1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅

3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1

5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)

6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅

7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅

8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1

9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)

\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)

\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

Câu 3, 4 tương tự nhé.

\(5.\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Leftrightarrow\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-5-1\right)\left(x-5+1\right)=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-6\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x=6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

Vậy \(x\in\left\{4;5;6\right\}\)

toi ko co the bt day nh vau ko dau

Bài 1: 

a) 0²⁰⁰² < 0²⁰²³

b) 2022⁰ = 2023⁰

c) 54⁹ < 55¹⁰

d) ( 4 + 5 )³ > 4² + 5²

đ) 9² - 3² > ( 9 - 3 )²

Bài 2:

a) 3² x 4³ - 3² + 333

= 9 x 64 - 9 + 333

= 576 - 9 + 333

= 567 + 333

= 900

b) 5 x 4³ + 24 x 5 + 41⁰

= 5 x 64 + 24 x 5 + 1

= 5 x ( 64 + 24 ) + 1

= 5 x 88 + 1

= 440 + 1

= 441

c) 2³ x 4² + 3² x 5 - 40 x 1²⁰²³

= 8 x 16 + 9 x 5 - 40 x 1

= 128 + 45 - 40

= 133

Bài 1 :

a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)

b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)

c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)

d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)

đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)

Bài 2 :

a) \(3^2.4^3-3^2+333=3^2\left(4^3-1\right)+9.37=9.63+9.37=9\left(63+37\right)=9.100=900\)

b) \(5.4^3+24.5+41^0=20.4^2+20.6+1=20\left(16+6\right)+1=20.22+1=441\)

c) \(2^3.4^2+3^2.5-40.1^{2023}=8.16+9.5-40.1=128+45-40=128+5=133\)

1.(x -5)^2 - 25 =0

=> (x - 5)^2 = 25

=> x - 5 = 5 hoặc x - 5 = -5

=> x = 10 hoặc x = 0

vậy_

2. (x -2)^3 =27

=> x - 2 = 3

=> x = 5

vậy_

3. 3(x -7) + 2x(x+2) = 2x^2

=> 3x - 21 + 2x^2 + 4x = 2x^2

=> 7x - 21 = 0

=> 7x = 21

=> x = 3

vậy_

4. (x^2 - 4) (x +8) =0

=> x^2 - 4 = 0 hoặc x + 8 = 0

=> x^2 = 4 hoặc x = -8

=> x = 2 hoặc x = -2 hoặc x = -8

vậy_

5. x^ 2 + 3x = 0

=> x(x + 3) = 0 

=> x = 0 hoặc x + 3 = 0

=> x = 0 hoặc x = -3

vậy_

6. 3x^3 - 3x = 0

=> 3x(x^2 - 1) = 0

=> 3x(x - 1)(x + 1) = 0

=> x = 0 hoặc x = 1 hoặc x = -1

vậy_

7. (x +1)^2 = ( 2x +3)^2

=> (x + 1 + 2x + 3)(x + 1 - 2x - 3) = 0

=> (3x + 3)(-x - 2) = 0

=> x = -1 hoặc x = -2

vậy_

Bài làm

1) ( x - 5 )² - 25 = 0

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

<=> x( x - 10 ) = 

<=> \(\orbr{\begin{cases}x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)

Vậy S = { 0; 10 }

2) \(\left(x-2\right)^3=27\)

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

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=5\)

Vậy x = 5 là nghiệm phương trình.

3) \(3\left(x-7\right)+2x\left(x+2\right)=2x^2\)

\(\Leftrightarrow3x+2x^2+4x-2x^2=21\)

\(\Leftrightarrow7x=21\)

\(\Leftrightarrow x=\frac{21}{7}=3\)

Vậy x = 3 là nghiệm phương trình

4) \(\left(x^2-4\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=\pm2\\x=-8\end{cases}}}\)

Vậy S = { 2; -2; -8 }

5) \(x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

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

Vậy S = { 0; -3 } 

6) \(3x^3-3x=0\)

\(\Leftrightarrow3x\left(x^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)

Vậy S = { +1; 0 }

7) \(\left(x+1\right)^2=\left(2x+3\right)^2\)

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

\(\Leftrightarrow\left(x+1-2x-3\right)\left(x+1+2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x-2=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}}\)

Vậy S = { -2; -4/3 }

# Học tốt #

Nên đợi ai đó giải hết 2 3 bài xong rồi mới đăng tiếp những bài còn lại, chứ dài vậy giải hơi nản =)))

Bài 1: 
1, \(13\frac{2}{5}-\left(\frac{18}{32}-2\frac{6}{10}\right)\)
\(=\frac{67}{5}-\left(\frac{9}{16}-\frac{13}{5}\right)\)(Chuyển hỗn số thành p/số và rút gọn hai số trong ngoặc luôn)
\(=\frac{67}{5}-\left(\frac{-163}{80}\right)\)
\(=\frac{246}{16}\)
2, \(22.4\frac{5}{7}-\left(8.91+1,09\right)\)(Phần 2 viết vầy có đúng không vậy ? Nếu sai thì kêu chị sửa nhé)
\(=22.\frac{33}{7}-10\)
\(=\frac{726}{7}-10\)
\(=\frac{656}{7}\)
3, Chỗ ''3 phần 10 phần 2'' là sao :v ?
4, \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)
\(=\frac{37}{7}.\frac{8}{11}+\frac{37}{7}.\frac{5}{11}-\frac{37}{7}.\frac{2}{11}\)(Chuyển hỗn số thành p/số)
\(=\frac{37}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)(Dùng tính chất phân phối)
\(=\frac{37}{7}.\frac{11}{11}\)
\(=\frac{37}{7}.1=\frac{37}{7}\)

Bài 1: 
2, \(22-4\frac{5}{7}-\left(8,91+1,09\right)\)
\(=22-\frac{33}{7}-10\)
\(=\frac{121}{7}-10\)
\(=\frac{51}{7}\)
3, \(\frac{3-\frac{1}{5}+\frac{3}{10}}{2+\frac{1}{4}-\frac{3}{5}}\)
\(=\frac{\frac{14}{5}+\frac{3}{10}}{\frac{9}{4}-\frac{3}{5}}\)
\(=\frac{\frac{31}{10}}{\frac{33}{20}}\)
\(=\frac{3,1}{1,65}\)
\(=\frac{62}{33}\)(Nếu muốn thì có thể để như vầy, còn không thì để như p/số có số thập phân ấy)

3.

a) \(\left(x-1\right)^3=125\)

=> \(\left(x-1\right)^3=5^3\)

=> \(x-1=5\)

=> \(x=5+1\)

=> \(x=6\)

Vậy \(x=6.\)

b) \(2^{x+2}-2^x=96\)

=> \(2^x.\left(2^2-1\right)=96\)

=> \(2^x.3=96\)

=> \(2^x=96:3\)

=> \(2^x=32\)

=> \(2^x=2^5\)

=> \(x=5\)

Vậy \(x=5.\)

c) \(\left(2x+1\right)^3=343\)

=> \(\left(2x+1\right)^3=7^3\)

=> \(2x+1=7\)

=> \(2x=7-1\)

=> \(2x=6\)

=> \(x=6:2\)

=> \(x=3\)

Vậy \(x=3.\)

Chúc bạn học tốt!

Giúp mk với nha các bạn

Bài 3 :

a) \(\left(x-1\right)^3=125=5^3\)

\(\Leftrightarrow x-1=5\)

\(\Leftrightarrow x=6\)

Vậy : \(x=6\)

b) \(2^{x+2}-2^x=96\)

\(\Leftrightarrow2^x.\left(2^2-1\right)=96\)

\(\Leftrightarrow2^x=96:3\)

\(\Leftrightarrow2^x=32=2^5\)

\(\Leftrightarrow x=5\)

Vậy : \(x=5\)

c) \(\left(2x+1\right)^3=343=7^3\)

\(\Leftrightarrow2x+1=7\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

Vậy : \(x=3\)

d) \(720:\left(4.\left(2x-5\right)\right)=2^3.5=40\)

\(\Leftrightarrow4.\left(2x-5\right)=18\)

\(\Leftrightarrow2x-5=\frac{18}{4}\)

\(\Leftrightarrow2x=\frac{18}{4}+5=\frac{19}{2}\)

\(\Leftrightarrow x=\frac{19}{4}\)

Vậy : \(x=\frac{19}{4}\)

Còn mấy bài kia để chiều nhé !! Giờ mình chỉ làm được vậy thôi ! Chúc bạn học tốt !

\(\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)

\(\Leftrightarrow x-\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy x = 1/2

\(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left(x-2\right)^2=1^2\)

\(\Leftrightarrow x-2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy x = 3 hoặc x = 1

\(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-2\)

<=> 2x = -1

<=> x = -0,5

Vậy x = -0,5

\(\left(x-\frac{1}{2}\right)^2=0\)

\(x-\frac{1}{2}=0\)

\(x=\frac{1}{2}\)

\(\left(x-2\right)^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)

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

\(2x-1=-2\)

\(2x=\left(-2\right)+1\)

\(2x=-1\)

\(x=-1\times2\)

\(x=-2\)

\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)

\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)

\(49\times7^x=2401\)

   \(7^x=2401:49\)

\(7^x=49\)

\(7^x=7^2\)

\(\Rightarrow x=2\)

\(2x^x-15=17\)

\(2x^x=17+15\)

\(2x^x=32\)

\(2x^x=2^5\)

\(\Rightarrow x\times x=5\)

\(\Rightarrow x^2=5\)

\(2^x+3\times2^x=32\)

\(2^x\times\left(3+1\right)=32\)

\(2^x\times4=32\)

\(2^x=32:4\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

\(2x\left(x-\frac{1}{7}\right)=0\)

\(x\left(x-\frac{1}{7}\right)=0:2\)

\(x\left(x-\frac{1}{7}\right)=0\)

\(x\times x-x\times\frac{1}{7}=0\)

\(x^2-x\frac{1}{7}=0\)

\(x-x=0:\frac{1}{7}\)

\(x-x=0\)

\(\Rightarrow x=0\)