a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)

b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)

c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)

d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)

Mong bạn xem lại đề bài.

\(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\Leftrightarrow\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}\Rightarrow x=\dfrac{1}{10}\)

\(\dfrac{1}{2}.x+\dfrac{1}{2}=\dfrac{5}{2}\Leftrightarrow\dfrac{1}{2}.x=\dfrac{5}{2}-\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}.x=2\Leftrightarrow x=4\)

\(\dfrac{1}{2}-\dfrac{2}{3}.x=\dfrac{7}{12}\Leftrightarrow\dfrac{2}{3}.x=\dfrac{1}{2}+\dfrac{7}{12}\Leftrightarrow\dfrac{2}{3}.x=\dfrac{13}{12}\Leftrightarrow x=\dfrac{13}{12}:\dfrac{2}{3}\Leftrightarrow x=\dfrac{13}{8}\)

a) \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

b) \(B=2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{13}{12}.\dfrac{-11}{5}=-\dfrac{65}{12}\)

c) \(C=\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\left(\dfrac{-2}{5}\right)=\dfrac{-11}{50}\)

A = 2/3 + -1/3

    = 1/3

B = 25/11 . 13/12 . (-2,2)

    = 325/132 . (-2,2)

    = -65/12

C = 11/20 . -2/5

    = -11/50

Chúc bạn học tốt!! ^^

    = -

a, A = \(\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)\)

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

       =\(\dfrac{1}{3}\)

b, B = \(2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)\)

       = \(\dfrac{23}{11}.\dfrac{13}{12}.\dfrac{-11}{5}\)

       = \(\dfrac{23.13.\left(-11\right)}{11.12.5}\)

       = \(\dfrac{-299}{60}\)

c, C =\(\left(\dfrac{3}{4}-0,2\right)\)\(.\left(0,4\right)-\dfrac{4}{5}\)

        = \(\left(\dfrac{3}{4}-\dfrac{1}{2}\right)\)\(.\dfrac{2}{5}-\dfrac{4}{5}\)

        = \(\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)\(.\dfrac{-2}{5}\)

        = \(\dfrac{1}{4}.\dfrac{-2}{5}\)

        = \(\dfrac{-2}{20}=\dfrac{-1}{10}\)

Chúc bạn học tốt!

a) ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

b) Ta có: \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\dfrac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+4x-5}{2\left(x+5\right)}\)

\(=\dfrac{x^2+5x-x-5}{2\left(x+5\right)}\)

\(=\dfrac{x\left(x+5\right)-\left(x+5\right)}{2\left(x+5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}\)

\(=\dfrac{x-1}{2}\)

Để B=0 thì \(\dfrac{x-1}{2}=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(nhận)

Để \(B=\dfrac{1}{4}\) thì \(\dfrac{x-1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow x-1=\dfrac{1}{2}\)

hay \(x=\dfrac{3}{2}\)(nhận)

Vậy: Để B=0 thì x=1 và Để \(B=\dfrac{1}{4}\) thì \(x=\dfrac{3}{2}\)

Chọn C

`a)a/b<c/d`
Nhân 2 vế cho `bd>0` ta có:
`(abd)/b<(bcd)/d`
`<=>ad<bc`
`b)ad<bc`
Chia 2 vế cho `bd>0` ta có:
`(ad)/(bd)<(bc)/(bd)`
`<=>a/b<c/d`.

Mình ko phải thầy việt lâm thì mình làm có được ko nhỉ kk :v

\(A,B\in d\Rightarrow\left\{{}\begin{matrix}A\left(t_A+2;-4t_A+1;-t_A+3\right)\\B\left(t_B+2;-4t_B+1;-t_B+3\right)\end{matrix}\right.\)

\(\overrightarrow{AM}=\left(-1-t_A;4t_A-2;-2+t_A\right);\overrightarrow{BM}=\left(-1-t_B;4t_B-2;-2+t_B\right)\)

\(\overrightarrow{AM}.\overrightarrow{BM}=0\Leftrightarrow\left(1+t_A\right)\left(1+t_B\right)+\left(4t_A-2\right)\left(4t_B-2\right)+\left(t_A-2\right)\left(t_B-2\right)=0\)

\(\left|\overrightarrow{AM}\right|=\left|\overrightarrow{BM}\right|\Leftrightarrow\left(t_A+1\right)^2+\left(4t_A-2\right)^2+\left(t_A-2\right)^2=\left(t_B+1\right)^2+\left(4t_B-2\right)^2+\left(t_B-2\right)^2\)

hệ phương trình 2 ẩn, đến đây là việc của bạn r :v

xác định để thầy việt lâm lm òi, cj ráng chờ nghe 

tịt