Bài 2:

a, |x-1| -x +1=0

|x-1| = 0-1+x

|x-1| = -1 + x

 \(\orbr{\begin{cases}x-1=-1+x\\x-1=1-x\end{cases}}\)

 \(\orbr{\begin{cases}x=-1+x+1\\x=1-x+1\end{cases}}\)

 \(\orbr{\begin{cases}x=x\\x=2-x\end{cases}}\)

x = 2-x

2x = 2

x = 2:2

x=1

b, |2-x| -2 = x

|2-x| = x+2

\(\orbr{\begin{cases}2-x=x+2\\2-x=2-x\end{cases}}\)

2-x = x+2

x+x = 2-2

2x = 0

x = 0

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\((x-6)(3x-9)>0\)
TH1:
\(\orbr{\begin{cases}x-6< 0\\3x-9< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< 6\\x< 3\end{cases}}\)\(\Rightarrow x< 3\)
TH2:
\(\orbr{\begin{cases}x-6>0\\3x-9>0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>6\\x>3\end{cases}}\)\(\Rightarrow x>6\)
Vậy \(x< 3\) hoặc \(x>6\)thì \((x-6)(3x-9)>0\)
Học tốt!

20.
\((2x-1)(6-x)>0\)
TH1:
\(\orbr{\begin{cases}2x-1>0\\6-x>0\end{cases}\Rightarrow\orbr{\begin{cases}x< \frac{1}{2}\\x< 6\end{cases}}\Rightarrow x< 6}\)
TH2
\(\orbr{\begin{cases}2x-1< 0\\6-x< 0\end{cases}\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x>6\end{cases}}\Rightarrow x>\frac{1}{2}}\)
Vậy \(x< 6\)hoặc \(x>\frac{1}{2}\)thì \((2x-1)(6-x)>0\)
 

21.
\((2-x)(x+7)< 0\)
TH1.
\(\orbr{\begin{cases}2-x>0\\x+7< 0\end{cases}\Rightarrow\orbr{\begin{cases}x< 2\\x>-7\end{cases}}\Rightarrow-7< x< 2}\)
TH2.
\(\orbr{\begin{cases}2-x< 0\\x+7>0\end{cases}\Rightarrow\orbr{\begin{cases}x>2\\x< -7\end{cases}}\Rightarrow2< x< -7}\)(vô lí)
Vậy \(-7< x< 2\) thì \((2-x)(x+7)< 0\)
 

tu lam

bạn ơi cho mình hỏi chút GTTĐ là gì

 Nguyễn Phương Uyên  là giá trị tuyệt đối nhé

Bài 1:

<=>7[3(-x)]-12(x-5)=-3(11x-20)

=>-3(11x-20)=5

=>-33x=-55

=>-11.3x=-11.5 (rút gọn -11)

=>3x=5

\(\Rightarrow x=\frac{5}{3}\)

Đã duyệt

bài 1:

<=>7[3(-x)]-12(x-5)=-3(11x-20)

=>-3(11x-20)=5

=>-33x=-55

=>-11.3x=-11.5 (rút gọn -11)

=>3x=5

=>x=\(\frac{5}{3}\)

bài 1:

<=>7[3(-x)]-12(x-5)=-3(11x-20)

=>-3(11x-20)=5

=>-33x=-55

=>-11.3x=-11.5 (rút gọn -11)

=>3x=5

=>x=\(\frac{5}{3}\)

đậu phộng thằng Phan Bá Lộc chép bài ông

a,

\(\left|x+\dfrac{9}{2}\right|\ge0\forall x\\ \left|y+\dfrac{4}{3}\right|\ge0\forall y\\ \left|z+\dfrac{7}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy \(x=\dfrac{-9}{2};y=\dfrac{-4}{3};z=\dfrac{-7}{2}\)

d,

\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{1}{5}\right|\ge0\forall y\\ \left|x+y+z\right|\ge0\forall x,y,z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-3}{4}+\dfrac{1}{5}+z=0\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-11}{20}+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)

Bạn mới hỏi ở dưới rồi :v

b,

\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{2}{5}\right|\ge0\forall y\\ \left|z+\dfrac{1}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\forall x,y,z\\ \)

Mà \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{2}{5}\right|=0\\\left|z+\dfrac{1}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{2}{5}=0\\z+\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{2}{5}\\z=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy ...

c,

\(\left|x+\dfrac{19}{5}\right|\ge0\forall x\\ \left|y+\dfrac{1890}{1975}\right|\ge0\forall y\\ \left|z-2004\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{1890}{1975}\right|=0\\\left|z-2004\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-1890}{1975}=\dfrac{-378}{395}\\z=2004\end{matrix}\right. \)

Vậy ...