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Tìm x biết:
\(\left(-\frac{3}{4}\right)^x=\frac{81}{256}\)
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\(5\frac{4}{7}\): [ x : 1,3 + 8,4 . \(\frac{6}{7}\). ( 6 - \(\frac{\left(2,3+5\div6,25\right)\times7}{8\times0,-125+6,9}\)) ] = \(1\frac{1}{14}\)
\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - \(\frac{\left(2,3+0,8\right).7}{0,1+6,9}\)) ] = \(\frac{15}{14}\)
\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - \(\frac{3,1.7}{7}\)) ] = \(\frac{15}{14}\)
\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - 3,1 ) ] = \(\frac{15}{14}\)
x : 1,3 + \(\frac{36}{5}\). 2,9 = \(\frac{39}{7}\): \(\frac{15}{14}\)
x : 1,3 + 20,88 = 5,2
x : 1,3 = - 15,68
x = - 15,68 . 1,3
x = - 20,384
ta có
\(5\frac{4}{7}:\left\{x:1,3+8,4.\frac{6}{7}.\left[6-\frac{\left(2,3+5:6,25\right).7}{8.0,0125+6,9}\right]\right\}=1\frac{1}{14}\)
\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.\left[6-\frac{\left(2,3+0,8\right).7}{0,1+6,9}\right]\right\}=\frac{15}{14}\)
\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.\left[6-\frac{3,1.7}{7}\right]\right\}=\frac{15}{14}\)
\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.2,9\right\}=\frac{15}{14}\Leftrightarrow\left\{x:1,3+7,2.2,9\right\}=\frac{39}{7}:\frac{15}{14}\)
\(\Leftrightarrow x:1,3+20,88=5,2\Leftrightarrow x:1,3=-15,68\Leftrightarrow x=-20,384\)
(2/3×x-1/3)=2/3+1/3
(2/3×x-1/3)=3/3
2/3×x=3/3+1/3
2/3×x=4/3
x=4/3:3/2
x=4/3×2/3
x=8/9
\(\frac{3}{2}+\frac{3}{14}+\frac{3}{15}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)
\(\frac{6}{\left(1.2\right).2}+\frac{6}{\left(2.7\right).2}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)
\(\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{\left(x-3\right).x}=\frac{96}{49.2}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{\left(x-3\right)}-\frac{1}{x}=\frac{96}{98}\)
=> \(1-\frac{1}{x}=\frac{48}{49}\)
=> \(\frac{1}{x}=\frac{1}{49}\)
=> \(x=49\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne3\\x\ne\pm2\end{cases}}\)
b) \(D=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right)\div\left(\frac{x-3}{2-x}\right)\)
\(\Leftrightarrow D=\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2-x}{x-3}\)
\(\Leftrightarrow D=\frac{4+4x+x^2-4+4x-x^2+4x^2}{\left(2+x\right)\left(x-3\right)}\)
\(\Leftrightarrow D=\frac{4x^2+8x}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow D=\frac{4x}{x-3}\)
c) Để D = 0
\(\Leftrightarrow\frac{4x}{x-3}=0\)
\(\Leftrightarrow4x=0\)
\(\Leftrightarrow x=0\)
Vậy để D = 0 \(\Leftrightarrow\)x = 0
d) Khi \(\left|2x-1\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=5\\1-2x=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\left(ktm\right)\\x=-2\left(ktm\right)\end{cases}}\)
Vậy khi \(\left|2x-1\right|=5\Leftrightarrow D\in\varnothing\)
=>x - 3 0 = 6001 hoặc x - 30 = -6001
=> x = 6031 hoặc x = -5971
BÀi nay không khó lắm
Dễ thấy vế bên phải bằng 0 vì \(\frac{3^6}{9}-81=0\)
=> lx - 30 l - 6001 = 0
=> lx - 30 l = 6001
Tự làm tiếp