\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+4}{96}+\frac{x+1}{99}=-4\)

\(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+1}{99}+1\right)=-4+4\)

\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{96}+\frac{x+100}{99}=0\)

\(\left(x+100\right).\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\right)=0\)

=> \(\orbr{\begin{cases}x+100=0\\\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}=0\end{cases}}\)

Mà \(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\ne0\)

=> x + 100 = 0

=> x = -100

Vậy x = -100

Câu b trừ mỗi số đi 1 tức là trừ cả cụm đó cho 3 rùi lm tương tự câu a

trời ??? này mà là toán lớp 5 what???

\(\frac{x+5}{95}+\frac{x+3}{97}+\frac{x+1}{99}=\frac{x+15}{85}+\frac{x+20}{80}+\frac{x+25}{75}.\)

\(\frac{x+5}{95}+1+\frac{x+3}{97}+1+\frac{x+1}{99}+1-\frac{x+15}{85}-1-\frac{x+20}{80}-1-\frac{x+25}{75}-1=0\)

\(\frac{x+100}{95}+\frac{x+100}{97}+\frac{x+100}{99}-\frac{x+100}{85}-\frac{x+100}{80}-\frac{x+100}{75}=0\)

\(\left(x+100\right).\left(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\right)=0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

\(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\ne0\)

câu 2: 2x^3-1=5 

=>x=2

=>......=>.....

y=57; z=41

x+y+z=100

a) (1-1/2)(1-1/3)...(1-1/100)=lx-1 99/100l

=> (1-1/2)(1-1/3)...(1-1/100)=1/2.2/3.3/4...99/100

=> (1-1/2)(1-1/3)...(1-1/100)=1.2.3.4....99/2.3.4....100

=>(1-1/2)(1-1/3)...(1-1/100)=1/100      (1)

từ (1)=>1/100= l x-1 99/100 l

TH₁:x-1 99/100 =1/100                 TH₂ : x-1 99/100= -1/100

=>x- 199/100 =1/100                           =>x- 199/100= -1/100

=>x=1/100+199/100                            =>x=-1/100+199/100

=>x=200/100                                       =>x=198/100

=>x=2                                                  =>x=99/50

Vậy x=2 hoặc x=99/50

1. A = 75(4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 25

    A = 25 . [3 . (4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 1]

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 3 + 1)

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 4)

    A = 25 . 4 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1)

    A =100 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1) \(⋮\) 100

3a) |x| = 1/2 

=> x = 1/2 hoặc x = -1/2

với x = 1/2:

A = \(3.\left(\frac{1}{2}\right)^2-2.\frac{1}{2}+1\)

\(A=\frac{3}{4}-1+1=\frac{3}{4}\)

với x = -1/2

A = \(3.\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)+1\)

\(A=\frac{3}{4}+1+1=\frac{3}{4}+2=\frac{11}{4}\)

2.

A=\(\frac{\left(1+2+3+.....+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+......+99-100}\)\

A=0

mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn 

a) (5 - x) +12 = -25

<-> 5 - x + 12 = -25

<-> 17 - x = - 25

<-> x = 42

b) 12 - 4(x - 2) = -4

<-> 12 - 4x + 8 = -4

<-> 20 - 4x = -4

<-> 4x = 24

<-> x = 6

a) (5 - x) + 12 = -25

<=> -x = -25 - 12 - 5

<=> -x = -42

<=> x = 42

b) 12 - 4(x - 2) = -4

<=> 12 - 4x + 8 = -4

<=> -4x = -4 - 8 - 12

<=> -4x = -24

<=> x = 6

c) -15 - |3 - x| = -19

<=> -|3 - x| = -4

<=> 3 - x = 4 hoặc 3 - x = -4

<=> x = -1 hoặc x = 7

Bài 3 : 

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)

Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)

           \(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)

            \(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)

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

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

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

\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)

Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)

            \(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)

           \(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)

           \(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)

\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)

\(B=\frac{1}{10}-\frac{1}{40}\)

\(B=\frac{3}{40}\) 

3. 

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

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

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

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

\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)

\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)

\(B=\frac{1}{10}-\frac{1}{40}\)

\(B=\frac{3}{40}\)

Yến Nhi ơi !Bài B=............Ở phần mâu số là chư X hay là dấu *