=>5(4x-1)-2+x<=3(10x-3)

=>20x-5+x-2<=30x-9

=>21x-7<=30x-9

=>-9x<=-2

=>x>=2/9

a/ A =  \(2x-5\ge0\)

\(\Leftrightarrow\)\(2x\ge5\)

\(\Leftrightarrow\)\(x\ge\frac{5}{2}\)

b/ \(\frac{4x-1}{3}\)-  \(\frac{2-x}{15}\)\(\le\)\(\frac{10x-3}{5}\)

\(\Leftrightarrow\)5(4x + 1) - 2 - x \(\le\)3(10x - 3)

\(\Leftrightarrow\)20x + 5 - 2 -x \(\le\)30x - 9

\(\Leftrightarrow\)9x - 30x \(\le\)-9 + 2

\(\Leftrightarrow\)-21x \(\le\)-7

\(\Leftrightarrow\)x \(\ge\)\(\frac{1}{3}\)

Kết luận và biểu diễn tập nghiệm nha

Ta có: \(\dfrac{x-1}{3}-\dfrac{3x+5}{2}\ge1-\dfrac{4x+5}{6}\)

\(\Leftrightarrow2\left(x-1\right)-3\left(3x+5\right)\ge6-4x-5\)

\(\Leftrightarrow2x-2-9x-15-6+4x+5\ge0\)

\(\Leftrightarrow-3x\ge18\)

hay \(x\le-6\)

1:

a: 2x-3=5

=>2x=8

=>x=4

b: (x+2)(3x-15)=0

=>(x-5)(x+2)=0

=>x=5 hoặc x=-2

2:

b: 3x-4<5x-6

=>-2x<-2

=>x>1

2:

a: =>x-4>=0

=>x>=4

b: =>x+1>0

=>x>-1

b, ĐK: \(x\ne8\)

\(A=\dfrac{x-5}{x-8}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5>0\\x-8>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5< 0\\x-8< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>5\\x>8\end{matrix}\right.\\\left\{{}\begin{matrix}x< 5\\x< 8\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>8\\x< 5\end{matrix}\right.\)

a/ -4 + 2x < 0 

2x < 4

x < 2 

b) Để A dương 

\(\left[{}\begin{matrix}x< 5\\x>8\end{matrix}\right.\)

a. 

A, 3X+6>0

(=)3X>-6

(=)X>-2

VẬY ...

B,10-2X≥-4

(=)-2X≥-4-10

(=)-2X≥-14

(=)X≤7

VẬY....

C,

(=)

(=) -15X+10>-3+3X

(=)-15X-3X>-3-10

(=)-18X>-13

(=)X<

1.

|x-9|=2x+5

x<9; x-9=-2x-5

3x=4=>x=4/3(n)

x≥9; x-9=2x+5=> x=-14(l)

2.a

A=2x-5≥0<=>2x≥5; x≥5/2

1. a) / x - 9 / = 2x + 5

Do : / x - 9 / ≥ 0 ∀x

⇒2x + 5 ≥ 0

⇔ x ≥ \(\dfrac{-5}{2}\)

Bình phương cả hai vế của phương trình , ta được :

( x - 9)² = ( 2x + 5)²

⇔ ( x - 9)² - ( 2x + 5)² = 0

⇔ ( x - 9 - 2x - 5)( x - 9 + 2x + 5) = 0

⇔ ( - x - 14)( 3x - 4) = 0

⇔ x = - 14 ( KTM) hoặc : x = \(\dfrac{4}{3}\) ( TM)

KL....

b) Mạn phép làm luôn , ko chép lại đề :

\(\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-5}{\left(x-3\right)\left(x+3\right)}\) ( x # 3 ; x # - 3)

⇔ 5x + 15 + 4x - 12 = x - 5

⇔ 9x + 3 = x - 5

⇔ 8x = - 8

⇔ x = -1 ( TM)

KL....

2a) Bạn kia làm rùi

b) Mạn phép ko chép lại đề , làm luôn :

\(\dfrac{4x-1}{3}-\dfrac{2-x}{15}\) ≤ \(\dfrac{10x-3}{5}\)

⇔ \(\dfrac{5\left(4x-1\right)+x-2}{15}\) ≤ \(\dfrac{3\left(10x-3\right)}{15}\)

⇔ 20x - 5 + x - 2 ≤ 30x - 9

⇔ 21x - 7 ≤ 30x - 9

⇔ 9x ≥ 2

⇔ x ≥ \(\dfrac{2}{9}\)

KL.....