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a) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\Leftrightarrow\left(\frac{x-45}{55}-1\right)+\left(\frac{x-47}{53}-1\right)=\left(\frac{x-55}{45}-1\right)+\left(\frac{x-53}{47}-1\right)\)
\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
Vì \(\hept{\begin{cases}\frac{1}{55}< \frac{1}{45}\\\frac{1}{53}< \frac{1}{47}\end{cases}}\Rightarrow\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}< 0\)
\(\Rightarrow x-100=0\Rightarrow x=100\)
Vậy x = 100
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a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)
b, \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)
a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền
b, - 1 hoặc + 1 vs mỗi phân số nha
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\(\left(\dfrac{x+1}{55}+\dfrac{x+2}{56}+\dfrac{x+3}{57}+\dfrac{x+4}{58}\right)-4=0\)
<=>\(\dfrac{x+1}{55}+\dfrac{x+2}{56}+\dfrac{x+3}{57}+\dfrac{x+4}{58}=4\)
<=>\(\dfrac{x+1}{55}-1+\dfrac{x+2}{56}-1+\dfrac{x+3}{57}+\dfrac{x+4}{58}-1=4-4\)
<=>\(\dfrac{x+1}{55}-\dfrac{55}{55}+\dfrac{x+2}{56}-\dfrac{56}{56}+\dfrac{x+3}{57}-\dfrac{57}{57}+\dfrac{x+4}{58}-\dfrac{58}{58}=0\)
<=>\(\dfrac{x-54}{55}+\dfrac{x-54}{56}+\dfrac{x-54}{57}+\dfrac{x-54}{58}=0\)
<=>\(\left(x-54\right)\left(\dfrac{1}{55}+\dfrac{1}{56}+\dfrac{1}{57}+\dfrac{1}{58}\right)=0\)
<=>x-54=0
<=>x=54
vậy phương trình có tập nghiệm là S={54}
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a
Ta có \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) ( đúng )
\(\Rightarrow x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=\frac{3^2}{3}=3\)
Dấu "=" xảy ra tại a=b=c=1
b
\(P=\frac{x}{\left(x+10\right)^2}\)
Đặt \(y=\frac{1}{x+10}\Rightarrow x=\frac{1}{y}-10\)
\(\Rightarrow P=\left(\frac{1}{y}-10\right)\cdot y^2=-10y^2+y\)
\(=-10\left(y^2-2\cdot y\cdot\frac{1}{20}\cdot y+\frac{1}{400}\right)+\frac{1}{40}\)
\(=-10\left(y-\frac{1}{2}\right)^2+\frac{1}{40}\le\frac{1}{40}\)
Dấu "=" xảy ra tại \(y=\frac{1}{2}\Leftrightarrow x=10\)
Vậy...............................
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Lời giải:
$x^{99}+x^{55}+x^n+x-7=(x^{99}+x)+(x^{55}+x)+x^n-x-7$
$=x(x^{98}+1)+x(x^{54}+1)+x^n-x-7$
Hiển nhiên: $x^{98}+1=(x^2)^{49}+1\vdots x^2+1$
$x^{54}+1=(x^2)^{27}+1\vdots x^2+1$
Xét các TH sau:
TH1: $n=4k$ thì $x^n-1=x^{4k}-1\vdots x^4-1\vdots x^2+1$. Khi đó đa thức dư là $-x-6$
TH2: $n=4k+1$ thì $x^{n}-x=x(x^{4k}-1)\vdots x^2+1$. Khi đó đa thức dư là $-7$
TH3: $n=4k+2$ thì: $x^n+1=x^{4k+2}+1=(x^2)^{2k+1}+1\vdots x^2+1$. Khi đó đa thức dư là $-x-8$
TH4: $n=4k+3$ thì $x^n+x=x^{4k+3}+x=x(x^{4k+2}+1)\vdots x^2+1$. Khi đó đa thức dư là $-2x-7$
Ta có: \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=55\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=55\)
\(\Leftrightarrow x^4-10x^2-4x^2+40=55\)
\(\Leftrightarrow x^4-14x^2-15=0\)
Đặt \(t=x^2\left(t\ge0\right)\), ta có: \(t^2-14t-15=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=15\\t=-1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{15}\\x=-\sqrt{15}\end{matrix}\right.\)
<=>\(\left(x^2-4\right)\left(x^2-10\right)=55\)
<=>\(x^4-14x^2+40=55\)
<=>\(x^4-14x^2-15=0\)
<=>\(\left(x^2-15\right)\left(x^2+1\right)=0\)
<=>\(x^2-15=0\)(cái kia lun lớn hơn 0)
<=>\(x^2=15\)
<=>\(x=\pm\sqrt{15}\)