Bài 1: Giải các phương trình sau:

a) 3(2,2-0,3x)=2,6 + (0,1x-4)

<=> 6.6 - 0.9x = 2,6 + 0,1x - 4

<=> - 0.9x - 0,1x = -6.6 -1,4

<=> -x = -8

<=> x = 8

Vậy x = 8

b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)

<=> 3,6 - x - 0,5 = x - 5,5 + x

<=> - x - 3,1 = -5,5

<=> - x = -2.4

<=> x = 2.4

Vậy  x = 2.4

g.\(\dfrac{1-3x}{6}+x-1=\dfrac{x+2}{2}\)

\(\Leftrightarrow\dfrac{\left(1-3x\right)+6\left(x-1\right)}{6}=\dfrac{3\left(x+2\right)}{6}\)

\(\Leftrightarrow\left(1-3x\right)+6\left(x-1\right)=3\left(x+2\right)\)

\(\Leftrightarrow1-3x+6x-6=3x+6\)

\(\Leftrightarrow-5=6\left(vô.lí\right)\)

Vậy pt vô nghiệm

h.\(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)

\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)

\(\Leftrightarrow30x+15-100-6x-4=24x-8\)

\(\Leftrightarrow-89=-8\left(vô.lí\right)\)

Vậy pt vô nghiệm

i.\(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)

\(\Leftrightarrow\dfrac{21\left(4x+3\right)-15\left(6x-2\right)}{105}=\dfrac{35\left(5x+4\right)+215}{105}\)

\(\Leftrightarrow21\left(4x+3\right)-15\left(6x-2\right)=35\left(5x+4\right)+215\)

\(\Leftrightarrow84x+63-90x+30=175x+140+215\)

\(\Leftrightarrow-181=262\)

\(\Leftrightarrow x=-\dfrac{262}{181}\)

a: Ta có: \(3x-\left(3x+2\right)=x+3\)

\(\Leftrightarrow x+3=-2\)

hay x=-5

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

\(\Leftrightarrow15x-3+8x-4=18x\)

\(\Leftrightarrow5x=7\)

hay \(x=\dfrac{7}{5}\)

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

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

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

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

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)

a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-2}\le0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow-3\le x< 2\)

d) \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)

\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5\right)^2-\left(3x+2\right)^2}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{-\left(5x-3\right)\left(x+7\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)

\(a.ĐK:x\ne3;1\)

\(\Rightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)+2\left(3x-10\right)}{2\left(x-1\right)\left(x-3\right)}=\dfrac{7\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow x-1+6x-20=7\left(x^2-4x+3\right)\)

\(\Leftrightarrow7x-21=7x^2-28x+21\)

\(\Leftrightarrow7x^2-35x+42=0\)

\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow x^2-5x+6=0\)

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

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

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

b.\(ĐK:x\ne2;4\)

\(\Rightarrow\dfrac{x-1}{x-2}-\dfrac{x+3}{4-x}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(4-x\right)}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)

\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)

\(\Leftrightarrow4x-x^2-4+x-x^2+2x-3x+6-2=0\)

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

\(\Leftrightarrow-2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

a: \(\Leftrightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)

\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow7\left(x^2-4x+3\right)=x-1+6x-20=7x-21\)

\(\Leftrightarrow\left(x-3\right)\left(7x-7\right)-7\left(x-3\right)=0\)

=>(x-3)(7x-14)=0

=>x=3(loại) hoặc x=2(nhận)

b: \(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=-2\)

\(\Leftrightarrow x^2-5x+4+x^2+x-6=-2\)

\(\Leftrightarrow2x^2-4x=0\)

=>2x(x-2)=0

=>x=0(nhận) hoặc x=2(loại)

a: \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)

=>\(9x^2+12x+4-\left(9x^2-12x+4\right)-5x-38=0\)

=>\(9x^2+7x-34-9x^2+12x-4=0\)

=>19x-38=0

=>19x=38

=>x=38/19=2

b: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

=>\(x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)

=>\(x^3+3x^2+12x-9=x^3+3x^2+3x+1\)

=>12x-9=3x+1

=>12x-3x=1+9

=>9x=10

=>x=10/9