Tìm \(x\), biết :
a) \(x+13=32-76\)
b) \(\left(-15\right)+x=\left(-14\right)-\left(-57\right)\)
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20 tháng 9 2020
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
20 tháng 9 2020
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: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
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14 tháng 8 2023
a/
\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)
\(\Rightarrow x=12\)
14 tháng 8 2023
\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)
\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)
Vậy x = 2023
CC
7 tháng 2 2020
\(A=\left|x-13\right|+\left|x-14\right|+\left|x-15\right|+\left|x-16\right|+\left|x-17\right|-10\)
\(=\left(\left|x-13\right|+\left|x-16\right|\right)+\left(\left|x-14\right|+\left|x-17\right|\right)-10+\left|x-15\right|\)
\(=\left(\left|x-13\right|+\left|16-x\right|\right)+\left(\left|x-14\right|+\left|17-x\right|\right)-10+\left|x-15\right|\)
\(\Rightarrow A\ge\left|x-13+16-x\right|+\left|x-14+17-x\right|-10+\left|x-15\right|\)
\(=\left|3\right|+\left|3\right|-10+\left|x-15\right|\)\(=3+3-10+\left|x-15\right|=-6+\left|x-15\right|\)
Vì \(\left|x-15\right|\ge0\forall x\)\(\Rightarrow A\ge-6\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-13\right)\left(16-x\right)\ge0\\\left(x-14\right)\left(17-x\right)\ge0\\x-15=0\end{cases}}\Leftrightarrow\hept{\begin{cases}13\le x\le16\\14\le x\le17\\x=15\end{cases}}\Leftrightarrow x=15\)
Vậy \(minA=-6\Leftrightarrow x=15\)
a) x + 13 = 32 - 76
=> x + 13 = -44
=> x = (-44) - 13
=> x = -57
b) ( -15) + x = ( -14 ) - ( -57 )
=> (-15) + x = (-14) + 57
=> (-15) + x = 43
=> x = 43 - (-15)
=> x = 43 + 15
=> x = 58
Tìm x
a) x+13=32-76
x+13= -44
x= -44-13
x= -57
b) (-15)+x=(-14)-(-57)
(-15)+x= (-14)+57
(-15)+x= 43
x= 43-(-15)
x=43+15
x=58