a) Ta có: \(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)

a)

b) câu b, bạn vt hơi khó hiểu

`a)5/8x+2/5=1/5`

`=>5/8x=1/5-2/5`

`=>5/8x=-1/5`

`=>x=-1/5:5/8=-8/25`

`b)5/7:x+11/7=18/7`

`=>5/7:x=18/7-11/7`

`=>5/7:x=1`

`=>x=5/7`

`c)(-1,2).(-3/24)+(0,4-1 4/15):1 2/3`

`=(-6/5).(-1/8)+(2/5-19/15):5/3`

`=3/20+(-13/15)*3/5`

`=3/20-13/25=-37/100`

a)5/8.x+2/5=1/5

5/8.x=1/5 - 2/5

5/8.x=-1/5

x=(-1/5):5/8

x=(-1/5).8/5

x=-8/25.      Vậy x=-8/25

b)5/7:x +11/7=18/7

5/7:x=1

x=5/7:1

x=5/7.      Vậy x=5/7

a) Ta có: \(\dfrac{5}{8}x+\dfrac{2}{5}=\dfrac{1}{5}\)

\(\Leftrightarrow x\cdot\dfrac{5}{8}=\dfrac{-1}{5}\)

\(\Leftrightarrow x=\dfrac{-1}{5}:\dfrac{5}{8}=\dfrac{-1}{5}\cdot\dfrac{8}{5}=\dfrac{-8}{25}\)

b) Ta có: \(\dfrac{5}{7}:x+\dfrac{11}{7}=\dfrac{18}{7}\)

\(\Leftrightarrow\dfrac{5}{7}:x=1\)

hay \(x=\dfrac{5}{7}\)

lỗi phép tính r nha e

(3x - 4)(x - 1)³ = 0

=> 3x - 4 = 0 hoặc (x - 1)³ = 0

=> 3x = 4 hoặc x - 1 = 0

=> x = 4/3 hoặc x = 1

Vậy...

x¹⁷ = x

=> x¹⁷ - x = 0

=> x(x¹⁶ - 1) = 0

=> x = 0 hoặc x¹⁶ - 1 = 0

=> x = 0 hoặc x¹⁶ = 1

=> x 0 hoặc x = 1 hoặc x = -1

Vậy..

(3x-4)x(x-1)³=0

=>3x-4=0                                                            x-1=0

3x=4                                                                   x=1

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

Vậy x thuộc {\(\frac{4}{3}\);1}

x¹⁷=x

=>x thuộc {0;1;-1}

Vậy x thuộc {0;1;-1}

\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)

\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)

\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)

\(\Rightarrow2x+2=0\)

\(\Rightarrow x=-1\left(loai\right)\)

Vậy \(S=\varnothing\)

\(4,\dfrac{x}{x-3}-\dfrac{1}{x+2}=0\left(dkxd:x\ne3;-2\right)\)

\(\Rightarrow x\left(x+2\right)-\left(x-3\right)=0\)

\(\Rightarrow x^2+3x-x+3=0\)

\(\Rightarrow x^2+2x+3=0\)

\(\Rightarrow S=\varnothing\)

giúp em tl những câu tl trên vs

\(x^4+x^2y^2+y^4=\left(x^4+2x^2y^2+y^4\right)-x^2y^2\)\(=\left(x^2+y^2\right)^2-\left(xy\right)^2=\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)\).

\(C=-\left[\dfrac{1}{3}\cdot\dfrac{\left(3+1\right)\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{\left(4+1\right)\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{\left(50+1\right)\cdot50}{2}\right]\\ C=-\left(\dfrac{1}{3}\cdot\dfrac{4\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{5\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{51\cdot50}{2}\right)\\ C=-\left(2+\dfrac{5}{2}+...+\dfrac{51}{2}\right)\\ C=-\dfrac{4+5+...+51}{2}=-\dfrac{\dfrac{\left(51+4\right)\left(51-4+1\right)}{2}}{2}=-\dfrac{55\cdot48}{4}=-660\)