Tính và so sánh các cặp kết quả sau:
\(\left( { - 1} \right) + \left( { - 3} \right)\) và \(\left( { - 3} \right) + \left( { - 1} \right)\)
\(\left( { - 7} \right) + \left( { + 6} \right)\) và \(\left( { + 6} \right) + \left( { - 7} \right)\)
Tính và so sánh các cặp kết quả sau:
\(\left( { - 1} \right) + \left( { - 3} \right)\) và \(\left( { - 3} \right) + \left( { - 1} \right)\)
\(\left( { - 7} \right) + \left( { + 6} \right)\) và \(\left( { + 6} \right) + \left( { - 7} \right)\)
\(\left( { - 1} \right) + \left( { - 3} \right) = - \left( {1 + 3} \right) = - 4\)
\(\left( { - 3} \right) + \left( { - 1} \right) = - \left( {3 + 1} \right) = - 4\)
\( \Rightarrow \left( { - 1} \right) + \left( { - 3} \right) = \left( { - 3} \right) + \left( { - 1} \right)\)
\(\left( { - 7} \right) + \left( { + 6} \right) = - \left( {7 - 6} \right) = - 1\)
\(\left( { + 6} \right) + \left( { - 7} \right) = - \left( {7 - 6} \right) = - 1\)
\( \Rightarrow \left( { - 7} \right) + \left( { + 6} \right) = \left( { + 6} \right) + \left( { - 7} \right)\)
\(\dfrac{3}{\left(x-4\right).\left(x-7\right)}+\dfrac{6}{\left(x-7\right).\left(x-13\right)}+\dfrac{15}{\left(x-13\right).\left(x-28\right)}-\dfrac{1}{x-28}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{1}{x-7}+\dfrac{1}{x-7}-\dfrac{1}{x-13}+\dfrac{1}{x-13}-\dfrac{1}{x-28}-\dfrac{1}{x-28}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{2}{x-28}=-\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{x-28-2x+8}{\left(x-4\right)\left(x-28\right)}=\dfrac{-5}{2}\)
\(\Leftrightarrow-5\left(x^2-32x+112\right)=2\left(-x-20\right)\)
\(\Leftrightarrow-5x^2+160x-560=-2x-40\)
\(\Leftrightarrow-5x^2+162x-520=0\)
\(\text{Δ}=162^2-4\cdot\left(-5\right)\cdot\left(-520\right)=15844\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{162-2\sqrt{3961}}{10}\\x_2=\dfrac{162+2\sqrt{3961}}{10}\end{matrix}\right.\)
\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(2-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{1}{20}\)
\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(2-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{1}{20}\)
ghi sai để nên mik ko giải dc
\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(2-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{1}{20}\)
Tính rồi so sánh từng cặp kết quả sau:
a) \( - \left( {4 + 7} \right)\) và \( - 4 - 7\)
b) \( - \left( {12 - 25} \right)\) và \(\left( { - 12 + 25} \right)\)
c)\( - \left( { - 8 + 7} \right)\) và \(\left( {8 - 7} \right)\)
d) \( + \left( { - 15 - 4} \right)\) và \(\left( { - 15 - 4} \right)\)
e) \( + \left( {23 - 12} \right)\) và \(\left( {23 - 12} \right)\).
a) \( - \left( {4 + 7} \right) = - 11\)
\(\begin{array}{l}\left( { - 4 - 7} \right) = \left( { - 4} \right) + \left( { - 7} \right)\\ = - \left( {4 + 7} \right) = - 11\\ \Rightarrow \left( { - 4 - 7} \right) = - \left( {4 + 7} \right)\end{array}\)
b)
\(\begin{array}{l} - \left( {12 - 25} \right) = - \left[ {12 + \left( { - 25} \right)} \right]\\ = - \left[ { - \left( {25 - 12} \right)} \right] = - \left( { - 13} \right) = 13\end{array}\)
\(\begin{array}{l}\left( { - 12 + 25} \right) = 25 - 12 = 13\\ \Rightarrow - \left( {12 - 25} \right) = \left( { - 12 + 25} \right)\end{array}\)
c)
\(\begin{array}{l} - \left( { - 8 + 7} \right) = - \left[ { - \left( {8 - 7} \right)} \right] = - \left( { - 1} \right) = 1\\\left( {8 - 7} \right) = 1\\ \Rightarrow - \left( { - 8 + 7} \right) = \left( {8 - 7} \right)\end{array}\)
d)
\(\begin{array}{l} + \left( { - 15 - 4} \right) = + \left[ {\left( { - 15} \right) + \left( { - 4} \right)} \right]\\ = + \left[ { - \left( {15 + 4} \right)} \right] = + \left( { - 19} \right) = - 19\\\left( { - 15 - 4} \right) = \left( { - 15} \right) + \left( { - 4} \right)\\ = - \left( {15 + 4} \right) = - 19\\ \Rightarrow + \left( { - 15 - 4} \right) = \left( { - 15 - 4} \right)\end{array}\)
e)
\(\begin{array}{l} + \left( {23 - 12} \right) = + 11 = 11\\\left( {23 - 12} \right) = 11\\ \Rightarrow + \left( {23 - 12} \right) = \left( {23 - 12} \right)\end{array}\)
tìm x biết:
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}\)\(+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
B)\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{5}{2}\)
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)
\(=\frac{1}{x+3}-\frac{1}{x+34}\)
\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)
\(\Rightarrow x=31\)
Vậy, x = 31
Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với \(x,k\inℝ;x\ne0;x\ne-k\)
Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
B) \(\frac{\left(x-4\right)-\left(x-7\right)}{\left(x-7\right)\left(x-4\right)}+\frac{\left(x-7\right)-\left(x-13\right)}{\left(x-13\right)\left(x-7\right)}+\frac{\left(x-13\right)-\left(x-28\right)}{\left(x-28\right)\left(x-13\right)}\)
\(=\frac{1}{x-7}-\frac{1}{x-4}+\frac{1}{x-13}-\frac{1}{x-7}+\frac{1}{x-28}-\frac{1}{x-13}\)
\(=\frac{1}{x-28}-\frac{1}{x-4}=-\frac{5}{2}+\frac{1}{x-28}\)
\(\Leftrightarrow\frac{1}{x-28}-\frac{1}{x-4}-\frac{1}{x-28}=-\frac{5}{2}\)
\(\Leftrightarrow\frac{1}{x-4}=\frac{5}{2}\)
=> 5x - 20 = 2
=> 5x = 22
\(\Rightarrow x=\frac{22}{5}=4,4\)
Vậy, x = 4,4
Tìm x: \(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}+\frac{1}{x-28}=\frac{-1}{20}\)
Mình cần gấp bài này !!!
1/Tìm x, biết :
a/ \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}+\frac{x}{\left(x+3\right)\left(x+34\right)}\)
b/ \(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=\frac{-1}{20}\)
Tìm x, biết :
\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}\)
\(+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=\frac{-1}{20}\)
giúp mình với các bạn !!!!!!!!!!!!