\(\frac{x+5}{x+1} - \frac{x-4}{x+6} = \frac{20}{x^{2}+7x+6} \)
Điền số thích hợp:
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\(\Leftrightarrow5\left(7x-1\right)+60x>6\left(16-x\right)\)
=>35x-5+60x>96-6x
=>95x+6x>96+5
=>101x>101
hay x>1
Vậy: S={x|x>1}
\(\dfrac{7x-1}{6}+2x>\dfrac{16-x}{5}\\ \Leftrightarrow\dfrac{5.\left(7x-1\right)}{30}+\dfrac{60x}{30}>\dfrac{6.\left(16-x\right)}{30}\\ \Leftrightarrow35x-5+60x>96-6x\\ \Leftrightarrow35x+60x+6x>96+5\\ \Leftrightarrow101x>101\\ \Leftrightarrow x>1\)
Em tự biểu diễn trục số nha!
\(\Rightarrow\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{6}\)
ĐK:\(x\ne-2;-3;-4;-5\)
MTC:\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right).6\)
Quy đồng khử mẫu:
Đk x khác -2;-3;-4;-5
pt <=> 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) = 1/6
<=> 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 = 1/6
<=> 1/x+2 - 1/x+5 = 1/6
<=> x+5-x-2/(x+2).(x+5) = 1/6
<=> 3/(x+2).(x+5) = 1/6
<=> (x+2).(x+5) = 3 : 1/6 = 18
<=> x^2+7x+10 = 18
<=> x^2+7x-8=0
<=> (x-1).(x+8) = 0
<=> x1=0 hoặc x+8=0
<=> x=1 hoặc x=-8
k mk nha
pt <=> 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/(x+5).(x+6) = 1/8
<=> 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 - 1/x+6 = 1/8
<=> 1/x+2 - 1/x+6 = 1/8
<=> (x+6-x-2)/(x+2).(x+6) = 1/8
<=> 4/(x+2).(x+6) = 1/8
<=>(x+2).(x+6) = 4 : 1/8 = 32
<=>x^2 + 8x + 12 = 32
<=> x^2+8x+12-32=0
<=>x^2+8x-20=0
<=>(x-2).(x+10)=0
<=> x-2 =0 hoặc x+10 = 0
<=> x=2 hoặc x=-10
giang sinh an lanh $%###Xuyen gam cu chuoi###%$
Ko có cách nào hết
\(\frac{x+5}{x+1}-\frac{x-4}{x+6}=\frac{20}{x^2+7x+6}\left(x\ne-1;x\ne-6\right)\)
\(\Leftrightarrow\frac{x+5}{x+1}-\frac{x-4}{x+6}-\frac{20}{x^2+7x+6}=0\)
\(\Leftrightarrow\frac{x+5}{x+1}-\frac{x-4}{x+6}-\frac{20}{\left(x+1\right)\left(x+6\right)}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x+6\right)}{\left(x+1\right)\left(x+6\right)}-\frac{\left(x-4\right)\left(x+1\right)}{\left(x+1\right)\left(x+6\right)}-\frac{20}{\left(x+1\right)\left(x+6\right)}=0\)
\(\Leftrightarrow\frac{x^2+11x+30}{\left(x+1\right)\left(x+6\right)}-\frac{x^2-3x-4}{\left(x+1\right)\left(x+6\right)}-\frac{20}{x^2+7x+6}=0\)
\(\Leftrightarrow\frac{x^2+11x+30-x^2+3x+4-20}{\left(x+1\right)\left(x+6\right)}=0\)
\(\Leftrightarrow\frac{14x+14}{\left(x+1\right)\left(x+6\right)}=0\)
=> 14x+14=0
<=> x=-1 (ktm)
Vậy pt vô nghiệm