Bài 8:
a: \(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
=>\(\dfrac{4}{5}x+\dfrac{3}{5}-\dfrac{6}{7}x+\dfrac{2}{7}=\dfrac{5}{3}x+\dfrac{4}{3}+3\)
=>\(x\left(\dfrac{4}{5}-\dfrac{6}{7}\right)+\dfrac{31}{35}=\dfrac{5}{3}x+\dfrac{13}{3}\)
=>\(x\cdot\dfrac{-2}{35}-\dfrac{5}{3}x=\dfrac{13}{3}-\dfrac{31}{35}\)
=>\(x\cdot\left(-\dfrac{2}{35}-\dfrac{5}{3}\right)=\dfrac{362}{105}\)
=>\(x\cdot\dfrac{-181}{105}=\dfrac{362}{105}\)
=>x=-2
b: \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
=>\(\dfrac{5}{6}x+\dfrac{1}{3}-\dfrac{8}{3}x+\dfrac{1}{3}=\dfrac{4}{5}x+\dfrac{2}{5}-5\)
=>\(x\left(\dfrac{5}{6}-\dfrac{8}{3}\right)+\dfrac{2}{3}=\dfrac{4}{5}x-\dfrac{13}{5}\)
=>\(x\cdot\dfrac{-11}{6}+\dfrac{2}{3}=\dfrac{4}{5}x-\dfrac{13}{5}\)
=>\(x\cdot\left(-\dfrac{11}{6}-\dfrac{4}{5}\right)=-\dfrac{13}{5}-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{-79}{30}=\dfrac{-49}{15}\)
=>\(x=\dfrac{49}{15}:\dfrac{79}{30}=\dfrac{49}{15}\cdot\dfrac{30}{79}=\dfrac{98}{79}\)
f: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
=>\(\dfrac{3\left(3x+2\right)-3x-1}{6}=\dfrac{12x+10}{6}\)
=>3(3x+2)-3x-1=12x+10
=>12x+10=9x+6-3x-1
=>12x+10=6x+5
=>6x=-5
=>\(x=-\dfrac{5}{6}\)
g: \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
=>\(\dfrac{15\left(2x-1\right)}{20}-\dfrac{6x+2}{20}+\dfrac{20}{20}=\dfrac{8\left(3x+2\right)}{20}\)
=>15(2x-1)-6x-2+20=8(3x+2)
=>30x-15-6x+18=24x+16
=>3=16(vô lý)
bài 7:
a: \(\left(x+5\right)\left(x-1\right)=2x\left(x-1\right)\)
=>\(2x\left(x-1\right)-\left(x+5\right)\left(x-1\right)=0\)
=>(x-1)(2x-x-5)=0
=>(x-1)(x-5)=0
=>\(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
b: \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)
=>\(\left(x-1\right)\left(6x-3\right)=\left(x-1\right)\left(5x+40\right)\)
=>\(\left(x-1\right)\left(6x-3-5x-40\right)=0\)
=>(x-1)(x-43)=0
=>\(\left[{}\begin{matrix}x=1\\x=43\end{matrix}\right.\)
c: \(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)
=>\(\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)
=>\(\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)
=>(x-3)(x-4)(x-1)(x-2)=0
=>\(\left[{}\begin{matrix}x=3\\x=4\\x=1\\x=2\end{matrix}\right.\)
d: \(x^3-5x^2+6x=0\)
=>\(x\left(x^2-5x+6\right)=0\)
=>x(x-2)(x-3)=0
=>\(\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
e: \(\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\)
=>\(\left(x+3\right)\left(x-5+3x-4\right)=0\)
=>\(\left(x+3\right)\left(4x-9\right)=0\)
=>\(\left[{}\begin{matrix}x=-3\\x=\dfrac{9}{4}\end{matrix}\right.\)
f: \(\left(x+7\right)\left(3x-1\right)=49-x^2\)
=>\(\left(x+7\right)\left(3x-1\right)+\left(x-7\right)\left(x+7\right)=0\)
=>\(\left(x+7\right)\left(3x-1+x-7\right)=0\)
=>(x+7)(4x-8)=0
=>4(x-2)(x+7)=0
=>(x-2)(x+7)=0
=>\(\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)
g: \(3x^2-7x+4=0\)
=>\(3x^2-3x-4x+4=0\)
=>3x(x-1)-4(x-1)=0
=>(x-1)(3x-4)=0
=>\(\left[{}\begin{matrix}x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)
h: \(2x^3+3x^2-32x=48\)
=>\(2x^3+3x^2-32x-48=0\)
=>\(x^2\left(2x+3\right)-16\left(2x+3\right)=0\)
=>(2x+3)(x^2-16)=0
=>(2x+3)(x-4)(x+4)=0
=>\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=4\\x=-4\end{matrix}\right.\)
Bài 6
a) 5 - (x - 6) = 4(2x - 3)
5 - x + 6 = 8x - 12
8x + x = 11 + 12
9x = 23
x = 23/9
Vậy S = {23/9}
b) 3 - 4x + 24 + 6x = x + 27 + 3x
x + 3x + 4x - 6x = 3 + 24 - 27
2x = 0
x = 0
Vậy S = {0}
c) (x + 5)(x - 1) = 2x(x - 1)
(x + 5)(x - 1) - 2x(x - 1) = 0
(x - 1)(x + 5 - 2x) = 0
(x - 1)(-x + 5) = 0
x - 1 = 0 hoặc -x + 5 = 0
*) x - 1 = 0
x = 1
*) -x + 5 = 0
x = 5
Vậy S = {1; 5}
d) 5(3x - 2) - 4(5 - 3x) = 1
15x - 10 - 20 + 12x = 1
27x - 30 = 1
27x = 1 - 30
27x = -29
x = -29/27
Vậy S = {-29/27}
