Đặt A = 1 + ( - 2 ) + ( - 3 ) + 4 + 5 + ( - 6 ) + ( - 7 ) + 8 + ... + 99 - 100 - 101 - 102 + 103

=> A = [ 1 + ( - 2 ) + ( - 3 ) + 4 ] + [ 5 + ( - 6 ) + ( - 7 ) + 8 ] + .... + [ 99 - 100 - 101 + 102 ] + 103

=> A = 0 + 0 + 0 + .... + 103

=> A = 103

Vậy A = 103

làm bừa cho rồi

Bài 1:

\(A=\frac{8}{7}+\frac{4}{11}(\frac{-6}{7}-\frac{5}{11})=\frac{8}{7}+\frac{-404}{847}=\frac{564}{847}\)

\(B=\frac{1}{5}.10-\frac{1}{3}.\frac{-21}{20}-\frac{1}{8}=2+\frac{7}{20}-\frac{1}{8}=\frac{89}{40}\)

Bài 2:

a.

$\frac{3}{4}+\frac{1}{4}:x=-3$

$\frac{1}{4}:x =-3-\frac{3}{4}=\frac{-15}{4}$
$x=\frac{1}{4}: \frac{-15}{4}=\frac{-1}{15}$

b.

$(x-\frac{1}{3})^2=1-\frac{5}{9}=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2$
$\Rightarrow x-\frac{1}{3}=\frac{2}{3}$ hoặc $x-\frac{1}{3}=\frac{-2}{3}$

$\Rightarrow x=1$ hoặc $x=\frac{-1}{3}$

1/ Ta có :\(-43.92-46.27+46.41\)

            \(=\left(-86\right).46-46.27+46.41\) 

            \(=46.\left(-86-27+41\right)\)

            \(=46.\left(-72\right)\) 

            \(=-3312\)

a) = 54

b) = ( - 315 )

c) = 132

d) = 2400

Bạn có thể giải cách làm đc k ?

\(a,\dfrac{x^2+2}{x^3+1}-\dfrac{1}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ =\dfrac{x^2+2-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\\ c,\dfrac{1}{2-2x}-\dfrac{3}{2+2x}+\dfrac{2x}{x^2-1}\\ =\dfrac{-1}{2\left(x-1\right)}-\dfrac{3}{2\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm1\right)\\ =\dfrac{-1\left(x+1\right)-3\left(x-1\right)+2x.2}{2\left(x+1\right)\left(x-1\right)}\\ =\dfrac{-x-1-3x+3+4x}{2\left(x+1\right)\left(x-1\right)}=\dfrac{2}{2\left(x+1\right)\left(x-1\right)}=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\) (ĐK: \(x\ne0;x\ne\pm2\) )
\(=\dfrac{x}{x\left(x-2\right)}-\dfrac{x\left(x+4\right)}{x\left(x^2-4\right)}-\dfrac{2}{x\left(x+2\right)}\)

\(=\dfrac{1}{x-2}-\dfrac{x+4}{\left(x+2\right)\left(x-2\right)}-\dfrac{2}{x\left(x+2\right)}\)

\(=\dfrac{x\left(x+2\right)}{x\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+4\right)}{x\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{-4x+4}{x\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{4-4x}{x^3-4x}\) 

\(b,\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\\ =\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x^2-4\right)}-\dfrac{2}{x\left(x+2\right)}\left(ĐKXĐ:x\ne0;x\ne\pm2\right)\\ =\dfrac{x\left(x+2\right)-\left(x^2+4x\right)-2\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2-x^2+2x-4x-2x+4}{x\left(x+2\right)\left(x-2\right)}\\ =\dfrac{-4x+4}{x\left(x+2\right)\left(x-2\right)}\)

mik ko hiểu chỗ 1 19/60 và 1 23/24 là s bạn có f vầy ko :( sau thì ib để mik sửa ạ

\(B=\frac{13}{15}\cdot0,25\cdot3+\left(\frac{8}{15}-1\frac{19}{60}\right):1\frac{23}{24}\)

\(B=\frac{13}{15}\cdot0,25\cdot3+\left(\frac{8}{15}-\frac{79}{60}\right):\frac{47}{24}\)

\(B=\frac{13}{15}\cdot\frac{1}{4}\cdot\frac{3}{1}+\left(\frac{8}{15}-\frac{79}{60}\right):\frac{47}{24}\)

\(B=\frac{39}{60}+\left(\frac{8}{15}-\frac{79}{60}\right):\frac{47}{24}\)

\(B=\frac{13}{20}+\left(\frac{32}{60}-\frac{79}{60}\right):\frac{47}{24}\)

\(B=\frac{13}{20}+\left(-\frac{47}{60}\right):\frac{47}{24}\)

\(B=\frac{13}{20}+\left(-\frac{2}{5}\right)\)

\(B=\frac{13}{20}+\left(-\frac{8}{20}\right)\)

\(B=\frac{5}{20}=\frac{1}{4}\)

\(\frac{x+1}{x+2}:\left(\frac{x+2}{x+3}:\frac{x+3}{x+1}\right)=\frac{x+1}{x+2}:\frac{\left(x+2\right)\left(x+1\right)}{\left(x+3\right)^2}=\frac{\left(x+3\right)^2}{\left(x+2\right)^2}\)