Ôn tập: Phương trình bâc nhất một ẩn

\(\left(x+y\right)^3+\left(x-y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\)

\(=2x^3+6xy^2=2x\left(x^2+3y^2\right)\)

\(y^2\left(x^2+y\right)-2x^2-2y\\= y^2\left(x^2+y\right)-\left(2x^2+2y\right)\\ =y^2\left(x^2+y\right)-2\left(x^2+y\right)\\ =\left(x^2+y\right)\left(y^2-2\right)\)

=\(y^2\)\(\left(x^2+y\right)-z\left(x^2+y\right)\)

\(=\left(x^2+y\right)\left(y^2-z\right)\)

ĐKXĐ:\(x\ne-3\)

\(-\dfrac{4}{3+x}+5=\dfrac{4x+7}{x+3}\\ \Leftrightarrow\dfrac{-4}{x+3}+\dfrac{5\left(x+3\right)}{x+3}-\dfrac{4x+7}{x+3}=0\\ \Leftrightarrow\dfrac{-4+5x+15-4x-7}{x+3}=0\\ \Rightarrow x+4=0\\ \Leftrightarrow x=-4\left(tm\right)\)

ĐKXĐ:\(x\ne1\)

\(\dfrac{1}{x-1}+2=\dfrac{3-2x}{x-1}\\ \Leftrightarrow\dfrac{1}{x-1}+\dfrac{2\left(x-1\right)}{x-1}-\dfrac{3-2x}{x-1}=0\\ \Leftrightarrow\dfrac{1+2x-2-3+2x}{x-1}=0\\ \Rightarrow4x-4=0\\ \Leftrightarrow x=1\left(ktm\right)\)

\(\dfrac{1}{x-1}+2=\dfrac{3-2x}{x-1}\left(đk:x\ne1\right)\)

\(\dfrac{1+2x-2}{x-1}=\dfrac{3-2x}{x-1}\)

\(2x-1=3x-2\)

\(x=1\)\(\left(L\right)\)

Gọi V xe máy là x(x>0) km/h

V ô tô là x+10 km/h

tg Ô tô đi từ A-B là 300/(x+10) h

tg xe máy đi từ a-b là 300/x 

vì  ô tô đến B sớm hơn se máy 1h nên ta có pt:

300/x-300/(x+10)=1 <=> x=50(tm) hoặc x=-60ktm 

tự kết luận 

1: \(2x^2+x-3=0\)

\(\Leftrightarrow2x^2+3x-2x-3=0\)

=>(2x+3)(x-1)=0

=>x=-3/2 hoặc x=1

Thay x=-3/2 vào B, ta được:

\(B=\left(\dfrac{9}{4}+\dfrac{3}{2}\right):\left(2\cdot\dfrac{-3}{2}+1\right)=\dfrac{15}{4}:\left(-2\right)=-\dfrac{15}{8}\)

Thay x=1 vào B, ta được:

\(B=\dfrac{1^2-1}{2\cdot1+1}=0\)

2: \(A=\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\)

3: \(M=A\cdot B=\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x\left(x-1\right)}{2x+1}=\dfrac{x}{x+1}\)

Để M là số nguyên thì \(x+1-1⋮x+1\)

\(\Leftrightarrow x+1\in\left\{1;-1\right\}\)

hay \(x\in\left\{0;-2\right\}\)

a, Với x khác 0 ; -1 

\(M=\dfrac{x^2+2x+3x+3-3}{x\left(x+1\right)}=\dfrac{x^2+5x}{x\left(x+1\right)}=\dfrac{x+5}{x+1}\)

b, Ta có \(x^2-1=0\Leftrightarrow x=1;x=-1\left(ktmđk\right)\)

Thay x = 1 vào M ta được \(M=\dfrac{1+5}{1+1}=\dfrac{6}{2}=3\)

c, Ta có \(\dfrac{M}{N}=\dfrac{x+5}{x+1}:\dfrac{x}{x+1}=\dfrac{x+5}{x}=1+\dfrac{5}{x}\Rightarrow x\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Kết hợp đk vậy x = 1 ; 5 ; -5

a.

\(M=\dfrac{x+2}{x+1}+\dfrac{3}{x}-\dfrac{3}{x\left(x+1\right)}\)

\(M=\dfrac{x\left(x+2\right)+3\left(x+1\right)-3}{x\left(x+1\right)}\)

\(M=\dfrac{x^2+2x+3x+3-3}{x\left(x+1\right)}\)

\(M=\dfrac{x^2+5x}{x\left(x+1\right)}\)

\(M=\dfrac{x\left(x+5\right)}{x\left(x+1\right)}\)

\(M=\dfrac{x+5}{x+1}\)

b.

\(N=\dfrac{x}{x+1}\)

\(N=\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(N=\dfrac{x^2-x}{x^2-1}\)

\(N=\dfrac{x^2-x}{0}\)

Vậy N ko có giá trị khi \(x^2-1=0\)

c.\(P=\dfrac{x+5}{x+1}:\dfrac{x}{x+1}\)

\(P=\dfrac{\left(x+5\right)\left(x+1\right)}{x\left(x+1\right)}\)

\(P=\dfrac{x+5}{x}\)

\(P=\dfrac{x}{x}+\dfrac{5}{x}\)

\(P=1+\dfrac{5}{x}\)

Để P nhận giá trị nguyên dương thì \(5⋮x\) hay \(x\in U\left(5\right)=\left\{1;5\right\}\) ( vì để P dương )

\(\Rightarrow x=\left\{1;5\right\}\)

Vậy \(x=\left\{1,5\right\}\) thì P nhận giá trị nguyên dương

a, \(5x=11\Leftrightarrow x=\dfrac{11}{5}\)

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

c, \(2\left(x-3\right)+10x=5x\Leftrightarrow12x-6=5x\Leftrightarrow7x=6\Leftrightarrow x=\dfrac{6}{7}\)

d, đk : x khác -2 ; 2 

\(x\left(x-2\right)-4\left(x+2\right)=8\Leftrightarrow x^2-2x-4x-8=8\)

\(\Leftrightarrow x^2-6x-16=0\Leftrightarrow x^2-6x+9-25=0\Leftrightarrow\left(x-3\right)^2-25=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\Leftrightarrow x=8\left(tm\right);x=-2\left(ktm\right)\)

\(a,2x+1+3x=12\\ \Leftrightarrow5x=11\\ \Leftrightarrow x=\dfrac{11}{5}\\ b,x^2-9+x-3=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)+\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3+1\right)=0\\ \Leftrightarrow\left(x+4\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=3\end{matrix}\right.\)

\(c,\dfrac{x-3}{5}+x=\dfrac{x}{2}\\ \Leftrightarrow\dfrac{2\left(x-3\right)}{10}+\dfrac{10x}{10}-\dfrac{5x}{10}=0\\ \Leftrightarrow2x-6+10x-5x=0\\ \Leftrightarrow7x=6\\ \Leftrightarrow x=\dfrac{6}{7}\)

d, ĐKXĐ:\(x\ne\pm2\)

\(\dfrac{x}{x+2}+\dfrac{4}{2-x}=\dfrac{8}{x^2-4}\\ \Leftrightarrow\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{8}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2-2x-4x-8-8}{\left(x+2\right)\left(x-2\right)}=0\\ \Rightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x^2-8x\right)+\left(2x-16\right)=0\\ \Leftrightarrow x\left(x-8\right)+2\left(x-8\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=8\left(tm\right)\end{matrix}\right.\)

\(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30x}{30}+\dfrac{120}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\\ \Leftrightarrow6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\\ \Leftrightarrow6x+24-30x+120=10x-15x+30\\ \Leftrightarrow-24x+144=-5x+30\\ \Leftrightarrow-5x+30+24x-144=0\\ \Leftrightarrow19x-114=0\\ \Leftrightarrow x=6\)

\(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)

=>6x+24-30x+120=10x-15x+30

=>-24x+144=-5x+30

=>-19x=-114

hay x=6

d) \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)

\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30x}{30}+\dfrac{30.4}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)

\(\Leftrightarrow6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)

\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)

\(\Leftrightarrow6x-30x-10x+15x=30-24-120\)

\(\Leftrightarrow-19x=-114\)

\(\Leftrightarrow x=\dfrac{-114}{-19}\)

\(\Leftrightarrow x=6\)

Vậy : \(S=\left\{6\right\}\)