\(K=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9999}{10000}=\left(-1\right)^{99}.\frac{1.3.2.4...99.101}{2.2.3.3.4.4...100.100}=-\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}=-\frac{1}{100}.\frac{101}{2}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)

\(\Leftrightarrow\left(\frac{3}{4}x-\frac{9}{16}\right)\left(\frac{1}{3}-\frac{3}{5}.\frac{1}{x}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{4}x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)

Vậy \(x\in\left\{\frac{3}{4};\frac{9}{5}\right\}\)

\(\left(\frac{x}{y}-1\right).\left(\frac{y}{z}+1\right).\left(\frac{z}{x}-1\right)\)=\(\left(\frac{x-y}{y}\right).\left(\frac{y+z}{z}\right).\left(\frac{z-x}{x}\right)\)

ta có:x-y-z=0

     \(\rightarrow\)x-y=z

     \(\rightarrow\)y+z=x

     \(\rightarrow\)z-x=-y

thay các số trên vào bt,ta đc:

\(\frac{z}{y}.\frac{x}{z}.\frac{-y}{x}\)= -1

=>\(A=\frac{5-2}{5}.\frac{7-2}{7}.\frac{9-2}{9}....\frac{99-2}{99}\)

=>\(A=\frac{3}{5}.\frac{5}{7}.\frac{7}{9}.....\frac{97}{99}=\frac{3.5.7.....97}{5.7.9.....99}=\frac{3.\left(5.7.9.....97\right)}{99.\left(5.7.9.....97\right)}=\frac{3}{99}=\frac{1}{3}\)

a) chịu

b)bó tay

c)ko biết 

d)làm bừa

a)7^2x+7^2x.49=2450

7^2x.50=2450

7^2x=2450:50=49

7^2x=7²

2x=2

x=1

b)(33:11)^x=81

3^x=81

3^x=3⁴

x=4

c)1/6=2/3:8x

8x=2/3:1/6

8x=4

x=1/2

d)(x+1)³=64

(x+1)³=4³

x+1=4

x=3

minh chỉ lam đc vậy thôi nha !hi hi

Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)

\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)

Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)

      \(\left|x-\frac{3}{7}\right|\ge0\forall x\)

Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)

a) \(\left|\frac{1}{2}+x\right|+\left|x+y+z\right|+\left|\frac{1}{3}+y\right|=0\)

=> \(\left|\frac{1}{2}+x\right|=\left|x+y+z\right|=\left|\frac{1}{3}+y\right|=0\)

1/2 + x = 0 => x = -1/2

1/3 + y = 0 => y = -1/3

-1/2 + -1/3 + z = 0 

=> z = 5/6