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Câu 2 đây:
\(|x^2+|x-1||=x^2+2\)
\(\Rightarrow\orbr{\begin{cases}x^2+\left|x-1\right|=x^2+2\\x^2+\left|x-1\right|=-x^2-2\left(l\right)\end{cases}}\)
\(\Rightarrow\left|x-1\right|=2\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
a) \(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+0,5}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{2}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2012}{2013}\)
\(=\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\right):\frac{2012}{2013}\)
\(=\left(\frac{2}{7}-\frac{2}{7}\right):\frac{2012}{2013}\)
\(=0\)
Tìm x biết: \(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}.\)
\(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}\)
\(\Rightarrow\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}+3=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}+3\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+4}{6}+1\right)+\left(\frac{x+5}{5}+1\right)=\left(\frac{x+2}{8}+1\right)\)\(+\left(\frac{x+3}{7}+1\right)+\left(\frac{x+6}{4}\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}\right)=\left(x+10\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{4}\right)\)
\(\Rightarrow\left(x+10\right)\frac{43}{90}=\left(x+10\right)\frac{29}{56}\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
cộng 3 vào cả hai vế nên phương trình vẫn bằng nhau
Ta có \(\frac{x+1}{9}+1+\frac{x+4}{6}+1+\frac{x+5}{5}+1=\frac{x+2}{8}+1+\frac{x+3}{7}+1+\frac{x+6}{4}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}-\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{4}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
mà \(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x+10=0\)
\(\Leftrightarrow x=-10\)
\(\begin{array}{l}a)x + 0,25 = \frac{1}{2}\\x = \frac{1}{2} - 0,25\\x = \frac{1}{2} - \frac{1}{4}\\x = \frac{2}{4} - \frac{1}{4}\\x = \frac{1}{4}\end{array}\)
Vậy \(x = \frac{1}{4}\)
\(\begin{array}{l}b)x - \left( { - \frac{5}{7}} \right) = \frac{9}{{14}}\\x = \frac{9}{{14}} + \left( { - \frac{5}{7}} \right)\\x = \frac{9}{{14}} + \left( { - \frac{{10}}{{14}}} \right)\\x = \frac{{ - 1}}{{14}}\end{array}\)
Vậy \(x = \frac{{ - 1}}{{14}}\)
\(\frac{7}{x-1}=\frac{x+1}{9}\)
=> 7.9 = (x-1)(x+1)
=> x(x-1)+x-1=63
=> x2-x+x-1=63
=> x2-1=63
=> x2=64
=> x = -8 ; 8
\(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)=7.9=63\)
Vì \(\left(x+1\right)-\left(x-1\right)=2\)Mà (x+1) và (x-1) đều là Ư(63) nên
\(\hept{\begin{cases}x-1=7\\x+!=9\end{cases}\Rightarrow x=8}\)
Ai k cho tôi sẽ may mắn cả năm!
bạn ơi trả lời được câu này kông
( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35
\(\frac{x+1}{29}+\frac{x+4}{13}=\frac{x+9}{7}\)
\(\Leftrightarrow\frac{x+1}{29}+1+\frac{x+4}{13}+2=\frac{x+9}{7}+3\)
\(\Leftrightarrow\frac{x+30}{29}+\frac{x+30}{13}=\frac{x+30}{7}\)
\(\Leftrightarrow\frac{x+30}{29}+\frac{x+30}{13}-\frac{x+30}{7}=0\)
\(\Leftrightarrow\left(x+30\right)\left(\frac{1}{29}+\frac{1}{13}-\frac{1}{7}\right)=0\)
\(\Leftrightarrow x+30=0\)( vì \(\frac{1}{29}+\frac{1}{13}-\frac{1}{7}\ne0\))
\(\Leftrightarrow x=-30\)
Vậy...
\(\frac{x+1}{29}+\frac{x+4}{13}=\frac{x+9}{7}\)
\(\Leftrightarrow\frac{x+1}{29}+1+\frac{x+4}{13}+2=\frac{x+9}{7}+3\)
\(\Leftrightarrow\frac{x+30}{29}+\frac{x+30}{13}=\frac{x+30}{7}\)
\(\Leftrightarrow\frac{x+30}{29}+\frac{x+30}{13}-\frac{x+30}{7}=0\)
\(\Leftrightarrow\left(x+30\right)\left(\frac{1}{29}+\frac{1}{13}-\frac{1}{7}\right)=0\)
Vì \(\frac{1}{29}+\frac{1}{13}-\frac{1}{7}\ne0\)
Nên \(x+30=0\)
\(\Leftrightarrow x=-30\)
\(\frac{x-1}{7}=\frac{9}{x+1}\)
\(\Leftrightarrow\left(x-1\right).\left(x+1\right)=9.7\)
Tới đây VT ta áp dụng hằng đẳng thức số 3 .
\(x^2-1^2=63\)
\(x^2=63+1\)
\(x^2=64\)
\(\Rightarrow x=\pm\sqrt{63}\)
\(\frac{x-1}{7}=\frac{9}{x+1}\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)=9.7\)
\(\Rightarrow x-1=7;x+1=9\Rightarrow x=8\)