đúng bởi vì máy tính ra kq đó !

\(\frac{x+3}{4}\)  = \(\frac{-7}{2}\) 

(x+3 ) .2 = -7 . 4

(x+3) . 2 = - 28

x+3 = -28 : 2 

x+3 = -14

x = -14 + 3 = -17

Ta có: (x+3).2 = (-7).4

x+3 = \(\dfrac{-7.4}{2}\)

x+3 = \(\dfrac{-28}{2}\) = = -14

x = -14 - 3 = -17

Cách của bn cũng đúng nhưng cách này mik nghĩ nó sẽ dễ hiểu và đúng vs chương trình học hơn

a)

\(\frac{1}{x^2+x+1}dx=\frac{1}{\left(x-\frac{1}{4}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx\)

Đặt

\(\left(x-\frac{1}{4}\right)=\frac{\sqrt{3}}{2}tant\) => dx=\(\frac{\sqrt{3}}{2}\left(1+tan^2t\right)dt\) =>\(\frac{1}{x^2+x+1}dx=\frac{1}{\frac{3}{4}\left(1+tan^2t\right)+\frac{3}{4}}\left(1+tan^2t\right)dt=\frac{3}{4}dt=\frac{3}{4}t+C\) 

Với \(\left(x-\frac{1}{4}\right)=\frac{\sqrt{3}}{2}tant=>t=\left(\frac{2\sqrt{3}}{4x-1}\right)\)

Câu b nhá :

\(\frac{1}{x^2+2x+2}dx=\frac{1}{\left(x+1\right)^2+\left(\sqrt{2^2}\right)}dx\)

Đặt

 \(x+1=\sqrt{2}tant=>dx=\sqrt{2}\left(1+tan^2t\right)dt\)

=> \(\frac{1}{x^2+2x+3}dx=\frac{1}{2\left(tan^2t+1\right)}.\left(1+tan^2t\right)dt=\frac{1}{2}dt=\frac{1}{2}t+C\)

Với

\(x+1=\sqrt{2}tant=>tant=\frac{x+1}{\sqrt{2}}<=>t=arctan\left(\frac{x+1}{\sqrt{2}}\right)\)

Bài 2 :

1) \(x-70=-45\) 2) \(\frac{4}{7}:x=\frac{12}{28}\)

\(\Rightarrow\) \(x=-45+70\) \(\Rightarrow x=\frac{4}{7}:\frac{12}{28}\)

\(\Rightarrow\) \(x=25\) \(\Rightarrow x=\frac{4}{3}\)

Vậy \(x=25\) Vậy \(x=\frac{4}{3}\)

3) Giống câu c) ở bài 1

4) \(x-50=-35\) 5) \(\frac{4}{7}.x=\frac{11}{18}\)

\(\Rightarrow x=-35+50\) \(\Rightarrow x=\frac{11}{28}:\frac{4}{7}\)

\(\Rightarrow x=15\) \(\Rightarrow x=\frac{77}{72}\)

Vậy \(x=15\) Vậy \(x=\frac{77}{72}\)

6) \(\left(\frac{2}{3}x+2,5\right):2\frac{2}{6}=6\)

\(\Rightarrow\)\(\left(\frac{2}{3}x+2,5\right):\frac{14}{6}=6\)

\(\Rightarrow\) \(\frac{2}{3}x+2,5=6.\frac{14}{6}\)

\(\Rightarrow\frac{2}{3}x+2,5=14\)

\(\Rightarrow\frac{2}{3}x=\frac{23}{2}\)

\(\Rightarrow x=\frac{23}{2}:\frac{2}{3}\)

\(\Rightarrow x=\frac{69}{4}\)

Vậy \(x=\frac{69}{4}\)

Bài 1:

1) \(\frac{7}{5}+\frac{-8}{5}=-\frac{1}{5}\)

2) \(-\frac{6}{5}.\frac{15}{24}=-\frac{3}{4}\)

3) \(\left(\frac{2}{3}+1,5\right)-3,5:7\frac{1}{2}=\)\(\frac{13}{6}-\frac{7}{15}=\frac{17}{10}\)

4) \(\frac{5}{8}-\frac{-7}{9}=\frac{5}{8}+\frac{7}{9}=\frac{101}{72}\)

5)\(\frac{-7}{3}.\frac{12}{28}=-1\)

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

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)

\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

\(A=\frac{B}{6}=\frac{100}{2}=50\)

Vậy \(A=50\)

\(\frac{x}{2}-\frac{1}{x}=\frac{1}{12}\Rightarrow\frac{x^2}{2}-\frac{x}{x}=\frac{1}{6}\)

\(\Rightarrow\frac{x^2}{2}-1=\frac{1}{6}\Rightarrow\frac{x^2}{2}=\frac{1}{6}+1=\frac{7}{6}\)

\(\Rightarrow x^2=\frac{7}{6}.2=\frac{7}{3}\)

\(\Rightarrow x=1.5275252317\)

\(\text{Mình nhầm :}\)

\(\frac{x}{2}-\frac{1}{x}=\frac{1}{12}\Rightarrow\frac{x^2}{2}-\frac{x}{x}=\frac{1}{6}\)

\(\Rightarrow\frac{x^2}{2}-1=\frac{1}{6}\Rightarrow\frac{x^2}{2}=\frac{1}{6}+1=\frac{7}{6}\)

\(\Rightarrow x^2=\frac{7}{6}.2=\frac{7}{12}\)

\(\Rightarrow x = 0.76376261583\)

x/2-1/x=1/12

x.x-2/2x=1/12

12x²-24=2x

12x²-2x-24=0

x²-x/6=2

x²-x/6+1/144=2+1/144

(x-1/12)²=289/144

x-1/12=\(\pm17\)/12

nếu x-1/12=-\(17\)/12

x=-4/3

nếu x-1/12=17/12

x=3/2

vậy x=3/2 và -4/3

Bạn khử như thế đúng rồi đó nha !

đúng rồi đó

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)

=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)

=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)

a)=768/512+192/512+48/512+12/512+3/512

=768+192+48+12+3/512

=1023/512 

b)=405/81+135/81+45/81+15/81+5/81

=405+135+45+15+5/81

=595/81

c)=256/192+64/192+16/192+4/192+1/192

=256+64+16+4+1/192

=341/192

cai do ham

\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)

=> \(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)

=> \(\left(x+2\right)\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)

Vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}\ne\frac{1}{12^{12}}+\frac{1}{13^{13}}\Rightarrow\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)

=> \(x+2=0\Rightarrow x=-2\)