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\(\dfrac{100}{x}-\dfrac{100}{x+10}=\dfrac{30}{60}=0,5\left(ĐKXĐ:x\ne0;x\ne-10\right)\\ \Leftrightarrow\dfrac{100\left(x+10\right)-100x}{x\left(x+10\right)}=\dfrac{0,5x\left(x+10\right)}{x\left(x+10\right)}\\ \Leftrightarrow100x-100x+1000=0,5x^2+5x\\ \Leftrightarrow0,5x^2+5x-1000=0\\ \Leftrightarrow0,5x^2-20x+25x-1000=0\\ \Leftrightarrow0,5x.\left(x-40\right)+25.\left(x-40\right)=0\\ \Leftrightarrow\left(0,5x+25\right)\left(x-40\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}0,5x+25=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-50\\x=40\end{matrix}\right.\\ Vậy:S=\left\{-50;40\right\}\)
ĐKXĐ: \(x\notin\left\{10;-10\right\}\)
Ta có: \(\dfrac{720}{x+10}+4=\dfrac{720}{x-10}\)
\(\Leftrightarrow\dfrac{720\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}+\dfrac{4\left(x^2-100\right)}{\left(x+10\right)\left(x-10\right)}=\dfrac{720\left(x+10\right)}{\left(x+10\right)\left(x-10\right)}\)
Suy ra: \(720x-7200+4x^2-400-720x-7200=0\)
\(\Leftrightarrow4x^2=14800\)
\(\Leftrightarrow x^2=3700\)
hay \(x\in\left\{10\sqrt{37};-10\sqrt{37}\right\}\)
ĐKXĐ: \(x\ne\pm10\)
\(\Leftrightarrow\dfrac{180}{x-10}-\dfrac{180}{x+10}=1\)
\(\Leftrightarrow\dfrac{180\left(x+10-x+10\right)}{\left(x-10\right)\left(x+10\right)}=1\)
\(\Leftrightarrow\dfrac{3600}{x^2-100}=1\)
\(\Rightarrow x^2-100=3600\)
\(\Leftrightarrow x^2=3700\)
\(\Leftrightarrow x=\pm10\sqrt{37}\) (thỏa mãn)
Lời giải:
Để pt có 2 nghiệm $x_1,x_2$ thì:
$\Delta'=1+(3+m)=4+m\geq 0\Leftrightarrow m\geq -4$ (chứ không phải với mọi m như đề bạn nhé)!
Áp dụng định lý Viet: \(\left\{\begin{matrix} x_1+x_2=-2\\ x_1x_2=-(m+3)\end{matrix}\right.\)
$x_1, x_2\neq 0\Leftrightarrow -(m+3)\neq 0\Leftrightarrow m\neq -3$
$\frac{x_1}{x_2}-\frac{x_2}{x_1}=\frac{-8}{3}$
$\Leftrightarrow \frac{x_1^2-x_2^2}{x_1x_2}=\frac{-8}{3}$
$\Leftrightarrow \frac{-2(x_1-x_2)}{-(m+3)}=\frac{-8}{3}$
$\Leftrightarrow x_1-x_2=\frac{4}{3}(m+3)$
$\Rightarrow (x_1-x_2)^2=\frac{16}{9}(m+3)^2$
$\Leftrightarrow (x_1+x_2)^2-4x_1x_2=\frac{16}{9}(m+3)^2$
$\Leftrightarrow 4+4(m+3)=\frac{16}{9}(m+3)^2$
$\Leftrightarrow m+3=3$ hoặc $m+3=\frac{-3}{4}$
$\Leftrightarrow m=0$ hoặc $m=\frac{-15}{4}$ (đều thỏa mãn)
`1/10x+1/15(11-x)=1`
`<=>1/10x+11/15-1/15x=1`
`<=>1/30x=1-11/15=4/15`
`<=>x=4/15*30=8`
Vậy `x=8`
\(\dfrac{x}{10}+\dfrac{11-x}{15}=1< =>\dfrac{3x+22-2x}{30}=1\)
\(< =>\dfrac{3x+22-2x}{30}=1=>x+22=30< =>x=30-22< =>x=8\)
\(\dfrac{120}{x}+\dfrac{120}{x-10}=\dfrac{3}{5}\left(dkxd:x>0,x\ne10\right)\)
\(\Leftrightarrow\dfrac{120}{x}+\dfrac{120}{x-10}-\dfrac{3}{5}=0\)
\(\Leftrightarrow\dfrac{120.5\left(x-10\right)+5.120x-3x\left(x-10\right)}{5x\left(x-10\right)}=0\)
\(\Leftrightarrow600x-6000+600x-3x^2+30x=0\)
\(\Leftrightarrow-3x^2+1230x-6000=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\approx405\\x\approx5\end{matrix}\right.\)\(\left(tmdk\right)\)
Vậy ...
\(\dfrac{x^2-26}{10}+\dfrac{x^2-25}{11}\ge\dfrac{x^2-24}{12}+\dfrac{x^2-23}{13}\)
\(\Leftrightarrow\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)
\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)
\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)
\(\Leftrightarrow\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)
Vì \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\Rightarrow x^2-36\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-6\\x\ge6\end{matrix}\right.\)
Bất phương trình đó tương đương với:
\(\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)
⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)
⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)
⇔ \(\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)
+)Vì \(\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}\) nên \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\)
⇔ \(x^2-36\ge0\)
⇔ \(x^2\ge36\)
⇔ \(\sqrt{x^2}\ge6\)
⇔ \(\left|x\right|\ge6\)
⇔ \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)
➤ Vậy \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)
a: =>2/x+2/y=2 và 4/x-2/y=1
=>6/x=3 và 1/x+1/y=1
=>x=2 và 1/y=1-1/2=1/2
=>x=2; y=2
b: Đặt 1/x=a; 1/y=b
=>1/3a+1/3b=1/4 và 5/6a+b=2/3
=>a=1/2; b=1/4
=>x=2; y=4
Đặt \(\dfrac{1}{y-1}=a\), hpt tở thành
\(\left\{{}\begin{matrix}\dfrac{5}{x+1}+a=10\\\dfrac{1}{x-2}+3a=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15}{x+1}+3a=30\left(1\right)\\\dfrac{1}{x-1}+3a=18\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\), ta được:
\(\dfrac{15}{x+1}-\dfrac{1}{x-1}=12\\ \Leftrightarrow\dfrac{15x-15-x-1}{\left(x-1\right)\left(x+1\right)}=12\\ \Leftrightarrow12x^2-12=14x-16\\ \Leftrightarrow12x^2-14x+4=0\\ \Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Với \(x=\dfrac{1}{2}\Leftrightarrow\dfrac{10}{3}+\dfrac{1}{y-1}=10\Leftrightarrow\dfrac{10y-7}{3\left(y-1\right)}=10\)
\(\Leftrightarrow30y-30=10y-7\Leftrightarrow y=\dfrac{23}{20}\)
Với \(x=\dfrac{2}{3}\Leftrightarrow3+\dfrac{1}{y-1}=10\Leftrightarrow\dfrac{1}{y-1}=7\Leftrightarrow7y-7=1\Leftrightarrow y=\dfrac{8}{7}\)
Vậy \(\left(x;y\right)=\left\{\left(\dfrac{1}{2};\dfrac{23}{20}\right);\left(\dfrac{2}{3};\dfrac{8}{7}\right)\right\}\)
$\begin{cases}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac16\\\dfrac{10}{3x}+\dfrac{10}{y}=1\\\end{cases}$
`<=>` $\begin{cases}\dfrac{10}{x}+\dfrac{10}{y}=\dfrac53\\\dfrac{10}{3x}+\dfrac{10}{y}=1\\\end{cases}$
`<=>` $\begin{cases}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac16\\\dfrac{20}{3}x=\dfrac23\\\end{cases}$
`<=>` $\begin{cases}x=\dfrac{1}{10}\\y=\dfrac{1}{15}\\\end{cases}$
Vậy `(x,y)=(1/10,1/15)`
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{10}{3x}+\dfrac{10}{y}=1\end{matrix}\right.\left(x,y\ne0\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{10}{3}.\dfrac{1}{x}+10.\dfrac{1}{y}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{10}{x}+\dfrac{10}{y}=\dfrac{5}{3}\left(1\right)\\\dfrac{10}{3}.\dfrac{1}{x}+\dfrac{10}{y}=1\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\Rightarrow\dfrac{20}{3}.\dfrac{1}{x}=\dfrac{2}{3}\Rightarrow\dfrac{1}{x}=\dfrac{1}{10}\Rightarrow x=10\)
\(\Rightarrow\dfrac{1}{y}=\dfrac{1}{6}-\dfrac{1}{10}=\dfrac{1}{15}\Rightarrow y=15\)
x.10=(x-10).30
=>10x=30x-300
=>10x-30x+300=0
=>-20x-300=0
=>x=15
\(\Leftrightarrow\dfrac{x}{30}-\dfrac{x-10}{10}=0\)
\(\Leftrightarrow\dfrac{x-3\left(x-10\right)}{30}=0\)
\(\Leftrightarrow x-3x+30=0\)
\(\Leftrightarrow-2x+30=0\)
\(\Leftrightarrow-2x=-30\)
\(\Leftrightarrow x=15\)