Ôn tập: Bất phương trình bậc nhất một ẩn

\(\left(x+1\right)^2+\left|x-7\right|+6=\left(x+2\right)^2\)

\(< =>x^2+2x+1+\left|x-7\right|+6=x^2+4x+4\)

\(< =>\left|x-7\right|=x^2-x^2+4x-2x+4-1-6\)

\(< =>\left|x-7\right|=2x-3\)

\(< =>\left[{}\begin{matrix}x-7=2x-3\\x-7=-2x+3\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x-2x=-3+7\\x+2x=3+7\end{matrix}\right.\\ < =>\left[{}\begin{matrix}-x=4\\3x=10\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-4\\x=\dfrac{10}{3}\end{matrix}\right.\)

a: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)

=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)

=>2x-2018<0

=>x<2019

b: \(\Leftrightarrow\left(\dfrac{3-x}{100}+\dfrac{4-x}{101}\right)>\dfrac{5-x}{102}+\dfrac{6-x}{103}\)

=>\(\dfrac{x-3}{100}+\dfrac{x-4}{101}-\dfrac{x-5}{102}-\dfrac{x-6}{103}< 0\)

=>\(x+97< 0\)

=>x<-97

\(5-\left|2x-1\right|=x\)

\(\Leftrightarrow\left|2x-1\right|=5-x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5-x\\2x-1=-\left(5-x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5-x\\2x-1=-5+x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+x=5+1\\2x-x=-5+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\x=-4\end{matrix}\right.\)

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

=>|2x-1|=5-x

TH1: x>=1/2

=>2x-1=5-x

=>3x=6

=>x=2(nhận)

TH2: x<1/2

=>1-2x=5-x

=>-x=4

=>x=-4(nhận)

\(a,\) Thay \(x=-3\) vào \(Q=\dfrac{x+1}{x}\) \(\left(x\ne0\right)\)

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

\(b,P=\dfrac{x^2-2}{x^2+2x}+\dfrac{1}{x+2}\left(x\ne0,x\ne-2,x\ne-1\right)\)

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

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

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

\(c,P:Q=\dfrac{5}{2}\Leftrightarrow\dfrac{x^2+x-1}{x^2+2x}:\dfrac{x+1}{x}=\dfrac{5}{2}\)

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

\(\Leftrightarrow\dfrac{x^2+x-1}{\left(x+2\right)\left(x+1\right)}-\dfrac{5}{2}=0\)

\(\Leftrightarrow\dfrac{x^2+x-1}{x^2+3x+2}-\dfrac{5}{2}=0\)

\(\Leftrightarrow2\left(x^2+x-1\right)-5\left(x^2+3x+2\right)=0\)

\(\Leftrightarrow2x^2+2x-2-5x^2-15x-10=0\)

\(\Leftrightarrow-3x^2-13x-12=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x=-3\end{matrix}\right.\)

a: Xét ΔHAD vuông tại H và ΔABD vuông tại A có

góc ADB chung

=>ΔHAD đồng dạng với ΔABD

b: ΔHAD đồng dạng vơi ΔABD

=>DH/DA=DA/DB

=>DA^2=DH*DB

khoảng cách của oto và xe máy lúc 11h là 
    310-(40x1)=270km
 2 xe gặp nhau sau số giờ là 
    270:(40+50)=3h
 điểm gặp nhau cách B là 
    50x3=150km

Chọn B

=> \(\left[{}\begin{matrix}x+9=0\\x-3=0\\2x-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-9\\x=3\\x=\dfrac{4}{2}=2\end{matrix}\right.\)

=>(x+9)(x-3)(x-2)=0

hay \(x\in\left\{3;2;-9\right\}\)

`12(x+9)(x-3)(2x-4)=0`

`=>` \(\left[{}\begin{matrix}x+9=0\\x-3=0\\2x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-9\\x=0+3\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-9\\x=3\\x=2\end{matrix}\right.\)

Vậy `S = {-9;3;2}`

`(x-9)(x+1)=0`

`<=>` $\left[\begin{matrix} x-9=0\\ x+1=0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=9\\ x=-1\end{matrix}\right.$

Vậy `S={9;-1}`

`(x-9)(x+1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)

vậy : `S={9;-1}`

=>x-9=0 hoặc x+1=0

=>x=9 hoặc x=-1

` (-x+2)(-3x-15)=0`

\(\Leftrightarrow\left[{}\begin{matrix}-x+2=0\\-3x-15=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=-2\\-3x=15\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Vậy `S={2;-5}`

`(-x+2)(-3x-15)=0`

`<=>` $\left[\begin{matrix} -x+2=0\\ -3x-15=0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=2\\ x=-5\end{matrix}\right.$

Vậy `S={2;-5}`

\(\Leftrightarrow\left[{}\begin{matrix}-x+2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)