Question

tìm x biết :

a, 3-2 | 4x-5 | = 2 phần 6

b, | 4 - 7x | - 3 phần 2 : 5 = giá trị tuyệt đối -1 và 1 phần 3

c, ( 7-3x ) ( 2x + 1 ) = 0

d, 2x ( 5-3x ) > 0

e, ( 4-2x) ( 5x+ 3 ) < 0

g, | 3x + 1 | + | 1- 3x | = 0

mọi người giúp mình với nhé nhanh lên mình đang cần gấp

Answer 1

a) * Nếu 4x - 5 \(\ge\) 0 thì x \(\ge\) \(\dfrac{5}{4}\)

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

\(\Leftrightarrow\) \(-8x=-3-10+\dfrac{2}{6}\)

\(\Leftrightarrow\) x = \(\dfrac{19}{12}\) (t/m)

* Nếu 4x - 5 < 0 thì x < \(\dfrac{5}{4}\)

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

\(\Leftrightarrow\) \(3+8x-10=\dfrac{2}{6}\)

\(\Leftrightarrow\) x = \(\dfrac{11}{12}\) (t/m)

b) Không hiểu đề :v

c) \(\left(7-3x\right)\left(2x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d) \(2x\left(5-3x\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{5}{3}\end{matrix}\right.\)

\(\Rightarrow0< x< \dfrac{5}{3}\)

e) \(\left(4-2x\right)\left(5x+3\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-2x< 0\\5x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4-2x>0\\5x+3< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x< -\dfrac{3}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x>-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)

Loại TH1, nhận TH2

Vậy \(-\dfrac{3}{5}< x< 2\)

g) \(\left|3x+1\right|+\left|1-3x\right|=0\) (1)

* Nếu x < \(\dfrac{-1}{3}\)

PT (1) \(\Leftrightarrow-3x-1-1+3x=0\)

0x - 2 = 0

0x = 2 \(\Rightarrow\) PT vô nghiệm

* Nếu \(\dfrac{-1}{3}\le x\le\dfrac{1}{3}\)

PT (1) \(\Leftrightarrow3x+1-1+3x=0\)

6x = 0

x = 0 (t/m)

* Nếu x > \(\dfrac{1}{3}\)

PT (1) \(\Leftrightarrow3x+1+1-3x=0\)

0x + 2 = 0

0x = -2

PT vô nghiệm.

Vậy x = 0

Answer 2

a, \(3-2\left|4x-5\right|=\dfrac{2}{6}\)

\(\Rightarrow2\left|4x-5\right|=\dfrac{8}{3}\)

\(\Rightarrow\left|4x-5\right|=\dfrac{4}{3}\)

+) Xét \(x\ge\dfrac{5}{4}\) có:

\(4x-5=\dfrac{4}{3}\Rightarrow4x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{12}\) ( t/m )

+) Xét \(x< \dfrac{5}{4}\) có:

\(4x-5=\dfrac{-4}{3}\Rightarrow4x=\dfrac{11}{3}\Rightarrow x=\dfrac{11}{12}\) ( t/m )

Vậy...

b, tương tự

c, \(\left(7-3x\right)\left(2x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy...

d, \(2x\left(5-3x\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}2x< 0\\5-3x< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{3}{5}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x>\dfrac{3}{5}\end{matrix}\right.\) (loại )

Vậy \(0< x< \dfrac{3}{5}\)

e, tương tự

g, \(\left|3x+1\right|+\left|1-3x\right|=0\)

\(\Rightarrow\left|3x+1\right|+\left|3x-1\right|=0\)

+) Xét \(x\ge\dfrac{1}{3}\) có:

\(3x+1+3x-1=0\)

\(\Rightarrow6x=0\)

\(\Rightarrow x=0\) ( ko t/m )
+) Xét \(\dfrac{-1}{3}\le x< \dfrac{1}{3}\) có:

\(3x+1+1-3x=0\)

\(\Rightarrow2=0\) ( vô lí )

+) Xét \(x< \dfrac{-1}{3}\) có:

\(-3x-1+1-3x=0\)

\(\Rightarrow-6x=0\Rightarrow x=0\) ( ko t/m )

Vậy ko có giá trị x thỏa mãn đề bài