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\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne30\\x\ne24\end{cases}}\)
Ta có \(\frac{60}{\frac{120}{x}-4}+\frac{60}{\frac{120}{x}-5}=x\)
\(\Leftrightarrow\frac{60}{\frac{120-4x}{x}}+\frac{60}{\frac{120-5x}{x}}=x\)
\(\Leftrightarrow\frac{60x}{120-4x}+\frac{60x}{120-5x}=x\)
\(\Leftrightarrow\frac{60}{120-4x}+\frac{60}{120-5x}=1\left(Do\text{ }x\ne0\right)\)
\(\Leftrightarrow\frac{15}{30-x}=1-\frac{12}{24-x}\)
\(\Leftrightarrow\frac{15}{30-x}=\frac{24-x-12}{24-x}\)
\(\Leftrightarrow\frac{15}{30-x}=\frac{12-x}{24-x}\)
\(\Leftrightarrow360-15x=\left(12-x\right)\left(30-x\right)\)
\(\Leftrightarrow360-15x=360-42x+x^2\)
\(\Leftrightarrow x^2-27x=0\)
\(\Leftrightarrow x\left(x-27\right)=0\)
\(\Leftrightarrow x=27\left(Tm\text{ }ĐKXĐ\right)\)
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=>\(\frac{30x\left(x-6\right)}{x\left(x+10\right)\left(x-6\right)}+\frac{30x\left(x+10\right)}{x\left(x+10\right)\left(x-6\right)}=\frac{60\left(x+10\right)\left(x-6\right)}{x\left(x-6\left(x+10\right)\right)}\)
=>30x2-180x+30x2+300x=60x2-360x+600x-3600
=>60x2+120x=60x2+240x-3600
=>-120x=-3600
=>x=30
nhớ k mk........Đúng 100%
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Ta có: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
<=> \(\frac{6\left(x+4\right)-30x+120}{30}=\frac{10x-15x+30}{30}\)
<=> 6x + 24 - 30x + 120 = -5x + 30
<=> -24x + 5x = 30 - 144
<=> -19x = -114
<=> x = 6
Vậy S = {6}
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\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)-2=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)-2\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\times\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\right)\)
\(\Leftrightarrow x=-100\)
Vậy......
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\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
\(\Leftrightarrow6\left(x^2+8\right)\left(x^3+3\right)+12\left(x^2+2\right)\left(x^2+3\right)=3\left(x^2+2\right)\left(x^2+8\right)\left(x^2+3\right)-7\left(x^2+2\right)\left(x^2+8\right)\)
\(\Leftrightarrow18x^4+126x^2+216=3x^6+32x^4+68x^2+32\)
\(\Leftrightarrow18x^4+126x^2+216-3x^6-32x^4-68x^2-32=0\)
\(\Leftrightarrow-14x^4+58x^2+184-3x^6=0\)
\(\Leftrightarrow x=\pm2\)
Vậy: nghiệm phương trình là: \(\left\{\pm2\right\}\)
Giải như bạn trên cũng được, nhưng mình nghĩ làm cách này đỡ tốn sức hơn :
\(2,\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
\(\Rightarrow\frac{6}{x^2+2}-1+\frac{12}{x^2+8}-1+\frac{7}{x^2+3}-1=0\)
\(\Rightarrow\frac{6-x^2-2}{x^2+2}+\frac{12-x^2-8}{x^2+8}+\frac{7-x^2-3}{x^2+3}=0\)
\(\Rightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}+\frac{-x^2+4}{x^2+3}=0\)
\(\Rightarrow-\left(x^2-4\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)
Vì \(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\left(>0\forall x\right)\)
\(\Rightarrow x^2-4=0\Rightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x=\pm2\)
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\(5x-200=\frac{5x}{2}-300+2x+300\)0
\(3x-2,5x=200\)\(0,5x=200\)\(x=400\)
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\(x+\frac{1}{x}=1+\sqrt{6}\left(DK:x\ne0\right)\)
\(\Leftrightarrow x^2+1=x\left(1+\sqrt{6}\right)\)
\(\Leftrightarrow x^2-x\left(1+\sqrt{6}\right)+1=0\)
Xét \(\Delta=\left(1+\sqrt{6}\right)^2-4=3+2\sqrt{6}>0\)
\(\Rightarrow\hept{\begin{cases}x_1=\frac{1+\sqrt{6}-\sqrt{3+2\sqrt{6}}}{2}\\x_2=\frac{1+\sqrt{6}+\sqrt{3+2\sqrt{6}}}{2}\end{cases}}\)(TM)
Vậy tập nghiệm của phương trình : \(S=\left\{\frac{1+\sqrt{6}-\sqrt{3+2\sqrt{6}}}{2};\frac{1+\sqrt{6}+\sqrt{3+2\sqrt{6}}}{2}\right\}\)
Mình giải cách của lớp 9 nhé ^^
\(x+\frac{1}{x}=1+\sqrt{6}\)
=> \(\frac{x^2+1}{x}=\frac{x\left(1+\sqrt{6}\right)}{x}\)
=> \(x^2+1=x+x\sqrt{6}\)
=> \(x\)
Ko xác định
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\(\sqrt{\frac{42}{5-x}}+\sqrt{\frac{60}{7-x}}=6\)
\(\Leftrightarrow\sqrt{\frac{42}{5-x}}-\sqrt{\frac{126}{14}}+\sqrt{\frac{60}{7-x}}-\sqrt{\frac{45}{5}}=0\)
\(\Leftrightarrow\frac{\frac{42}{5-x}-\frac{126}{14}}{\sqrt{\frac{42}{5-x}}+\sqrt{\frac{126}{14}}}+\frac{\frac{60}{7-x}-\frac{45}{5}}{\sqrt{\frac{60}{7-x}}+\sqrt{\frac{45}{5}}}=0\)
\(\Leftrightarrow\frac{\frac{-3\left(3x-1\right)}{x-5}}{\sqrt{\frac{42}{5-x}}+\sqrt{\frac{126}{14}}}+\frac{\frac{-3\left(3x-1\right)}{x-7}}{\sqrt{\frac{60}{7-x}}+\sqrt{\frac{45}{5}}}=0\)
\(\Leftrightarrow-3\left(3x-1\right)\left(\frac{\frac{1}{x-5}}{\sqrt{\frac{42}{x-5}}+\sqrt{\frac{126}{14}}}+\frac{\frac{1}{x-7}}{\sqrt{\frac{60}{7-x}}+\sqrt{\frac{45}{5}}}\right)=0\)
Dễ thấy : \(\frac{\frac{1}{x-5}}{\sqrt{\frac{42}{5-x}}+\sqrt{\frac{126}{14}}}+\frac{\frac{1}{x-7}}{\sqrt{\frac{60}{7-x}}+\sqrt{\frac{45}{5}}}>0\)
\(\Rightarrow3x-1=0\Rightarrow x=\frac{1}{3}\)
Chúc bạn học tốt !!!