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\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\Rightarrow\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
\(\Rightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\Rightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
\(\Rightarrow x-100=0\).Do \(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\ne0\)
\(\Rightarrow x=100\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{x-45}{55}-1-\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
\(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
(x-100)(\(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}=0\)
-> x-100 = 0 -> x = 100
mà \(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\) khác 0
Vậy x = 100
<=>(x+2/13)+1+(2x+45/15)-1 = (3x+8/37)+1+(4x+69/9)-1
<=> x+15/13 + 2x+30/15 = 3x+45/37 + 4x+60/9
<=>(x+15)(1/13+2/15-3/37-4/9) = 0
<=> x+15=0( vì 1/13+2/15-3/37-4/9=0)
<=>x=-15......
dễ thôi mà
Áp dụng tỉ lệ thức, ta có:
\(\Leftrightarrow\frac{108x-4970}{2915}=\frac{92x-4970}{2115}\Rightarrow\left(108x-4970\right)2115=2915\left(92x-4970\right)\)
=>x=100
852+752+652+552−452−352−252-152
=(852-152)+(752-252)+(652-352)+(552-452)
=70.100+50.100+30.100+10.100
=160.100=16000
a) x+1/2004 + 1 + x+2/2003 +1 - x+3/2002 +1 - x+4/2001 +1
=> x+2005/2004 + x+2005/2003 - x+2005/2002 - x+2005/2001=0
=> (x + 2005) ( 1/2004+1/2003 - 1/2002 - 1/2001) =0
ta thấy 1/2004+1/2003-1/2002-1/2001 # 0
=> x+2005=0 => x=-2005
\(\dfrac{x-55}{45}+\dfrac{x-30}{35}+\dfrac{x-25}{25}+\dfrac{x-40}{15}=10\)
\(< =>\dfrac{x-55}{45}+\dfrac{x-30}{35}+\dfrac{x-25}{25}+\dfrac{x-40}{15}-10=0\)
\(< =>\dfrac{x-55}{45}-1+\dfrac{x-30}{35}-2+\dfrac{x-25}{25}-3+\dfrac{x-40}{15}-4=0\)
\(< =>\dfrac{x-100}{45}+\dfrac{x-100}{35}+\dfrac{x-100}{25}+\dfrac{x-100}{15}=0\)
\(< =>\left(x-100\right)\left(\dfrac{1}{45}+\dfrac{1}{35}+\dfrac{1}{25}+\dfrac{1}{15}\right)=0\)
\(< =>x-100=0\left(\dfrac{1}{45}+\dfrac{1}{35}+\dfrac{1}{25}+\dfrac{1}{15}\ne0\right)\)
\(< =>x=100\)
Giải phương trình
\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\)\(\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\)
\(\Leftrightarrow\)\(\dfrac{x+2}{13}+\dfrac{13}{13}+\dfrac{2x+45}{15}-\dfrac{15}{15}=\dfrac{3x+8}{37}+\dfrac{37}{37}+\dfrac{4x+69}{9}-\dfrac{9}{9}\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2x+30}{15}=\dfrac{3x+45}{37}+\dfrac{4x+60}{9}\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}=\dfrac{3\left(x+15\right)}{37}+\dfrac{4\left(x+15\right)}{9}\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}\right)=\left(x+15\right)\left(\dfrac{3}{37}+\dfrac{4}{9}\right)\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}\right)-\left(x+15\right)\left(\dfrac{3}{37}+\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=0\\\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-15\\\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\ne0\end{matrix}\right.\)
Do đó: \(x=-15\)
Vậy \(S=\left\{-15\right\}\)
Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
1. \(x\left(y-4\right)=35-5\left(y-4\right)\) với y= 4 không phải nghiệm y khác 4
\(x=\frac{35}{y-4}-1\)
y=4+35/n
x=n-1
\(\hept{\begin{cases}n=\left\{-7,-5,-1,1,5,7\right\}\\y=\left\{-1,-3,-31,39,11,9\right\}\\x=n-1=\left\{-8,-6,-2,0,4,6\right\}\end{cases}}\)
2.x^2+x+6=y^2
4x^2+4x+1=4y^2-23
(2x+1)^2=4y^2-23
=>4y^2-23=t^2
(2y)^2-t^2=23
=>\(\hept{\begin{cases}y=+-6\\t=+-11\end{cases}\Rightarrow\hept{\begin{cases}2x+1=11\\2x+1=-11\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\x=-6\end{cases}}}\)