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Sửa đề: (x-15)/17
=>\(\left(\dfrac{x-15}{17}-5\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-76}{12}-2\right)=0\)=>x-100=0
=>x=100
![](https://rs.olm.vn/images/avt/0.png?1311)
a) n(n+2)
b) (3n-2)3n
c) ( 1) 1 n n 2
d) 1+n2 e) n(n+5)
f) (3n-2)(3n+1)
g) n ( n 3) 2 n n
h) ( 1)( 2) 2
i) n ( n 1)( n 2)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(36^2-14^2\)
\(=\left(36+14\right)\left(36-14\right)\)
\(=50\cdot22\)
\(=1100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)ĐKXĐ:x\(\ne\)0 x\(\ne\)6
=>90(x-6)-36x=2x(x-6)
<=>90x-540-36x=2x2-12x
<=>2x2-12x=54x-540
<=>2x2-66x+540=0
<=>x2-33x+270=0
<=>(x2-15x)-(18x-270)=0
<=>(x-15)(x-18)=0
<=>x=15(tm) hoặc x=18(tm)
b)ĐKXĐ:x\(\ne\)0 x\(\ne\)3
sai đề
c)ĐKXĐ:x\(\ne\)-2 x\(\ne\)2
=>3(x-2)-2(x+2)+8=0
<=>3x-6-2x-4+8=0
<=>x-2=0
<=>x=2(L)
Vậy PT vô nghiệm
d)ĐKXĐ: x\(\ne\)-7
=>10+8=\(\dfrac{3}{2}\)(câu này hình như đề cũng sai)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>x = 12
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow x=12\)
Vậy \(x=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ Vì \(p\) là số nguyên tố
\(\Leftrightarrow p\in\left\{2;3;5;7;........\right\}\)
+) Với \(p=2\Leftrightarrow p+2=2+2=4\left(loại\right)\)
+) Với \(p=3\Leftrightarrow p+6=3+6=9\left(loại\right)\)
+) Với \(p=5\Leftrightarrow\left\{{}\begin{matrix}p+2=5+2=7\\p+6=5+6=11\\p+8=5+8=13\\p+14=5+14=19\end{matrix}\right.\) (Thỏa mãn)
+) Với \(p>5\) \(\Leftrightarrow\left[{}\begin{matrix}p=5k+1\\p=5k+2\\p=5k+3\\p=5k+4\end{matrix}\right.\)
+) \(p=5k+1\Leftrightarrow p+14=\left(5k+1\right)+14=5k+15⋮5\left(loại\right)\)
+) \(p=5k+2\Leftrightarrow p+8=\left(5k+2\right)+8=5k+10⋮5\left(loại\right)\)
+) \(p=5k+3\Leftrightarrow p+2=\left(5k+3\right)+2=5k+5⋮5\left(loại\right)\)
+) \(p=5k+4\Leftrightarrow p+6=\left(5k+4\right)+6=5k+10⋮5\left(loại\right)\)
Vậy \(p=5\)
Còn lại tương tự :vv
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{X-90}{10}+\frac{X-76}{12}+\frac{X-58}{14}+\frac{X-36}{16}+\frac{X-15}{17}=15\)
\(\frac{X-90}{10}+\frac{X-76}{12}+\frac{X-58}{14}+\frac{X-36}{16}+\frac{X-15}{17}-15=0\)
\(\frac{X-90}{10}-1+\frac{X-76}{12}-2+\frac{X-58}{14}-3+\frac{X-36}{16}-4+\frac{X-15}{17}-5=0\)
\(\frac{X-100}{10}+\frac{X-100}{12}+\frac{X-100}{14}+\frac{X-100}{16}+\frac{X-100}{17}=0\)
\(\left(X-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
=> X-100=0
=> X=100
vậy x=100
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: =11
=>\(\dfrac{x-15}{17}-5+\dfrac{x-36}{16}-4+\dfrac{x-76}{12}-2=11-5-4-2=0\)
=>x-100=0
=>x=100
36+14=8+14=2.75*8=22
36-14= 8+14 =2,75*8 =22