Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

mình làm theo cách lớp 12 nhé 

Lần sau gõ Latex cho dễ nhìn nhé em! :)

\(2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2019-8\\ \Leftrightarrow2^x\cdot\left(1+2+2^2+...+2^{2015}\right)=2011\)

Ta thấy vế trái chia hết cho 2 nhưng vế phải chia 2 dư một nên không tồn tại giá trị của x thỏa mãn đề bài.

Có thể trình bày giải từng bước ra cho mình ko

Thêm nữa câu a) Tính: M(x) + N(x)+ P(x)

B) Tính M(x) - N (x) - P(x)

ok rồi giúp mình với nha

  Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

Ta có : $2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-8$

$\iff 2^x\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-8$ (1)

Đặt : $A=1+2+2^2+...+2^{2015}$

$\Rightarrow 2A=2+2^2+2^3+...+2^{2016}$

$\Rightarrow 2A-A=\left(2+2^2+2^3+...+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)$

$\Rightarrow A=2^{2016}-1$

Khi đó (1) trở thành :

$2^x\left(2^{2016}-1\right)=2^{2019}-2^3$

$\iff 2^x\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)$

$\iff 2^x=2^3\left(2^{2016}-1\ne 0\right)$

$\iff x=3$

Vậy : $x=3$

$2^x+2^{x+1}+...+2^{x+2015}=2^{2019}-8$

$\to 2^x.1+2^x.2+....+2^x.2^{2015}=2^{2019}-8$

$\to 2^x.(1+2+...+2^{2015})=2^{2019}-8$

Đặt:

$A=1+2+...+2^{2015}$

$2A=2.(1+2+...+2^{2015})$

$2A=2+2^2+...+2^{2016}$

$2A-A=(2+2^2+...+2^{2016})-(1+2+...+2^{2015})$

$A=2+2^2+...+2^{2016}-1-2-...-2^{2015}$

$A=2^{2016}-1$

Nên:

$2^x.(1+2+...+2^{2015})=2^{2019}-8$

$\to 2^x.(2^{2016}-1)=2^{2019}-8$

$\to 2^x=(2^{2019}-8):(2^{2016}-1)$

$\to 2^x=\frac{2^{2019}-8}{2^{2016}-1}$

$\to 2^x=\frac{2^3.(2^{2016}-1)}{2^{2016}-1}$

$\to 2^x=2^3$

$\to x=3$

Vậy $x=3.$

Bài 2: 

a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)

=>(x+5)(x-6)=0

=>x=-5 hoặc x=6

b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

=>-4x+2=0

hay x=1/2

c: \(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)

=>x=1 hoặc x=-1

b. 1404 : [118 - (4x + 6)] = 27

118 - (4x + 6) = 52

4x + 6 = 66

4x = 60

x = 15

d) \(5x^2-3x=0\)

\(\Leftrightarrow x\left(5x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\5x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{5}\end{cases}}\)

e) \(3\left(x-1\right)+4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[3-4.\left(x-1\right)\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\3-4\left(x-1\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\4\left(x-1\right)=3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x-1=\frac{3}{4}\Rightarrow x=\frac{7}{4}\end{cases}}\)

f) \(2\left(x-2\right)^2=\left(x-2\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2\left(x-2\right)-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x-2=\frac{1}{2}\Rightarrow x=\frac{5}{2}\end{cases}}\)

g) \(\left(x-2020\right)^4=\left(x-2020\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-2020\right)^2=0\\\left(x-2020\right)^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=2019,x=2021\end{cases}}\)