Sửa lại câu a

ĐKXĐ: \(x\ne5;x\ne-2\)

Đề \(\Rightarrow3\left(x+2\right)=\left(-4\right)\left(x-5\right)\)

\(\Rightarrow3x+6=-4x+20\)

\(\Rightarrow-7x=-14\)

\(\Rightarrow x=2\left(n\right)\)

                                                      Vậy x = 2

c viết lại phân số đc k ,  k thể phân biệt đc giữa số và phân số

a/ ĐKXĐ: \(x\ne5;x\ne-2\)

  \(\frac{3}{x-5}=-\frac{4}{x+2}\Rightarrow3\left(x+2\right)=\left(-4\right)\left(x-5\right)\)

\(\Rightarrow3x+6=-4x+20\)

\(\Rightarrow7x=-14\Rightarrow x=-2\left(l\right)\)

                                          Vậy x thuộc rỗng

b/ 3^x + 3^x+1 + 3^x+2 = 1053

=> 3^x + 3^x.3 + 3^x . 3² = 1053

=> 3^x  (1 + 3 + 3²) = 1053

=> 3^x = 81

=> 3^x = 3⁴

=> x = 4

c/ (x - 1)³ = (x - 1)²

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

=> (x - 1)(x - 2) = 0 

=> x = 1 hoặc x = 2

                                 Vậy x = 1. x = 2

a) \(5^x:\left(5^2\right)^2=625\)

\(5^x:625=625\)

\(5^x=5^8\)=> x = 8

Mấy câu kia tương tự

d) \(\left(x-1\right)^4-\left(x-1\right)^4\cdot\left(x-1\right)^3=0\)

\(\left(x-1\right)^4\cdot\left[1-\left(x-1\right)^3\right]=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=1\\x-1=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy,..........

làm rõ ra đi mà

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Dấu chấm là nhân

a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)

c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)

\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)

CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

Bài 1: 

a) Ta có: \(x\left(x^2-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{0;2;-2\right\}\)

b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)

\(\Leftrightarrow2x-3-3x-x+5=40\)

\(\Leftrightarrow-2x+2=40\)

\(\Leftrightarrow-2x=38\)

hay x=-19

Vậy: x=-19

Bài 2: 

a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)

\(=-45\cdot\left(12+34+54\right)\)

\(=-45\cdot100\)

\(=-4500\)

b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)

\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)

\(=43\cdot57-33\cdot57\)

\(=57\cdot\left(43-33\right)\)

\(=57\cdot10=570\)

Bài 1:

a)12,5 x (-5/7) + 1,5 x (-5/7)

=-5/7*(12,5+1,5)

=-5/7*14

=-10

b)(-1/4) x (6|2/11) + 3|9/11 x (-1/4)

=-1/4*(68/11+42/11)

=-1/4*10

=-5/2

c tương tự

d)\(\frac{9^8\cdot4^3}{27^4\cdot6^5}=\frac{\left(3^2\right)^8\cdot\left(2^2\right)^3}{\left(3^3\right)^4\cdot\left(2\cdot3\right)^5}=\frac{3^{16}\cdot2^6}{3^{12}\cdot2^5\cdot3^5}=\frac{3^{16}\cdot2^5\cdot2}{3^{16}\cdot3^1\cdot2^5}=\frac{2}{3}\)

Bài 2:

a)Ta có:

2⁸⁰⁰=(2⁸)¹⁰⁰=256¹⁰⁰

8²⁰⁰=(8²)¹⁰⁰=64¹⁰⁰

Vì 256¹⁰⁰>64¹⁰⁰ =>2⁸⁰⁰>8²⁰⁰

b)Ta có:

12⁴⁵=(12³)¹⁵=1728¹⁵

Vì 625¹⁵<1728¹⁵ =>625¹⁵<12⁴⁵

a) -5/7x(12,5+1,5)=-10

b) -1/4x (6I2/11+3I9/11) = -1/4x 10=-5/2

c) (-7/5 + -3/8 + -3/5+5/8): 2009/2010 = -7/4:2009/2010=-1005/574

d)3^16x 4^3/3^12x 6^5=3^4x4^3x6^5=......

bài 2)

2^800=2^4x200= 15^200> 8^200

=> 2^800>8^200

B) quên cách làm

dell   

bai o ban noi the co the ban la c ho day 

1. a

\(\dfrac{8}{5}-\dfrac{5}{6}\cdot\dfrac{3}{4}\)

\(=\dfrac{8}{5}-\dfrac{5\cdot3}{3\cdot2\cdot4}\)

\(=\dfrac{8}{5}-\dfrac{5}{8}=\dfrac{39}{40}\)

1.b

\(=\dfrac{7}{8}+\dfrac{5}{6}\cdot\dfrac{3}{2}\)

\(=\dfrac{7}{8}+\dfrac{5\cdot3}{3\cdot2\cdot2}\)

\(=\dfrac{7}{8}+\dfrac{5}{4}=\dfrac{17}{8}\)

2.a

\(\dfrac{4}{5}+x=\dfrac{11}{10}\)

\(x=\dfrac{11}{10}-\dfrac{4}{5}=\dfrac{3}{10}\)

2.b

\(x-\dfrac{3}{4}=\dfrac{5}{7}\)

\(x=\dfrac{5}{7}+\dfrac{3}{4}=\dfrac{41}{28}\)

a) \(2^x.4=128\Rightarrow2^x=32=2^5\Rightarrow x=5\)

b) \(x^{17}=x\Rightarrow x^{17}-x=0\Rightarrow x\left(x^{16}-1\right)=0\Rightarrow x=0\) hay \(x=1\)

c) \(\left(2x-2\right)^3=8\Rightarrow\left(2x-2\right)^3=2^3\Rightarrow2x-2=2\Rightarrow2x=4\Rightarrow x=2\)

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

\(\Rightarrow\left(x-6\right)^2\left(x-6-1\right)=0\Rightarrow\Rightarrow\left(x-6\right)^2\left(x-7\right)=0\)

\(\Rightarrow x-6=0\) hay \(x-7=0\Rightarrow x=6\) hay \(x=7\)

e) \(\left(7x-11\right)^3=2^5.5^2+200\Rightarrow\left(7x-11\right)^3=32.25+200\)

\(\Rightarrow\left(7x-11\right)^3=1000=10^3\Rightarrow7x-11=10\Rightarrow7x=21\Rightarrow x=3\)

f) \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\Rightarrow2^{x-1}=24-\left[16-3\right]-3\)

\(\Rightarrow2^{x-1}=24-13-3\Rightarrow2^{x-1}=8=2^3\Rightarrow2x-1=3\Rightarrow2x=4\Rightarrow x=2\)

Cảm ơn bạn Nguyễn Đức Trí rất nhìu nha!

a,12,5x(-5/7)+1,5x(-5/7)

=-125/14+-15/14

=-10

2,2mu800>8 mu 200

6254 lon hon 12