\(a.97-7\left(x-8\right)=3\cdot2^4\)

\(97-7\left(x-8\right)=48\)

\(7\left(x-8\right)=97-48\)

\(7\left(x-8\right)=49\)

\(x-8=49:7=7\)

\(x=7+8=15\)

\(b.5x^2-74=51\)

\(5x^2=51+74=125\)

\(x^2=125:5=25\)

\(\Rightarrow x=\pm5\)

\(c.3x+2-3x=216\)

\(0x=214\)

⇒ \(x\in\) rỗng

\(5\left(3x-4\right)^3=40\)

\(\left(3x-4\right)^3=40:5=8\)

⇒ 3x - 4 = 2

⇒ 3x = 2 + 4 = 6

⇒ x = 6 : 3 = 2

a.97−7(x−8)=3⋅24

\(97 - 7 \left(\right. x - 8 \left.\right) = 48\)

\(7 \left(\right. x - 8 \left.\right) = 97 - 48\)

\(7 \left(\right. x - 8 \left.\right) = 49\)

\(x - 8 = 49 : 7 = 7\)

\(x = 7 + 8 = 15\)

\(b . 5 x^{2} - 74 = 51\)

\(5 x^{2} = 51 + 74 = 125\)

\(x^{2} = 125 : 5 = 25\)

\(\Rightarrow x = \pm 5\)

\(c . 3 x + 2 - 3 x = 216\)

\(0 x = 214\)

⇒ \(x \in\) rỗng

\(5 \left(\left(\right. 3 x - 4 \left.\right)\right)^{3} = 40\)

\(\left(\left(\right. 3 x - 4 \left.\right)\right)^{3} = 40 : 5 = 8\)

⇒ 3x - 4 = 2

⇒ 3x = 2 + 4 = 6

⇒ x = 6 : 3 = 2

a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)

b: =(-4)+(-4)+...+(-4)

=-4*25=-100

c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)

=10*53

=530

a, 5^x+7 = 125

<=> 5^x+7 = 5³

=> x+7 = 3

<=> x = 3-7

<=> x = -4

b,

<=> 7^-10-x = 50-1

<=> 7^-10-x = 49

<=> 7^-10-x = 7²

=> -10-x = 2

<=> x = -10-2

<=> x = -12

c,

<=> x = 35-45

<=> x = -10

d,

<=> 6x = -5,6

<=> x = -5,6 ( Vô lí vì -5,6 không chia hết cho 6)

Vậy không tồn tại giá trị của x thỏa mãn đề bài

e,

<=> 7-10 = 3x-2x

<=> x = -3

f, 

<=> -15-9 = 3x-2x

<=> x = -24

tìm 2 số nguyên a và b biết :a+b=-1 và a.b=-12.Giup mình nha

\(-3x+2.\left[45-x-3\left(3x+7\right)-2x\right]+4x=-55\)

\(-3x+2.\left[45-x-3.3x+7-2x\right]+4=-55\)

\(-3x+2.45-x-6x+7-2x+4x=-55\)

\(-3x+90-x-6x+7-2x+4x=-55\)

\(x.\left(-3+90-6+7-2+4\right)=-55\)

\(x.90=-55\)

\(x=-\frac{11}{18}\)

B=51ⁿ +12¹⁰²:10
= .....1 + .......6 :10
=..........7 không chia hết cho 10( vô lí)

Ta có:

\(A=51^n+47^{102}\)

\(\Rightarrow A=\overline{...1}+47^{100}.47^2\)

\(\Rightarrow A=\overline{...1}+\left(47^4\right)^{25}.\left(\overline{...9}\right)\)

\(\Rightarrow A=\overline{...1}+\left(\overline{...1}\right)^{25}.\left(\overline{...9}\right)\)

\(\Rightarrow A=\overline{...1}+\left(\overline{...1}\right).\left(\overline{...9}\right)\)

\(\Rightarrow A=\overline{...1}+\overline{...9}\)

\(\Rightarrow A=\overline{...0}\)

Vì \(\overline{....0}\text{⋮}10\) nên \(A\text{⋮}10\)

Vậy \(A\text{⋮}10\left(đpcm\right)\)

phần a dễ bạn tự làm đi tử thì bạn tính như bình thường còn mẫu thì:7.(\(\frac{1}{3.13}\)+\(\frac{1}{13.23}\)+\(\frac{1}{23.33}\))

\(\frac{7}{10}\).(\(\frac{1}{3}\)-\(\frac{1}{33}\))=\(\frac{7}{33}\)

b)(1+1/3+1/5+..+1/199)-(1/2+1/4+...+1/200)

(1+1/2+1/3+...+1/199+1/200)-(1/2+1/2+1/4+1/4+...+1/200+1/200)

=1+1/2+1/3+...+1/199+1/200-(1+1/2+1/3+...+1/100)

=1/101+1/102+...+1/200

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