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

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

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

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

a) \(\frac{x^7}{3^5}=9\)

=> \(\frac{x^7}{3^5}=3^2\)

=> x⁷ = 3² . 3⁵ = 3⁷

=> x = 3

b) \(\frac{32}{x^7}=4\)

=> \(\frac{2^5}{x^7}=2^2\)

=> \(2^2\cdot x^7=2^5\)

=> \(x^7=\frac{2^5}{2^2}=2^3\)

=> không tìm được x

2. \(\frac{2^{30}}{3^{20}}-\left(\frac{2}{3}\right)^{20}\cdot2^{10}\)

\(=\frac{2^{30}}{3^{20}}-\frac{2^{20}}{3^{20}}\cdot2^{10}\)

\(=\frac{2^{30}}{3^{20}}-\frac{2^{30}}{3^{20}}=0\)

bài 1 có ý d nha các bạn mình viết thiếu

Bài dái quá, bạn nên tách ra đi nhé!

* dài

\(a,A=\left(x+10\right)+\left(2x-15\right)-\left(x-20\right)\)

       \(=x+10+2x-15-x+20\)

       \(=2x+15\)

\(b,\)Thay \(x=15\)vào biểu thức A = 2x + 15

Ta được : \(A=2\times15+15\)

\(\Rightarrow A=45\)

a) A = (x + 10) + (2x - 15) - (x - 20)

A = x + 10 + 2x - 15 - x + 20

A = x + 10 + 2x + (-15) + (-x) + 20

A = x + (-x) + 2x + 10 + 20 + (-15)

A = 2x + 15

b) x = 15 thì A = 2x + 15 = 2.15 + 15 = 15.3 = 45

Đặt \(g(x)=10x\).

Ta có \(g\left(1\right)=10=f\left(1\right);g\left(2\right)=20=f\left(2\right);g\left(3\right)=30=f\left(3\right)\).

Từ đó \(\left\{{}\begin{matrix}f\left(1\right)-g\left(1\right)=0\\f\left(2\right)-g\left(2\right)=0\\f\left(3\right)-g\left(3\right)=0\end{matrix}\right.\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=Q\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\).

\(\Rightarrow f\left(x\right)=10x+Q\left(x\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

\(\Rightarrow f\left(8\right)+f\left(-4\right)=80+Q\left(x\right).7.6.5+\left(-40\right)+Q\left(x\right).\left(-5\right).\left(-6\right).\left(-7\right)=80-50=40\).

a)  (a-b)-(c-b)-a=a-b-c+b-a=(a-a)+(b-b)-c=0+0-c=-c

1)

a) (a-b) - (c-b) - a= a - b - c + b - a = -c

b) -(300 - 400) + (300 - 400) + 100 = 100

2) a) 4x - 20 =  50 - ( 30 -3x)

        4x - 20 = 50 - 30 + 3x

        4x - 3x = 50 - 30 + 20= 40

        x          = 40

b) 100 - (-5x) = 40 - (-4x +10)

    100 + 5x    = 40 + 4x -10

    5x - 4x       = 40 - 10 -100

    x                = -70

c) (-50) + (-7x) = 100 - 8x

     -7x + 8x      = 100 + 50

      x                = 150

Bài 1.

a) \(\left(a-b\right)-\left(c-b\right)-a\)

\(=a-b-c+b-a\)

\(=\left(a-a\right)+\left(-b+b\right)-c\)

\(=-c\)

b)\(-\left(300-400\right)+\left(300-400\right)+100\)

\(=-300+400+300-400+100\)

\(=\left(-300+300\right)+\left(400-400\right)+100\)

\(=100\)

Bài 2.

a)\(4x-20=50-\left(30-3x\right)\)

\(\Leftrightarrow4x-20=50-30+3x\)

\(\Leftrightarrow4x-20=20+3x\)

\(\Leftrightarrow x=40\)

Vậy x = 40

b)\(100-\left(-5x\right)=40-\left(-4x+10\right)\)

\(\Leftrightarrow100+5x=40+4x-10\)

\(\Leftrightarrow x=-70\)

Vậy x = -70

c)\(\left(-50\right)+\left(-7x\right)=100-8x\)

\(\Leftrightarrow-50-7x=100-8x\)

\(\Leftrightarrow x=150\)

Vậy x = 150

Linz

Đặt \(g\left(x\right)=f\left(x\right)-10\) (bậc 4)

\(\Leftrightarrow\left\{{}\begin{matrix}g\left(1\right)=0\\g\left(2\right)=0\\g\left(3\right)=0\end{matrix}\right.\Leftrightarrow g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right)\) (m là hằng số)

\(\Leftrightarrow f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right)-10\\ \Leftrightarrow f\left(9\right)=8\cdot7\cdot6\left(9-m\right)-10=336\left(9-m\right)-10\\ f\left(-5\right)=\left(-6\right)\left(-7\right)\left(-8\right)\left(-5-m\right)-10=336\left(m+5\right)-10\)

Vậy \(A=336\left(9-m\right)+336\left(m+5\right)-20=4684\)

Chúc bạn hok tốt <3

Bài 1:

6a+1 \(⋮\)2a-1

=> 2a-1\(⋮\)2a-1

=> (6a+1)- 3(2a-1) \(⋮\)2a-1

=> (6a+1) - ( 6a-3) \(⋮\)2a-1

=> 6a+1 -6a+3\(⋮\)2a-1

=> 4 \(⋮\)2a-1

=> 2a-1\(\in\)Ư(4)

Còn j bn làm nốt nhoaaa

1. Ta có \(6a+1⋮2a-1\)

\(\Rightarrow3\left(2a-1\right)+4⋮a-1\)

\(\Rightarrow4⋮a-1\)

\(\Rightarrow a-1\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow a\in\left\{-3;-1;0;2;3;5\right\}\)  ( thỏa mãn a nguyên )

Vậy \(a\in\left\{-3;-1;0;2;3;5\right\}\)

2. a,     x - y - ( - y + a + x)

= x - y + y - a - x

= - a

b, (-90) - (b + 10) + 100

= - 90 - b - 10 + 100

= ( - 90 - 10 +100) - b

= 0 - b

= - b

@@ Học tốt

Bài 1 t sai r 

Cậu tham khảo của bn kia nhá

2a - 1 \(\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow2a\in\left\{-3;-1;0;2;3;5\right\}\)

\(\Rightarrow a\in\left\{\frac{-3}{2};\frac{-1}{2};0;1;\frac{3}{2};\frac{5}{2}\right\}\)

Mà a nguyên 

\(\Rightarrow a\in\left\{0;1\right\}\)

Vậy \(a\in\left\{0;1\right\}\)

@@ Học tốt