\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\)

\(\Rightarrow x^2=-9.\left(\frac{-4}{49}\right)\)

\(\Rightarrow x^2=\frac{36}{49}\)

\(\Rightarrow x=\sqrt{\frac{36}{49}}\)

\(\Rightarrow x=\sqrt{\frac{6^2}{7^2}}\)

\(\Rightarrow x=\sqrt{\left(\frac{6}{7}\right)^2}\)

\(\Rightarrow x=\frac{6}{7}\)

vay \(x=\frac{6}{7}\)

\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\)

-9 x\(\frac{4}{49}\)=\(x\times\left(-x\right)\)

\(-\frac{36}{49}\)= -2x

\(\Rightarrow\)2x=\(\frac{36}{49}\)( cùng bớt dấu âm của 2 vế)

x=\(\frac{36}{49}:2\)

x=\(\frac{18}{49}\)

vậy x = \(\frac{18}{49}\)

\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\)

=> \(x\left(-x\right)=\left(-9\right)\frac{4}{49}\)

=> \(-x^2\)\(=\frac{-36}{49}\)

=> \(x^2=\frac{36}{49}\)

=> \(x=\frac{6}{7}\)

\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\)

\(\Rightarrow x^2=\left(-9\right)\left(\frac{-4}{49}\right)\)

\(\Rightarrow x^2=\frac{36}{49}\)

\(\Rightarrow x=\sqrt{\frac{36}{49}}\)

\(\Rightarrow x=\frac{\sqrt{36}}{\sqrt{49}}\)

\(\Rightarrow x=\frac{\sqrt{6^2}}{\sqrt{7^2}}\)

\(\Rightarrow x=\frac{6}{7}\)

vay \(x=\frac{6}{7}\)

\(\Leftrightarrow x^2=9\times\frac{4}{49}\)

\(\Leftrightarrow x^2=\frac{36}{49}\)

\(\Leftrightarrow x^2=\left(\frac{3}{7}\right)^2\)

\(\Rightarrow x=\frac{3}{7}\)

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

\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\Leftrightarrow-x.x=-9.\frac{4}{49}\Leftrightarrow-x^2=-\frac{36}{49}\)

\(\Leftrightarrow x^2=\frac{36}{49}\Leftrightarrow\sqrt{x}=\sqrt{\frac{36}{49}}\Leftrightarrow x=\hept{\begin{cases}\frac{6}{7}\\-\frac{6}{7}\end{cases}}\)

(Nhớ k cho mình với nhé!)

\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)

\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)

\(=\frac{1.9}{2.8}=\frac{9}{16}\)

Hê lô

không có giá trị tồn tại x :)

Vì : 

9.4/49=x.(-x)

=> 0.7346938776 = x.(-x)

Nhưng đề phải có thí dụ như x thuộc tập hợp số hữu tỉ hoặc gì đó :)

\(\frac{9}{x}=\frac{-x}{\frac{4}{49}}\Rightarrow x.\left(-x\right)=9.\frac{4}{49}\)\(\Leftrightarrow-x^2=\frac{36}{49}\Rightarrow x^2=-\frac{36}{49}\)

 vậy đề bài vô lý

mk thấy đề bài bạn đưa có chút vô lí nha mk nghĩ bn nên xem lại thì hơn 

a) x/-27=-3/x

suy ra: x.x=-3.(-27)

suy ra: x^2=81. Suy ra: x^2=9^2 hoặc (-9)^2

suy ra: x=9 hoặc x=-9

Vậy x=9 hoặc x=-9

\(\frac{x}{-27}=\frac{-3}{x}\)=>x²=(-27).(-3)=81

=>x=-9 hoặc x=9

\(\frac{-9}{x}=\frac{4}{49}\)=>4x=(-9).49=441

=> x = 441:4

=> x = 441/4

câu đầu bạn làm đúng rồi nhưng câu sau bạn làm sai , mink nghĩ ra kết quả là 42 là đúng nhưng lại ko bít cách làm

c) \(\dfrac{7x-1}{2}=5+\dfrac{9-5x}{6}\)

\(\Leftrightarrow\dfrac{6\left(7x-1\right)}{12}=\dfrac{5\cdot12}{12}+\dfrac{2\left(9-5x\right)}{12}\)

\(\Rightarrow42x-6=60+18-10x\)

\(\Leftrightarrow52x-84=0\)

\(\Leftrightarrow x=\dfrac{21}{13}\)

Vậy....

d) tương tự

a) \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)ĐKXĐ : \(x\ne2;4\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)

\(\Leftrightarrow\dfrac{2x^2-11x+16}{x^2-6x+8}=-1\)

\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)

\(\Leftrightarrow3x^2-17x+24=0\)

\(\Leftrightarrow3x^2-9x-8x+24=0\)

\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\)( thỏa mãn ĐKXĐ )

Vậy....

b) \(\dfrac{x+16}{49}+\dfrac{x+18}{47}=\dfrac{x+20}{45}-1\)

\(\Leftrightarrow\dfrac{x+16}{49}+1+\dfrac{x+18}{47}+1=\dfrac{x+20}{45}-1+1+1\)

\(\Leftrightarrow\dfrac{x+16+49}{49}+\dfrac{x+18+47}{47}=\dfrac{x+20+45}{45}\)

\(\Leftrightarrow\dfrac{x+65}{49}+\dfrac{x+65}{47}=\dfrac{x+65}{45}\)

\(\Leftrightarrow\left(x+65\right)\left(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\right)=0\)

Vì \(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\ne0\)

\(\Leftrightarrow x+65=0\)

\(\Leftrightarrow x=-65\)

Vậy....

Chứng minh rằng (9^2n+1994^93)chia heets cho 5

a) \(x\cdot\frac{1}{2}+x\cdot\frac{1}{4}+x\cdot\frac{1}{8}=\frac{21}{24}\)

\(x\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)=\frac{7}{8}\)

\(x\cdot\frac{7}{8}=\frac{7}{8}\)

\(\Rightarrow x=\frac{7}{8}\div\frac{7}{8}=1\)

b) \(\left(x+4\right)+\left(x+9\right)+\left(x+14\right)+.....+\left(x+44\right)+\left(x+49\right)=1430\)

\(\left(x+x+x+....+x+x\right)+\left(4+9+14+...+44+49\right)=1430\)

\(10x+265=1430\)

\(10x=1430-265\)

\(10x=1165\)

\(\Rightarrow x=\frac{1165}{10}=116,5\)

c) \(x\cdot0,25-0,5=1\)

\(x\cdot0,25=1+0,5\)

\(x\cdot0,25=1,5\)

\(\Rightarrow x=1,5\div0,25=6\)

a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)

\(\frac{1}{3}:2x=\frac{-21}{4}\)

\(2x=\frac{-4}{63}\)

\(x=\frac{2}{63}\)

b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)

Vậy.........

c) \(\left|x+\frac{1}{5}\right|-\frac{1}{2}=\frac{9}{10}\)

\(\left|x+\frac{1}{5}\right|=\frac{7}{5}\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{7}{5}\\x+\frac{1}{5}=\frac{-7}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=\frac{-8}{5}\end{cases}}\)

Vậy.........