\(\frac{-9}{x}\)=\(\frac{-x}{\frac{4}{49}}\)
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\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\)
\(\Rightarrow x^2=-9.\left(\frac{-4}{49}\right)\)
\(\Rightarrow x^2=\frac{36}{49}\)
\(\Rightarrow x=\sqrt{\frac{36}{49}}\)
\(\Rightarrow x=\sqrt{\frac{6^2}{7^2}}\)
\(\Rightarrow x=\sqrt{\left(\frac{6}{7}\right)^2}\)
\(\Rightarrow x=\frac{6}{7}\)
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 \(x=\frac{6}{7}\)
\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\Leftrightarrow-x.x=-9.\frac{4}{49}\Leftrightarrow-x^2=-\frac{36}{49}\)
\(\Leftrightarrow x^2=\frac{36}{49}\Leftrightarrow\sqrt{x}=\sqrt{\frac{36}{49}}\Leftrightarrow x=\hept{\begin{cases}\frac{6}{7}\\-\frac{6}{7}\end{cases}}\)
(Nhớ k cho mình với nhé!)
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)
\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)
\(=\frac{1.9}{2.8}=\frac{9}{16}\)
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
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
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....
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.........
\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\)
\(\Rightarrow x.x=\left(-9\right).\left(-\frac{4}{49}\right)\)
\(\Rightarrow x^2=\frac{36}{49}\)
\(\Rightarrow x^2=\hept{\begin{cases}\left(\frac{6}{7}\right)^2\\\left(-\frac{6}{7}\right)^2\end{cases}}\)
\(\Rightarrow x=\hept{\begin{cases}+\frac{6}{7}\\-\frac{6}{7}\end{cases}}\)
\(\frac{-9}{x}=\frac{-x}{\frac{4}{49}}\)
=> -x.x = -9.4/49
=> -x2 = \(\frac{-36}{49}\)
=> x2 = \(\frac{36}{49}\)
=> x2 = \(\left(\pm\frac{6}{7}\right)^2\)
=> x = \(\pm\frac{6}{7}\)