a) 24/35

b) 2/5

c) 1/6

d) 1/7

a) 24/35
b) 2/5
c) 1/6
d) 1/7

\(\left(\frac{3}{5}-x\right)+\frac{13}{20}=\frac{5}{6}\)

\(\frac{3}{5}-x=\frac{5}{6}-\frac{13}{20}\)

\(\frac{3}{5}-x=\frac{11}{60}\)

\(x=\frac{3}{5}-\frac{11}{60}\)

\(x=\frac{5}{12}\)

\(\left(x-\frac{1}{2}\right)\cdot\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)

\(\left(x-\frac{1}{2}\right)\cdot\frac{5}{3}=\frac{5}{4}\)

\(x-\frac{1}{2}=\frac{5}{4}:\frac{5}{3}\)

\(x-\frac{1}{2}=\frac{3}{4}\)

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

\(x=\frac{5}{4}\)

( 3/5 - x ) + 13/20 = 5/6

  3/5 - x                = 5/6 - 13/20

  3/5 - x                = 89/60

          x                = 3/5 - 89/60

          x                = 25/12

( x - 1/2 ) x 5/3 = 7/4 - 1/2

( x - 1/2 ) x 5/3 = 5/4

 x - 1/2             = 5/4 : 5/3

 x - 1/2             = 35/12

 x                     = 35/12 + 1/2

 x                     = 41/12

a: Cách 1: \(\left(\frac23\times\frac45\right)\times\frac56\)

\(=\frac{2\times4}{3\times5}\times\frac56\)

\(=\frac{8}{15}\times\frac56=\frac{40}{90}=\frac49\)

Cách 2:

\(\left(\frac23\times\frac45\right)\times\frac56\)

\(=\frac23\times\frac45\times\frac56\)

\(=\frac23\times\frac46=\frac23\times\frac23=\frac49\)

b: Cách 1:

\(\left(\frac12+\frac13\right)\times\frac15\)

\(=\frac{3+2}{6}\times\frac15\)

\(=\frac56\times\frac15\)

\(=\frac16\)

Cách 2:

\(\left(\frac12+\frac13\right)\times\frac15\)

\(=\frac12\times\frac15+\frac13\times\frac15\)

\(=\frac{1}{10}+\frac{1}{15}=\frac{3}{30}+\frac{2}{30}=\frac{5}{30}=\frac16\)

5 mũ 56,nha bạn

kết quả là 5 ^ ⁵⁶

1) = \(\frac{3}{5}\)

2) =\(\frac{6}{7}\)

3)\(\frac{9}{13}\)

4)\(\frac{4}{13}\)

a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)

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

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

=> x=-1  

với \(3x^2+x-2=0\)

ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)

Vậy  ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)

b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)

\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)

\(\Leftrightarrow3x^2=3\)

hay \(x\in\left\{1;-1\right\}\)

c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)

hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)

dễ thế