5.(1/5+1/17)-(2/5+2/17+9/15+12/68)

=5.22/85-22/17

=22/17-22/17

=0

Ta có : \(5\cdot\left(\frac{1}{5}+\frac{1}{17}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{9}{15}+\frac{12}{68}\right)\)

\(=\) \(5\cdot\frac{1}{5}+5\cdot\frac{1}{17}-\left(\frac{2}{5}+\frac{2}{17}+\frac{3}{5}+\frac{3}{17}\right)\)

\(=\) \(1+\frac{5}{17}-\left[\left(\frac{2}{5}+\frac{3}{5}\right)+\left(\frac{2}{17}+\frac{3}{17}\right)\right]\)

\(=\) \(1+\frac{5}{17}-\left(1+\frac{5}{17}\right)\)

\(=\) \(1+\frac{5}{17}-1-\frac{5}{17}\)

\(=\)\(0\)

     Vậy ... 

                  Tk ủng hộ mk nha các bn ❣❣ C.ơn nhiều ^^

a,3x-3+6x-12=5

9x-15=5

9x=20

x=20/9

b,3x-6+6x-6=6

9x-12=6

9x=18

x=2

c,4x+4+5x+10=3x+20

9x+14=3x+20

9x-3x=6

6x=6

x=1

a) 3.(x-1) + 6.(x-2) = 5

3.x - 3 + 6.x - 12 = 5

9.x - 15 = 5

9.x = 5 + 15

9.x = 20

x = \(\frac{20}{9}\)

Vậy x = \(\frac{20}{9}\)

b) 3.(x-2) + 6.(x-1) = 6

  3.x - 6 + 6.x - 6 = 6

9.x - 12 = 6

9.x = 6 + 12

9.x = 18

x = 18 : 9

x = 2

Vậy x = 2

c) 4.(x+1) + 5.(x+2) = 3.x +20

4.x +4 +5.x +10 = 3.x +20

9.x + 14 = 3.x +20

9.x - 3.x = 20 - 14

6.x = 6

x = 6 : 6

x = 1

Vậy x = 1

a: \(P=2x^2+3xy+y^2=\left(2x+y\right)\left(x+y\right)\)

\(=\left(2\cdot\dfrac{-1}{2}+\dfrac{2}{3}\right)\left(\dfrac{-1}{2}+\dfrac{2}{3}\right)\)

\(=\dfrac{-1}{3}\cdot\dfrac{1}{6}=-\dfrac{1}{18}\)

d: \(Q=\dfrac{-1}{3}x^4y^2=\dfrac{-1}{3}\cdot16\cdot\dfrac{1}{16}=-\dfrac{1}{3}\)

Ta có: \(1-\frac{1}{2}.\frac{2}{3}....\frac{98}{99}\)

 = \(1-\frac{1}{99}\)

=    \(\frac{98}{99}\)

P=\(1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).........\left(1-\frac{1}{99}\right)\)

P=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.........\frac{98}{99}=\frac{1}{99}\)

\(P=1-\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{99}=1-\frac{1}{99}=\frac{98}{99}\)

Nha ban Speed Ninja

tôi đã thử lòng các bạn nhưng ko có ai trả lời thì tớ giải cho nhé.

bài làm:  Đặt \(\frac{x}{1998}=\frac{y}{1999}=\frac{z}{2000}=k\Rightarrow\)x =1998k   ; y =1999k   ; z =2000k

ta có : \(\left(x-z\right)^3=\left(1999k-2000k\right)^3\)  = \(\left[k\cdot\left(1999-2000\right)\right]^3\)= \(k^3\cdot\left(-8\right)\)                                         (1)

\(8\cdot\left(x-y\right)^2\cdot\left(y-z\right)\) = \(8\cdot\left(1998k-1999k\right)^2\cdot\left(1999k-2000k\right)\) 

                                                  = \(8\cdot\left[k\cdot\left(1999-2000\right)\right]^2\cdot\left[k\cdot\left(1999-2000\right)\right]\)

                                                 = \(8\cdot k^2\cdot1\cdot k\cdot\left(-1\right)=k^3\cdot\left(-8\right)\)                                                                        (2)

từ (1)và (2) \(\Rightarrow\left(x-z\right)^3=8\cdot\left(x-y\right)^2\cdot\left(y-z\right)\)  

Bạn dựa vào công thức

a² - n² = (a + n)(a - n)

a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)

Vậy x = 8 hoặc x = -7

a: Ta có: \(x^4-x^2-56=0\)

\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)

\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)

\(\Leftrightarrow x^2-8=0\)

hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)