ngu như con lợn

a/ Ta có :

\(12^8=\left(12^2\right)^4=24^4\)

\(8^{12}=\left(8^3\right)^4=512^4\)

Vì \(24^4< 512^4\Leftrightarrow12^8< 8^{12}\)

b/ Ta có :

\(\left(-5\right)^{39}=\left[\left(-5\right)^3\right]^{13}=\left(-125\right)^{13}\)

\(\left(-2\right)^{91}=\left[\left(-2\right)^7\right]^{13}=\left(-128\right)^{13}\)

Vì \(\left(-125\right)^{13}>\left(-128\right)^{13}\Leftrightarrow\left(-5\right)^{39}>\left(-2\right)^{91}\)

a) Ta có: 12⁸ = (3.4)⁸ = 3⁸.4⁸ = 3⁸.(2²)⁸ = 3⁸.2¹⁶

8¹² = (2³)¹² = 2³⁶ = 2²⁰.2¹⁶ = (2²)¹⁰.2¹⁶ = 4¹⁰.2¹⁶

Vì 3⁸.2¹⁶ < 4¹⁰.2¹⁶

=> 12⁸ < 8¹²

b) (-5)³⁹ = -5³⁹ = -(5³)¹³ = -125¹³

(-2)⁹¹ = -2⁹¹ = -(2⁷)¹³ = -128¹³

Vì 125¹³ < 128¹³

=> -125¹³ > -128¹³

=> (-5)³⁹ > (-2)⁹¹

Ta có A = 2018.2020 + 2019.2021

= (2020 - 2).2020 + 2019.(2019 + 2) 

= 2020² - 2.2020 + 2019² + 2.2019

= 2020² + 2019² - 2(2020 - 2019) = 2020² + 2019² - 2 = B

=> A = B

b) Ta có B = 9⁶⁴ - 1= (9³²)² - 1² 

= (9³² + 1)(9³² - 1) = (9³² + 1)(9¹⁶ + 1)(9¹⁶ - 1) = (9³² + 1)(9¹⁶ + 1)(9⁸ + 1)(9⁸ - 1) 

= (9³² + 1)(9¹⁶ + 1)(9⁸ + 1)(9⁴ + 1)(9⁴ - 1) 

= (9³² + 1)(9¹⁶ + 1)(9⁸ + 1)(9⁴ + 1)(9² + 1)(9² - 1) 

  (9³² + 1)(9¹⁶ + 1)(9⁸ + 1)(9⁴ + 1)(9² + 1).80 

mà A =   (9³² + 1)(9¹⁶ + 1)(9⁸ + 1)(9⁴ + 1)(9² + 1).10

=> A < B

c) Ta có A = \(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x^2-y^2}{x^2+2xy+y^2}< \frac{x^2-y^2}{x^2+xy+y^2}=B\)

=> A < B

d) \(A=\frac{\left(x+y\right)^3}{x^2-y^2}=\frac{\left(x+y\right)^3}{\left(x+y\right)\left(x-y\right)}=\frac{\left(x+y\right)^2}{x-y}=\frac{x^2+2xy+y^2}{x-y}< \frac{x^2-xy+y^2}{x-y}=B\)

=> A < B

câu a tự quy đồng cùng  mẫu rồi làm thôi :"))

b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)

\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)

Đặt \(x^2-x=k\), ta có:

\(k.\left(k-2\right)=24\)

\(\Leftrightarrow k^2-2k+1=25\)

\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)

\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)

\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)

c)\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)

\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)

p/s: bn tự kết luận nha :))

Có gấp lắm ko ?

có gấp lắm

Bài này mình cũng chịu, mới học lớp 6

Để mình giải thử:

( -5 )³⁹ = ( - 5¹³ )³

( - 2 )⁹¹ = ( - 2¹³ )⁷

Sau đó giải tiếp

Bài1:

\(a,\left(-8\right)^9\) và \(\left(-32\right)^5\)

Ta có:

\(\left(-8\right)^9=-2^{27}\)

\(\left(-32\right)^5=\left(-8.4\right)^5=-2^{27}.2^{10}\)

Vì \(-2^{27}.10< -2^{27}\) nên \(\left(-8\right)^9>\left(-32\right)^5\)

Các câu sau tương tự

Bài2:

\(a,2\left|x-1\right|-3x=7\)

+)Xét \(x\ge1\Rightarrow\left|x-1\right|=x-1\)

Do đó:

\(2\left(x-1\right)-3x=7\\ \Leftrightarrow2x-2-3x=7\\ \Leftrightarrow-x=9\\ \Leftrightarrow x=-9\left(loại\right)\)

+)Xét \(x< 1\Rightarrow\left|x-1\right|=1-x\)

Do đó:

\(2\left(1-x\right)-3x=7\\ \Leftrightarrow2-2x-3x=7\\ \Leftrightarrow-5x=5\\ x=-1\left(chon\right)\)

Vậy x=-1

Câu b tương tự

Bài 1:

\(a,\left(-8\right)^9\) và \(\left(-32\right)^5\)

\(\left(-8\right)^9=\left[\left(-2\right)^3\right]^9=\left(-2\right)^{27}\)

\(\left(-32\right)^5=\left[\left(-2\right)^5\right]^5=\left(-2\right)^{25}\)

\(\left(-2\right)^{27}< \left(-2\right)^{25}\)

\(\Rightarrow\left(-8\right)^9< \left(-32\right)^5\)

\(b,2^{21}\) và \(3^{14}\)

\(2^{21}=\left(2^3\right)^7\)

\(3^{14}=\left(3^2\right)^7\)

\(2^3< 3^2\)\(\Rightarrow2^{21}< 3^{14}\)

\(c,12^8\) và \(8^{12}\)

\(12^8=\left(12^2\right)^4=144^4\)

\(8^{12}=\left(8^3\right)^4=512^4\)

\(144^4< 512^4\)\(\Rightarrow12^8< 8^{12}\)

\(d,\left(-5\right)^{39}\) và \(\left(-2\right)^{91}\)

\(\left(-5\right)^{39}=\left[\left(-5\right)^3\right]^{13}\)

\(\left(-2\right)^{91}=\left[\left(-2\right)^7\right]^{13}\)

\(\left(-5\right)^3>\left(-2\right)^7\)\(\Rightarrow\left(-5\right)^{39}>\left(-2\right)^{91}\)

Bài 2:

\(a,2.\left|x-1\right|-3x=7\)

\(\left|x-1\right|=\dfrac{7+3x}{2}\)

Ta có 2 trường hợp:

Th1:\(x-1=\dfrac{7-3x}{2}\)

\(\dfrac{2x-2}{2}=\dfrac{7+3x}{2}\)

\(\Rightarrow2x-2=7+3x\)

\(2x-3x=7+2\)

\(-x=9\Rightarrow x=-9\)

Th2:\(x+1=-\dfrac{7+3x}{2}\)

\(\dfrac{2x-2}{2}=\dfrac{-7-3x}{2}\)

\(\Rightarrow2x-2=-7-3x\)

\(2x+3x=-7+2\)

\(5x=-5\Rightarrow x=-1\)

Vậy \(x\in\left\{-9;-1\right\}\)

\(b,\left|5x-3\right|=\left|7-x\right|\)

Ta có: Th1: \(\left|7-x\right|=7-x\) khi \(7-x\ge0\)\(\Rightarrow x\le7\)

\(5x-3=7-x\)

\(5x+x=7+3\)

\(6x=10\Rightarrow x=\dfrac{10}{6}=\dfrac{5}{3}\)( thoả mãn )

vì x thoả mãn \(x\le7\)\(\Rightarrow\) th1 thoả mãn x

Ta có: Th2: \(\left|7-x\right|=-\left(7-x\right)\) khi \(7-x< 0\Rightarrow x>7\)

\(5x-3=-\left(7-x\right)\)

\(5x-3=-7+x\)

\(5x-x=-7+3\)

\(4x=-4\Rightarrow x=-1\) ( loại )

Vì x thoả mãn \(x>7\) mà \(x=-1\Rightarrow\)th2 loại

bài 1

a) (-8)⁹ và (-32)⁵

=[(-2)³]⁹ và [(-2)⁵]⁵

=(-2)²⁷ và (-2)²⁵

vì 27 >25

=>(-2)²⁷<(-2)²⁵ hay(-8)⁹ <(-32)⁵