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Phương trình đầu bài tương đương với
\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)\(\Leftrightarrow\frac{x+43+57}{57}+\frac{x+46+54}{54}=\frac{x+49+51}{51}+\frac{x+52+48}{48}\)\(\Leftrightarrow\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
\(\Leftrightarrow\orbr{\begin{cases}x+100=0\\\frac{1}{57}+\frac{1}{54}=\frac{1}{51}+\frac{1}{48}\left(sai\right)\end{cases}\Leftrightarrow x+100=0\Leftrightarrow x=-100}\)
Vậy phương trình có nghiệm duy nhất là x=-100
<=> \(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)
<=> \(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
<=> \(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
vi \(\frac{1}{57}< \frac{1}{51};\frac{1}{54}< \frac{1}{48}\Rightarrow\frac{1}{57}-\frac{1}{51}+\frac{1}{54}-\frac{1}{48}< 0\)
=> x+100=0 => x= -100
vay pt co nghiem \(x=-100\)
\(\frac{2}{x^2+1}+\frac{4}{x^2+3}+\frac{6}{x^2+5}=3+\frac{x^2-1}{x^2+6}\)
\(\Leftrightarrow\frac{x^2-1}{x^2+6}+1-\frac{2}{x^2+1}+1-\frac{4}{x^2+3}+1-\frac{6}{x^2+5}=0\)
\(\Leftrightarrow\frac{x^2-1}{x^2+6}+\frac{x^2-1}{x^2+1}+\frac{x^2-1}{x^2+3}+\frac{x^2-1}{x^2+5}=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(\frac{1}{x^2+6}+\frac{1}{x^2+1}+\frac{1}{x^2+3}+\frac{1}{x^2+5}\right)=0\)
\(\Rightarrow x=\pm1\)
Ta thấy : \(\hept{\begin{cases}AD\perp DC\\MP\perp AD\end{cases}}\) \(\Rightarrow PM//DC\)
\(\Rightarrow\frac{MP}{CD}=\frac{AM}{AC}\) ( định lý Talet )
Chứng minh tương tự ta có : \(MN//AB\)
\(\Rightarrow\frac{MN}{AB}=\frac{MC}{AC}\) ( định lý Talet )
Khi đó : \(\frac{MN}{AB}+\frac{MP}{CD}=\frac{AM}{AC}+\frac{MC}{AC}=\frac{AC}{AC}=1\) (ĐPCM)
ĐKXĐ: ...
\(\Leftrightarrow\frac{49}{\left(x-7\right)^2}+1=\frac{25}{x^2}\)
\(\Leftrightarrow\frac{49x^2}{\left(x-7\right)^2}+x^2=25\)
\(\Leftrightarrow\frac{49x^2}{\left(x-7\right)^2}+2.\frac{7x}{x-7}.x+x^2-\frac{14x^2}{x-7}=25\)
\(\Leftrightarrow\left(\frac{7x}{x-7}+x\right)^2-\frac{14x^2}{x-7}=25\)
\(\Leftrightarrow\left(\frac{x^2}{x-7}\right)^2-\frac{14x^2}{x-7}-25=0\)
Đặt \(\frac{x^2}{x-7}=a\)
\(\Rightarrow a^2-14a-25=0\)
Nghiệm xấu, bạn tự giải tiếp đoạn cuối
ĐKXĐ: ...
Đặt \(\frac{x}{3}-\frac{4}{x}=a\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}=a^2\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=a^2+\frac{8}{3}\)
\(\Rightarrow\frac{x^2}{3}+\frac{48}{x^2}=3a^2+8\)
\(3a^2+8=10a\Leftrightarrow3a^2-10a+8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}-\frac{4}{x}=2\\\frac{x}{3}-\frac{4}{x}=\frac{4}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-6x-12=0\\x^2-4x-12=0\end{matrix}\right.\)
ĐKXĐ: ...
Đặt \(\frac{x}{3}-\frac{4}{x}=a\Rightarrow a^2=\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=a^2+\frac{8}{3}\)
\(a^2+\frac{8}{3}=\frac{10}{3}a\Leftrightarrow3a^2-10a+8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{x}{3}-\frac{4}{x}=2\\\frac{x}{3}-\frac{4}{x}=\frac{4}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-6x-12=0\\x^2-4x-12=0\end{matrix}\right.\)
ĐKXĐ: ...
Đặt \(\frac{10}{x}-\frac{x}{6}=a\Rightarrow a^2=\frac{100}{x^2}+\frac{x^2}{36}-\frac{10}{3}\Rightarrow\frac{100}{x^2}+\frac{x^2}{36}=a^2+\frac{10}{3}\)
\(\Rightarrow\frac{900}{x^2}+\frac{x^2}{4}=9a^2+30\)
Phương trình trở thành:
\(9a^2+30=2+48a\)
\(\Leftrightarrow9a^2-48a+28=0\Rightarrow\left[{}\begin{matrix}a=\frac{14}{3}\\a=\frac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{10}{x}-\frac{x}{6}=\frac{14}{3}\\\frac{10}{x}-\frac{x}{6}=\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{x^2}{6}+\frac{14}{3}x-10=0\\\frac{x^2}{6}+\frac{2}{3}x-10=0\end{matrix}\right.\)
Đặt \(x^2-4x+5=a\) (\(a\ge1\))
\(\frac{21}{a}-a-1=0\)
\(\Leftrightarrow-a^2-a+21=0\)
Nghiệm xấu, bạn coi lại dề
\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Rightarrow\frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)
\(\Rightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Rightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
Dễ thấy\(\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)>0\Rightarrow x-50=0\Rightarrow x=50\)
Vậy x = 50
Ta có
\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Leftrightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)
\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
Mà : \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\)
\(\Rightarrow x-50=0\)
\(\Rightarrow x=50\)
Vậy : \(x=50\)