right)\sqrt[3]{81x-8}+\sqrt[3]{\left(81x-8\right)^2}}=0\)

\(\Leftrightarrow\left(x^3-2x^2-\dfrac{5}{3}x\right)\left(1+\dfrac{27}{\left(3x-2\right)^2+\left(3x-2\right)\sqrt[3]{81x-8}+\sqrt[3]{\left(81x-8\right)^2}}\right)=0\)

\(\Leftrightarrow x^3-2x^2-\dfrac{5}{3}x=0\)

c.

\(\Leftrightarrow\sqrt[3]{x-2}=\left(2x-5\right)^3+x-3\)

\(\Leftrightarrow\sqrt[3]{x-2}=\left(2x-5\right)^3+\left(2x-5\right)-\left(x-2\right)\)

Đặt \(\left\{{}\begin{matrix}2x-5=a\\\sqrt[3]{x-2}=b\end{matrix}\right.\)

\(\Rightarrow b=a^3+a-b^3\)

\(\Leftrightarrow a^3-b^3+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow2x-5=\sqrt[3]{x-2}\)

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

\(\Leftrightarrow\left(x-3\right)\left(8x^2-36x+41\right)=0\)

Bài 1:

a) 17-25+55-17

=(17-17)+(-25+55)

=0+30

=30

b.(-37+101)-(91+37)

= -37+101-91-37

=(-37-37)+(101-91)

= -74 + 10

=-64

c.(-4-6):(-5)

=(-10) : (-5)

=2

d)35-(-95)+372-(372+95)

=35+95+372-372-95

=35+(95-95) + ( 372-372)

=35

Bài 2 :

2x²-(-17)=35

2x² +17=35

2x²=35-17

2x²=18

x²=9

x=3

-5|x|=-70:2

-5|x|=-35

|x|==-35:-5

|x|=7

=>\(\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

Vậy \(x\in\left\{7;-7\right\}\)