Ôn tập: Phương trình bâc nhất một ẩn

1b
2e
3a
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6c

a: \(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{x-6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{-4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow x^2-8x+12=-32\)

\(\Leftrightarrow x^2-8x+16+28=0\)

\(\Leftrightarrow\left(x-4\right)^2+28=0\)(vô lý)

c: \(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)

=>x-59=0

hay x=59

a, ĐKXĐ:\(x\ne2,x\ne3,x\ne4,x\ne5,x\ne6\)

\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}-\dfrac{1}{8}=0\\ \Leftrightarrow\dfrac{8\left(x-6\right)-8\left(x-2\right)-\left(x-2\right)\left(x-6\right)}{8\left(x-2\right)\left(x-6\right)}=0\)

\(\Rightarrow8x-48-8x+16-x^2+8x-12=0\\ \Leftrightarrow-x^2+8x-44=0\\ \Leftrightarrow x^2-8x+44=0\\\left(x^2-8x+16\right)+28=0\\ \Leftrightarrow\left(x-4\right)^2+28=0\left(vô.lí\right)\)

a, ĐKXĐ:\(x\ne2,x\ne3,x\ne4,x\ne5,x\ne6\)

\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}-\dfrac{1}{8}=0\\ \Leftrightarrow\dfrac{8\left(x-6\right)-8\left(x-2\right)-\left(x-2\right)\left(x-6\right)}{8\left(x-2\right)\left(x-6\right)}=0\)

\(\Rightarrow8x-48-8x+16-x^2+8x-12=0\\ \Leftrightarrow-x^2+8x-44=0\\ \Leftrightarrow x^2-8x+44=0\\\left(x^2-8x+16\right)+28=0\\ \Leftrightarrow\left(x-4\right)^2+28=0\left(vô.lí\right)\)

\(b,\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\\ \Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\\ \Leftrightarrow\dfrac{x-259}{17}+\dfrac{x-259}{19}+\dfrac{x-259}{21}+\dfrac{x-259}{23}=0\\ \Leftrightarrow\left(x-259\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0\\ \Leftrightarrow x=259\left(vì.\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\ne0\right)\)\(c,\dfrac{x-29}{30}+\dfrac{x-30}{29}=\dfrac{29}{x-30}+\dfrac{30}{x-29}\\ \Leftrightarrow\left(\dfrac{x-29}{30}-1\right)+\left(\dfrac{x-30}{29}-1\right)=\left(\dfrac{29}{x-30}-1\right)+\left(\dfrac{30}{x-29}-1\right)\\ \Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{59-x}{x-30}+\dfrac{59-x}{x-29}\\ \Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=-\dfrac{x-59}{x-30}-\dfrac{x-59}{x-29}\\ \Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}+\dfrac{x-59}{x-30}+\dfrac{x-59}{x-29}=0\)

\(\Leftrightarrow\left(x-59\right)\left(\dfrac{1}{30}+\dfrac{1}{29}+\dfrac{1}{x-30}+\dfrac{1}{x-29}\right)=0\\ \Leftrightarrow x=59\left(vì.\dfrac{1}{30}+\dfrac{1}{29}+\dfrac{1}{x-30}+\dfrac{1}{x-29}\ne0\right)\)

a: =>x-2+2=x²+2x

\(\Leftrightarrow x^2+x=0\)

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

b: Bạn xem lại đề

c: \(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\)

\(\Leftrightarrow2\left(x^2+10x+25\right)-\left(x-5\right)^2=x^2+25x\)

\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)

=>5x+25=0

hay x=-5(loại)

a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x}+\dfrac{2}{x^2-2x}=\dfrac{x+2}{x-2}\\ \Leftrightarrow\dfrac{x-2+2-x\left(x-2\right)}{x\left(x-2\right)}=0\\ \Rightarrow x-x^2+2x=0\\ \Leftrightarrow-x^2+3x=0\\ \Leftrightarrow-x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

b, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-9\\x\ne\pm\dfrac{3}{4}\end{matrix}\right.\)

\(\dfrac{x+5}{x+9}+\dfrac{3x-7}{9-12x}=\dfrac{4x^2+x-7}{16x^2-9}\\ \Leftrightarrow\dfrac{3\left(x+5\right)\left(4x-3\right)\left(4x+3\right)}{3\left(x+9\right)\left(4x-3\right)\left(4x+3\right)}-\dfrac{\left(x+9\right)\left(3x-7\right)\left(4x+3\right)}{3\left(4x-3\right)\left(x+9\right)\left(4x+3\right)}-\dfrac{3\left(4x^2+x-7\right)\left(x+9\right)}{3\left(4x-3\right)\left(4x+3\right)\left(x+9\right)}=0\)

\(\Rightarrow48x^3+240x^2-27x-135-12x^3-89x^2+192x+189-12x^3-111x^2-6x+189=0\)

\(\Leftrightarrow24x^3+40x^2+159x+243=0\)

nghiệm xấu lắm bạn

c,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\)

\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\\ \Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}-\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}-\dfrac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}=0\\ \Leftrightarrow\dfrac{2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)-\left(x^2+25x\right)}{2x\left(x+5\right)\left(x-5\right)}=0\)

\(\Rightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\\ \Leftrightarrow5x+25=0\\ \Leftrightarrow x=-5\left(ktm\right)\)

a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x}+\dfrac{2}{x^2-2x}=\dfrac{x+2}{x-2}\\ \Leftrightarrow\dfrac{x-2+2-x\left(x+2\right)}{x\left(x-2\right)}=0\\ \Rightarrow x-x^2-2x=0\\ \Leftrightarrow-x^2-x=0\\ \Leftrightarrow-x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

b, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-9\\x\ne\pm\dfrac{3}{4}\end{matrix}\right.\)

\(\dfrac{x+5}{x+9}+\dfrac{3x-7}{9-12x}=\dfrac{4x^2+x-7}{16x^2-9}\\ \Leftrightarrow\dfrac{3\left(x+5\right)\left(4x-3\right)\left(4x+3\right)}{3\left(x+9\right)\left(4x-3\right)\left(4x+3\right)}-\dfrac{\left(x+9\right)\left(3x-7\right)\left(4x+3\right)}{3\left(4x-3\right)\left(x+9\right)\left(4x+3\right)}-\dfrac{3\left(4x^2+x-7\right)\left(x+9\right)}{3\left(4x-3\right)\left(4x+3\right)\left(x+9\right)}=0\)

\(\Rightarrow48x^3+240x^2-27x-135-12x^3-89x^2+192x+189-12x^3-111x^2-6x+189=0\)

\(\Leftrightarrow24x^3+40x^2+159x+243=0\)

nghiệm xấu lắm bạn

c,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\)

\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\\ \Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}-\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}-\dfrac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}=0\\ \Leftrightarrow\dfrac{2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)-\left(x^2+25x\right)}{2x\left(x+5\right)\left(x-5\right)}=0\)

\(\Rightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\\ \Leftrightarrow5x+25=0\\ \Leftrightarrow x=-5\left(ktm\right)\)

Lỗi

\(\left(x-4\right)\left(x+4\right)\left(x-3\right)=0\Leftrightarrow x=4;x=-4;x=3\)

\(\left(x^2-16\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-16=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=16\\x=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\x=3\end{matrix}\right.\)

Vậy \(S=\left\{4;-4;3\right\}\)

\(\left(x^2-16\right)=0;\left(x-3\right)=0\\x^2=16;x=3\\ x=-4;x=4;x=3\)

Vậy...

a, \(P=\dfrac{x}{2\left(x-1\right)}+\dfrac{x^2+1}{2\left(1-x^2\right)}=\dfrac{x\left(1-x\right)+x^2+1}{2\left(1-x^2\right)}=\dfrac{x+1}{2\left(1-x^2\right)}=\dfrac{1}{2\left(1-x\right)}\)

b, Thay x = -2 ta được \(P=\dfrac{1}{2\left(1+2\right)}=\dfrac{1}{4}\)

a: =>5x-5+17x=1-12x-4

=>22x-5=-12x-3

=>34x=2

hay x=1/17

b: =>\(\left(x-3\right)^2-4x\left(x-3\right)=0\)

=>(x-3)(-3x-3)=0

=>x=3 hoặc x=-1

c: =>(x-4)(x-6)=0

=>x=4 hoặc x=6

Bài 39:

a: Đặt \(\dfrac{5}{x-2}-\dfrac{1}{x+2}+\dfrac{4}{\left(x-2\right)\left(x+2\right)}=0\)

=>5x+10-x+2+4=0

=>4x+16=0

hay x=-4(nhận)

b: Đặt \(\dfrac{-2}{x^2-x+1}+x+1=0\)

\(\Leftrightarrow-2+x^3+1=0\)

\(\Leftrightarrow x^3=1\)

hay x=1(nhận)

\(=abx^2+aby^2+a^2xy+b^2xy\)

\(=ax\left(bx+ay\right)+by\left(ay+bx\right)\)

=(ay+bx)(ax+by)