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bạn viết sai đề bài nhé
(x+2)/11+(x+2)/12+(x+2)/13=(x+2)/14+(x+2)15
<=> (x+2)/11+(x+2)/12+(x+2)/13 - (x+2)/14 - (x+2)/15 = 0
<=> (x+2)(1/11+1/12+1/13 - 1/14 - 1/15 ) = 0
vì: (1/11+1/12+1/13 - 1/14 - 1/15 ) khác 0 nên x-2 = 0 => x=2
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\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)
\(\Leftrightarrow\dfrac{1680.\left(x-3\right)+1560.\left(x-3\right)-1456.\left(x-3\right)-1365.\left(x-3\right)}{21840}=0\)
\(\Leftrightarrow\left(x-3\right).\left(1680+1560-1456-1365\right)=0\)
\(\Leftrightarrow\left(x-3\right).419=0\)
\(\Leftrightarrow419x=1257\)
\(\Leftrightarrow x=3\)
Lời giải:
\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\((x-3)\left(\frac{1}{13}+\frac{1}{14}\right)=(x-3)\left(\frac{1}{15}+\frac{1}{16}\right)\)
\((x-3)\left[\left(\frac{1}{13}+\frac{1}{14}\right)-\left(\frac{1}{15}+\frac{1}{16}\right)\right]=0\)
Ta thấy:
\(\frac{1}{13}>\frac{1}{15}; \frac{1}{14}>\frac{1}{16}\Rightarrow \frac{1}{13}+\frac{1}{14}> \frac{1}{15}+\frac{1}{16}\)
Do đó biểu thức trong ngoặc vuông lớn hơn $0$ hay khác $0$
$\Rightarrow x-3=0$
$\Leftrightarrow x=3$
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\(\frac{x-1}{13}+\frac{x-2}{14}+\frac{x-3}{15}+3=0\)
\(\frac{x-1}{13}+1+\frac{x-2}{14}+1+\frac{x-3}{15}+1=0\)
\(\frac{x-1}{13}+\frac{13}{13}+\frac{x-2}{14}+\frac{14}{14}+\frac{x-3}{15}+\frac{15}{15}=0\)
\(\frac{x+12}{13}+\frac{x+12}{14}+\frac{x+12}{15}=0\)
\(\left(x+12\right)\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)=0\)
\(x+12=0\)hay \(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}=0\)(vô lý)
=>x+12=0
x=0-12
x=-12
Vậy x=-12
\(\frac{x-1}{13}\)+ \(\frac{x-2}{14}\) + \(\frac{x-3}{15}\)+ 3 = 0
=> (\(\frac{x-1}{13}\)+ 1 ) + (\(\frac{x-2}{14}\)+ 1) + (\(\frac{x-3}{15}\) + 1) = 0
=> \(\frac{x+12}{13}\)+ \(\frac{x+12}{14}\)+ \(\frac{x+12}{15}\)= 0
=> (x + 12) . (\(\frac{1}{13}\)+\(\frac{1}{14}\)+ \(\frac{1}{15}\)) = 0
=>(x + 12) . \(\frac{587}{2730}\)= 0
=> x + 12 = 0
=> x = -12
Vậy x = -12
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\(\frac{x-3}{53}+\frac{x+14}{36}=\frac{x+15}{35}-\frac{x}{50}\)
\(\left(1+\frac{x-3}{53}\right)+\left(1+\frac{x+14}{36}\right)=\left(\frac{x+15}{35}+1\right)-\left(1+\frac{x}{50}\right)\)
\(\frac{x+50}{53}+\frac{x+50}{36}-\frac{x+15}{35}+\frac{x+50}{50}=0\)
\(\left(x+50\right).\left(\frac{1}{53}+\frac{1}{36}-\frac{1}{35}+\frac{1}{50}\right)=0\)
\(\left(\frac{1}{53}+\frac{1}{36}-\frac{1}{35}+\frac{1}{50}\right)\ne0\)
\(\Rightarrow\left(x+50\right)=0\Rightarrow x=-50\)
Vậy x=-50
p/s: bn viết sai đề nên mk sửa lại
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\(x:\dfrac{13}{16}=\dfrac{5}{-8}\\ x=\dfrac{5}{-8}.\dfrac{13}{16}=\dfrac{65}{-128}\)
\(x.\dfrac{-1}{2}=\dfrac{-4}{5}\\ x=\dfrac{-4}{5}:\dfrac{-1}{2}=\dfrac{8}{5}\)
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a. 2(x-1) + (x+2) - (x+3) = 15 - (x+1)
=>2x-2+x+2-x-3=15-x-1
=>(2x+x-x)-2+2-3=15-1-x
=>2x-3=14-x
=>3x=17
=>x=17/3
b. x+1/15 + x+2/14 = x+4/12 + x+5/11
\(\Rightarrow\frac{x+1}{15}+1+\frac{x+2}{14}+1=\frac{x+4}{12}+1+\frac{x+5}{11}+1\)
\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{14}=\frac{x+16}{12}+\frac{x+16}{11}\)
\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+16}{12}-\frac{x+16}{11}=0\)
\(\Rightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{11}\right)=0\)
\(\Rightarrow x+16=0\).Do \(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{11}\ne0\)
=>x=-16
đặt: x-13=a thì: \(|a|^{15}+|a-1|^{15}=1\)
Nếu \(a=1\text{ hoặc }0\text{ thì thỏa. Nếu: }a>1\text{ thì VT}>\text{VP}\left(\text{loại}\right)\)
\(\text{Nếu: }0< a< 1\text{ thì: }a^{15}+\left(1-a\right)^{15}=1\text{ thấy ngay: }0< a< 1;0< 1-a< 1\)
do đó: \(VT< a+1-a=1=VP\left(\text{loại}\right)\)
Nếu: a<0 thì hiển nhiên VT>VP nên loại
Vậy: a=0 hoặc bằng 1 hay x=13 hoặc 14