giải pt
x/60+x/45=7
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\(\frac{60}{5}+x=\frac{60}{x-1}\) (x khác 1)
\(\Leftrightarrow12+x=\frac{60}{x-1}\)
\(\Leftrightarrow\frac{x^2-x-60}{x-1}=12\)\(\Leftrightarrow x^2-x-60=12x-12\)
\(\Leftrightarrow x^2+11x-48=0\)
ĐỂ PT CÓ NGHIỆM THÌ
\(\Delta=11^2+4.48=121+192=323>0\)
giải tiếp là ra
\(\frac{x}{40}-\frac{x}{45}=\frac{3}{2}\)
\(\Leftrightarrow\frac{9x-8x}{360}=\frac{3}{2}\)
\(\Leftrightarrow2x=3.360\)
\(\Leftrightarrow2x=1080\)
\(\Leftrightarrow x=540\)
\(\frac{x}{40}-\frac{x}{45}=\frac{3}{2}\)
\(\Leftrightarrow\frac{18x}{720}-\frac{16x}{720}=\frac{1080}{720}\)
\(\Rightarrow18x-16x=1080\)(KHỬ MẪU)
\(\Leftrightarrow2x=1080\)
\(\Leftrightarrow x=\frac{1080}{2}\)
\(\Leftrightarrow x=540\)
Vậy tập nghiệm của phương trình là \(S=\left\{540\right\}\)
Bài làm:
1) đk: \(x\ne0;x\ne-5\)
Ta có: \(\frac{30}{x}-\frac{30}{x+5}=1\)
\(\Leftrightarrow\frac{30\left(x+5\right)-30x}{x\left(x+5\right)}=1\)
\(\Leftrightarrow x^2+5x=150\)
\(\Leftrightarrow x^2+5x-150=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)
2) đk: \(x\ne0;x\ne-2\)
Ta có: \(\frac{60}{x}-\frac{60}{x+2}=1\)
\(\Leftrightarrow\frac{60\left(x+2\right)-60x}{x\left(x+2\right)}=1\)
\(\Leftrightarrow x^2+2x=120\)
\(\Leftrightarrow x^2+2x-120=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)
\(\frac{30}{x}-\frac{30}{x+5}=1\)( ĐKXĐ : \(x\ne0;x\ne-5\))
<=> \(30\left(\frac{1}{x}-\frac{1}{x+5}\right)=1\)
<=> \(30\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{x\left(x+5\right)}\right)=1\)
<=> \(30\left(\frac{5}{x\left(x+5\right)}\right)=1\)
<=> \(\frac{5}{x\left(x+5\right)}=\frac{1}{30}\)
<=> \(5\cdot30=x\left(x+5\right)\)
<=> \(x^2+5x-150=0\)
<=> \(x^2+15x-10x-150=0\)
<=> \(x\left(x+15\right)-10\left(x+15\right)=0\)
<=> \(\left(x-10\right)\left(x+15\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)( tmđk )
Vậy S = { 10 ; -15 }
\(\frac{60}{x}-\frac{60}{x+2}=1\)( ĐKXĐ : \(x\ne0;x\ne-2\))
<=> \(60\left(\frac{1}{x}-\frac{1}{x+2}\right)=1\)
<=> \(60\left(\frac{x+2}{x\left(x+2\right)}-\frac{x}{x\left(x+2\right)}\right)=1\)
<=> \(60\left(\frac{2}{x\left(x+2\right)}\right)=1\)
<=> \(\frac{2}{x\left(x+2\right)}=\frac{1}{60}\)
<=> \(2\cdot60=x\left(x+2\right)\)
<=> \(x^2+2x-120=0\)
<=> \(x^2+12x-10x-120=0\)
<=> \(x\left(x+12\right)-10\left(x+12\right)=0\)
<=> \(\left(x-10\right)\left(x+12\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)
Vậy S = { 10 ; -12 }
<=>x-45/55 -1 + x-47/53 -1=x-55/45 -1 + x-53/47-1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55 + 1/53 - 1/45 - 1/47 )=0
vi (1/55 + 1/53 - 1/45 - 1/47 ) luon khac 0 nen x-100=0 <=>x=100
\(\dfrac{x}{60}+\dfrac{x}{45}=7\)
\(\Leftrightarrow\dfrac{315x+420x-132300}{18900}=0\)
\(\Leftrightarrow735x=132300\)
\(\Leftrightarrow x=180\)