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1.
\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\left(ĐKXĐ:x\ne1\right)\\ \Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\\ \Leftrightarrow21x-9=2x-2\\ \Leftrightarrow19x=7\\ \Leftrightarrow x=\dfrac{7}{19}\left(TMĐK\right)\)
2.
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\left(ĐKXĐ:x\ne-\dfrac{2}{3};x\ne\dfrac{1}{3}\right)\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\\ \Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\\ \Leftrightarrow-8x+1=-11x-14\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\left(TMĐK\right)\)
3.
\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ \Leftrightarrow\left(\dfrac{1-x}{x+1}+3\right)\left(x+1\right)=2x+3\\ \Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}.\left(x+1\right)=2x+3\\ \Leftrightarrow\dfrac{4+2x}{x+1}\left(x+1\right)=2x+3\\ \Leftrightarrow4+2x=2x+3\\ \Leftrightarrow4=3\)
Vô nghiệm.
a: \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
=>(x+4)(x+7)=54
=>x^2+11x+28-54=0
=>(x+13)(x-2)=0
=>x=-13 hoặc x=2
b: \(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-...+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{3}\)
=>\(\dfrac{x+5-x-1}{\left(x+5\right)\left(x+1\right)}=\dfrac{1}{3}\)
=>x^2+6x+5=12
=>x^2+6x-7=0
=>(x+7)(x-1)=0
=>x=-7 hoặc x=1
\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{x+5-x}{x\left(x+5\right)}=\dfrac{5}{x\left(x+5\right)}\)
\(B=\dfrac{1}{x\left(x+1\right)}+\)\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
=>B=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
=>B=1/x - 1/x+5=5/x(x+5)
b) \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{1}{18}\\< =>\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{18}\\ < =>\dfrac{1}{x+1}-\dfrac{1}{x+5}=\dfrac{1}{18}\\ quyđồngmẫuvàkhửmẫu\\ x^{2^{ }}+6x-27=0\\ giảipttìmđược:x=3;x=-9\)
a) \(\frac{x-2015}{1}+\frac{x-2014}{2}+\frac{x-2013}{3}+...+\frac{x-1}{2015}+\frac{x}{2016}=0\\ \Leftrightarrow\frac{x-2015}{1}-1+\frac{x-2014}{2}-1+...+\frac{x-1}{2015}-1+\frac{x}{2016}-1=-2016\)
\(\Leftrightarrow\frac{\left(x-2016\right).1}{1}+\frac{\left(x-2016\right).1}{2}+\frac{\left(x-2016\right).1}{3}+...+\frac{\left(x-2016\right).1}{2015}+\frac{\left(x-2016\right).1}{2016}=-2016\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=-2016\)
tới đây mình chịu. mình nghĩ là phương trình bạn cho là bằng 2016 chứ, như thế giải mới được, còn như này thì mình bó tay
b)
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}=\frac{1}{8}\\ \Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{8}\\ \Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\\ \Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\)
\(\Leftrightarrow x^2+8x+12-32=0\\ \Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+10=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-10\\x=2\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={-10;2}
\(M=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{x+5}\)\(\Rightarrow M=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{x+4}{\left(x+4\right)\left(x+5\right)}\)\(\Rightarrow M=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{x+4}\)\(\Rightarrow M=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{x+3}\)
\(\Rightarrow M=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{x+2}\)
\(\Rightarrow M=\dfrac{1}{x+1}\)
M= \(\dfrac{1}{(x+1)(x+2)}\)+ \(\dfrac{1}{(x+2)(x+3)}\)+ \(\dfrac{1}{(x+3)(x+4)}\)+ \(\dfrac{1}{(x+4)(x+5)}\)+ \(\dfrac{1}{x+5}\)
M= \(\dfrac{1}{x+1}\)- \(\dfrac{1}{x+2}\)+ \(\dfrac{1}{x+2}\)- \(\dfrac{1}{x+3}\)+ \(\dfrac{1}{x+3}\)- \(\dfrac{1}{x+4}\)+ \(\dfrac{1}{x+4}\)+ \(\dfrac{1}{x+5}\)
M= \(\dfrac{1}{x+1} + \dfrac{1}{x+5}\)
M= \(\dfrac{x+5}{(x+1)(x+5)} + \dfrac{x+1}{(x+1)(x+5)} \)
M= \(\dfrac{x+5+x+1}{(x+1)(x+5)}\)
M= \(\dfrac{2x+6}{(x+1)(x+5)}\)
M= \(\dfrac{2(x+3)}{(x+1)(x+5)}\)
\(M=\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(M=\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)
\(M=\dfrac{1}{x-1}-\dfrac{1}{x-5}\)
\(M=\dfrac{x-5-x+1}{\left(x-5\right)\left(x-1\right)}=-\dfrac{4}{x^2-6x+5}\)
a.
\(\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-11x+28}+\dfrac{1}{x^2-19x+84}=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-7\right)}+\dfrac{1}{\left(x-7\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Rightarrow\dfrac{1}{x-3}-\dfrac{1}{x-12}=\dfrac{1}{4}\\ \Rightarrow\dfrac{-9}{\left(x-3\right)\left(x-12\right)}=\dfrac{1}{4}\\ \Rightarrow x^2-15x+36=-36\\ \)
Tự giải tiếp
Ta có : \(P=\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
=\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-...+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
= \(\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{5}{x\left(x+5\right)}\)
a, Với x=\(\dfrac{\sqrt{29}-5}{2}\Rightarrow A=\dfrac{5}{\dfrac{\sqrt{29}-5}{2}\left(\dfrac{\sqrt{29}-5}{2}+5\right)}\)
Mấy cái còn lại tương tự , bạn tự làm nha
a/ \(\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}\)
\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+4}\)
Vậy..
b/ \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+5}\)
Vậy..