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HT
12 tháng 7 2017
làm tiếp cái trước(ấn nhầm)
\(x=\frac{5}{42}-\frac{15}{28}\)
\(x=\frac{5.4}{6.4.7}-\frac{15.6}{4.7.6}\)
\(x=\frac{20}{168}-\frac{90}{168}\)
\(x=\frac{-70}{168}\)
\(x=\frac{-5}{12}\)
2.
HT
12 tháng 7 2017
1.
\(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)
\(\frac{11}{13}-\frac{5}{42}+x=-\frac{15}{28}+\frac{11}{13}\)
\(\frac{11}{13}-\frac{11}{13}-\frac{5}{42}+\frac{15}{28}=-x\)
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: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
HP
15 tháng 6 2016
a) Dễ thấy VT > 0;mà VT=VP
=>VP > 0 => 4x > 0=> x > 0
=>\(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>BT đầu tương đương \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{6}\right)=4x\)
\(=>3x+1=4x=>x=1\)
15 tháng 6 2016
a) Để đẳng thức xảy ra thì: x>0 (vì: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|>0\) )
Khi đó: \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)
<=>x=1
Vậy x=1
b)Điều kiện: \(x\ne-3;-10;-21;-34\)
\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
=>x+34-x-3=x
<=>x=31 (nhận)
Vậy x=31
5 tháng 8 2016
a) \(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right).\left(x+8\right)}+\frac{6}{\left(x+8\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14}{\left(x+2\right).\left(x+14\right)}-\frac{x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14-x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{16}{\left(x+2\right).\left(x+4\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow x=16\)
Vậy x = 16
5 tháng 8 2016
\(b,\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(vì\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
\(\text{Vậy }x=-1\)
1 tháng 8 2016
b. (x+1)(1/10+1/11+1/12-1/13-1/14)=0
x+1=0 (vì : 1/10+1/11+1/12-1/13-1/14>0)
x=-1
10 tháng 1 2016
bài 1
[(x+2)/1010]+ [(x+2)/1111]= [(x+2)/1212]+[(x+2)/1313]
=>[(x+2)/1010]+[(x+2)/1111] - [(x+2)/1212]-[(x+2)/1313] = 0
=>(x+2).[(1/1010)+(1/1111)-(1/1212)-(1/1313)=0
Vì [(1/1010)+(1/1111)-(1/1212)-(1/1313)] khác 0
=>x+2=0
=>x=-2
\(\left(x+\frac{1}{2}\right)^{13}=\left(x+\frac{1}{2}\right)^{11}\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^{13}-\left(x+\frac{1}{2}\right)^{11}=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^{11}\left[\left(x+\frac{1}{2}\right)^2-1\right]=0\)
TH1 : \(x=-\frac{1}{2}\)
TH2 : \(\left(x+\frac{1}{2}\right)^2=1\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=1\\x+\frac{1}{2}=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)