A, 7[x + 5] - 20 = 190

7x + 35 - 20 = 190

7x + 15 = 190

7x = 175

x = 25

B, 155 - 10[x + 1] = 55

155 - 10x - 10 = 55

-10x + 90 = 55

-10x = -35

x = 3.5

C, 6[x + 2^3] + 40 = 100

6[x + 8] + 40 = 100

6x + 48 + 40 = 100

6x + 88 = 100

6x = 1

2 x = 2

D, 15x - 133 = 17

15x = 150

x = 10

E, 90[x + 2] = 45

90x + 180 = 45

90x = -135

x = -1.5

F, 4x + 54 = 82

4x = 28

x = 7

G, 17x - 20 = 14

17x = 34

x = 2

a: =39-13-62-39=-75

b: =(-2)+(-2)+(-2)=-6

c: 92-55+8-45=100-100=0

d: M={-20;-19;-18;...;18;19;20}

Tổng là (-20)+(-19)+(-18)+...+18+19+20=0

giúp mik với

a) 2² ( x + 3² ) - 5 = 55

=>4(x+9)-5=55

=>4(x+9)=60

=>x+9=15

=>x=6

b)7 ( x + 5² ) - 20 = 190

=>7(x+25)=210

=>x+25=30

=>x=5

c)6 ( x + 2³ ) + 40 = 100

=>6(x+8)+40=100

=>6(x+8)=60

=>x+8=10

=>x=2

a)6

b)5

c)2

6;5;2

Bài 2:

a: \(387+\left(-224\right)+\left(-87\right)\)

\(=\left(387-87\right)+\left(-224\right)\)

=300-224

=76

b: \(39+\left(13-26\right)-\left(62+39\right)\)

\(=39+13-26-62-39\)

\(=\left(39-39\right)+\left(13-26-62\right)\)

=0-75

=-75

c: \(32-34+36-38+40-42\)

\(=\left(32-34\right)+\left(36-38\right)+\left(40-42\right)\)

\(=\left(-2\right)+\left(-2\right)+\left(-2\right)\)

=-6

d: \(92-\left(55-8\right)+\left(-45\right)\)

\(=92-55+8-45\)

\(=\left(92+8\right)-\left(55+45\right)\)

=100-100

=0

e: -20<=x<=20

mà x nguyên

nên \(x\in\left\{-20;-19;-18;...;18;19;20\right\}\)

=>M={-20;-19;-18;-17;...;18;19;20}

Tổng các phần tử của M là:

\(\left(-20\right)+\left(-19\right)+\left(-18\right)+\left(-17\right)+...+18+19+20\)

\(=\left(-20+20\right)+\left(-19+19\right)+...+\left(-2+2\right)+\left(-1+1\right)+0\)

=0+0+...+0

=0

a)55.(70 -87) - 85.(70-55)

  =55.70- 55.87 -85.70- 85.55

  =(55.70- 85.70)- (55.85- 85.55)

 =70(55-85)- 0

=70.(-30)=-2100

X+21+(-18)+(-40)=x+21-18-40=x-37

-55-(y+

a) x + 21 + (-18) + (-40)

= x + 21 - 18 - 40

= x - 37

b) -55 - (y + 20) - (-55)

= -55 -y - 20 + 55

= - y - 20

c) -(a - b) + ( a - c -d + b )

= - a + b + a -c - d +b 

= 2b - c - d

-55-(y+20)-(-55)=-55-y-20+55=0-y-20=-(y+20)

-(a-b)+(a-c-d+b)

=2b-c-d

mNaOH=11(g) -> nNaOH= 0,275(mol)

mH3PO4=9,8(g) -> nH3PO4=0,1(mol)

Ta có: 2< nNaOH/nH3PO4 = 0,275/0,1=2,75< 3 

=> P.ứ kết thúc thu được hỗn hợp dd Na3PO4 và Na2HPO4

PTHH: 3 NaOH + H3PO4 -> Na3PO4 + 3 H2O

3x_____________x________x(mol)

2 NaOH + H3PO4 -> Na2HPO4 +2 H2O

2y_____y__________y(mol)

mddX=55+24,5=79,5(g)

\(\left\{{}\begin{matrix}3x+2y=0,275\\x+y=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,075\\y=0,025\end{matrix}\right.\)

=> mNa3PO4=0,075.164=12,3(g)

mNa2HPO4=142.0,025=3,55(g)

=>C%ddNa3PO4=(12,3/79,5).100=15,472%

C%ddNa2HPO4=(3,55/79,5).100=4,465%

Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=\dfrac{55\cdot20\%}{40}=0,275\left(mol\right)\\n_{H_3PO_4}=\dfrac{24,5\cdot40\%}{98}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo HPO₄^2- và PO₄^3-

PTHH: \(2NaOH+H_3PO_4\rightarrow Na_2HPO_4+2H_2O\)

                     2a_______a___________a_______2a   (mol)

           \(3NaOH+H_3PO_4\rightarrow Na_3PO_4+3H_2O\)

                     3b_______b_________b______3b    (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}2a+3b=0,275\\a+b=0,1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,025\\b=0,075\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2HPO_4}=\dfrac{0,025\cdot142}{55+24,5}\cdot100\%\approx4,47\%\\C\%_{Na_3PO_4}=\dfrac{0,075\cdot164}{55+24,5}\cdot100\%\approx15,47\%\end{matrix}\right.\)