Ôn tập: Bất phương trình bậc nhất một ẩn

a: x>2

y>2

=>x+y>2+2=4

x>y>2

=>xy>2^2=4

b: x^2-xy=x(x-y)

x-y>0; x>0

=>x(x-y)>0

=>x^2-xy>0

y>2

=>y-2>0

=>y(y-2)>0

=>y^2-2y>0

x>y và y>2

=>y>0 và x-y>0

=>y(x-y)>0

=>xy-y^2>0

ĐKXĐ:\(9-2x\ge0\Leftrightarrow x\le\dfrac{9}{2}\)

\(\left|x-3\right|=9-2x\\ \Leftrightarrow\left[{}\begin{matrix}x-3=9-2x\\x-3=2x-9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=12\\x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=6\left(ktm\right)\end{matrix}\right.\)

\(1.\\ a.\left(x-5\right)\left(x-11\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x-11=0.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=11.\end{matrix}\right.\)

\(b.8x-3=5x+12.\\ \Leftrightarrow3x=15.\\\Leftrightarrow x=5.\\ c.7+4x=3\left(x+1\right)-2x-2.\\ \Leftrightarrow7+4x=3x+3-2x-2.\\ \Leftrightarrow3x=-6.\\ \Leftrightarrow x=-2.\)

\(d.\dfrac{1-x}{x+1}+2=\dfrac{2x+3}{x+1}.\\ \left(x\ne-1\right).\\ \Leftrightarrow\dfrac{1-x+2x+2-2x-3}{x+1}=0.\\ \Rightarrow-x=0.\\ \Leftrightarrow x=0\left(TM\right).\)

\(2.\\ a.4x^2\left(x+3\right)-x-3=0.\\ \Leftrightarrow4x^2\left(x+3\right)-\left(x+3\right)=0.\\ \Leftrightarrow\left(4x^2-1\right)\left(x+3\right)=0.\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x+3\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}.\\x=\dfrac{-1}{2}.\\x=-3.\end{matrix}\right.\)

\(b.x^4-10x^2+9=0.\)

Đặt \(t=x^2\left(t\ge0\right).\)

\(\Rightarrow t^2-10t+9=0.\\ \Leftrightarrow\left(t-9\right)\left(t-1\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}t=9.\\t=1.\end{matrix}\right.\)

\(c.\dfrac{x+2}{2}-\dfrac{2x-3}{5}=\dfrac{10x+13}{10}.\\ \Rightarrow5x+10-4x+6-10x-13=0.\\ \Leftrightarrow-9x=-3.\\ \Leftrightarrow x=\dfrac{1}{3}.\)

\(d.\dfrac{2}{x^2-x+1}=\dfrac{1}{x+1}+\dfrac{2x-1}{x^3+1}.\\ \left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2}{x^2-x+1}-\dfrac{1}{x+1}-\dfrac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}=0.\\ \Rightarrow2x+2-x^2+x-1-2x+1=0.\\ \Leftrightarrow-x^2+x+2=0.\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x=2.\\x=-1.\end{matrix}\right.\)

a: =>|2x+2|=2x-10

=>2x+2=2x-10(x>=-1) hoặc 2x+2=10-2x(x<-1)

=>4x=8

hay x=2(loại)

b: =>x-6=3-2x hoặc x-6=2x-3

=>3x=9 hoặc -x=3

=>x=3 hoặc x=-3

c: \(\Leftrightarrow\left\{{}\begin{matrix}x^2-9=0\\2x-6=0\end{matrix}\right.\Leftrightarrow x=3\)

Áp dụng bđt cauchy, ta có 

\(\dfrac{2a}{b+c}+\dfrac{b+c}{2a}\ge2.\sqrt{\dfrac{2a}{b+c}\dfrac{b+c}{2a}}=2\)(đpcm)

Dấu ''='' xảy ra <=>2a = b + c

Theo bđt cauchy schwarz dạng engel 

\(a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{1+1+1}=\dfrac{1}{3}\)

Dấu ''='' xảy ra khi a = b = c 

a: \(\Leftrightarrow\dfrac{x-51}{9}-1+\dfrac{x-52}{8}-1=\dfrac{x-53}{7}-1+\dfrac{x-54}{6}-1\)

=>x-60=0

hay x=60

b: \(\Leftrightarrow\left(x-2\right)^2-3\left(x+2\right)=x-14\)

\(\Leftrightarrow x^2-4x+4-3x-6-x+14=0\)

\(\Leftrightarrow x^2-8x+12=0\)

=>(x-2)(x-6)=0

=>x=2(loại) hoặc x=6(nhận)

\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)

=>x+2011=0

hay x=-2011

6x2−2xy=3y−11x+2

→2xy+3y=6x2+11x−2

→y(2x+3)=(6x2+9x)+(2x+3)−5

→y(2x+3)=3x(2x+3)+(2x+3)−5

→y=3x+1−52x+3

→5⋮2x+3

→2x+3∈{1,5,−1,−5}

→2x∈{−2,2,−4,−8}

→x∈{−1,1,−2,−4}

→y∈{−7,3,0,−10}

Chúc em học tốt

Ta có :

\(6x^2-2x=3y-11x+2\)

\(\rightarrow2xy+3y=6x^2+11x-2\)

\(\rightarrow y\left(2x+3\right)=\left(6x^2+9x\right)+\left(2x+3\right)-5\)

\(\rightarrow y\left(2x+3\right)=3x\left(2x+3\right)+\left(2x+3\right)-5\)

\(\rightarrow y=3x+1-\dfrac{5}{2x+3}\)

\(\rightarrow5⋮2x+3\)

\(\rightarrow2x+3\in\left\{1,5,-1,-5\right\}\)

\(\rightarrow x\in\left\{-1;1;-2;-4\right\}\)

\(\rightarrow y\in\left\{-7,3,0,-10\right\}\)