a,(12x-5) (4x-1) + (3x-7) (1-16x)=81
b,5x(1/5 x-2)+3(6-1/3 x²)=14
c,3x(4/3 x+1)-4x(x-2)=10
d,(x-3) (x+7)-(x-5) (x-1)=0
e,3x(12x-4)-9(4x-3)=30
f,(x-3) (x²+3x+9)+x(5-x²=6x
g,8x(x-1)-8(x-1) (x+1)=20
a,(12x-5) (4x-1) + (3x-7) (1-16x)=81
b,5x(1/5 x-2)+3(6-1/3 x²)=14
c,3x(4/3 x+1)-4x(x-2)=10
d,(x-3) (x+7)-(x-5) (x-1)=0
e,3x(12x-4)-9(4x-3)=30
f,(x-3) (x²+3x+9)+x(5-x²=6x
g,8x(x-1)-8(x-1) (x+1)=20
`a) (12x-5)(4x-1) + (3x-7)(1-16x) = 81`
`=> 48x^2 – 12x – 20x + 5 + 3x – 48x^2 – 7 + 112x = 81`
`=> 83x -2 =81`
`=> 83x = 83`
`=> x= 1`
Vậy `x=1`
`b) 5x (1/5x- 2) + 3(6 – 1/3x^2) = 14`
`=> x^2 – 10x + 18 – x^2 = 14`
`=> -10x = 14 – 18`
`=> -10x = -4`
`=> x = 2/5`
Vậy `x=2/5`
`c) 3x (4/3x + 1)- 4x (x-2) = 10`
`=> 4x^2 + 3x- 4x^2 + 8x = 10`
`=> 11x = 10`
`=> x = 10/11`
Vậy `x = 10/11`
`d) (x-3)(x+7) – (x-5)(x-1) = 0`
`=> x^2 + 7x – 3x – 21 – x^2 + x + 5x – 5 =0`
`=> 10x – 26 =0`
`=>10x = 26`
`=> x = 13/5`
Vậy `x=13/5`
`e) 3x (12x – 4) – 9(4x-3) = 30`
`=> 36x^2 – 12x -36x + 27 = 30`
`=> 36x^2 – 48x – 3 = 0`
`=> 3 . (12x^2 – 16x – 1) =0`
`=> 12x^2 – 16x – 1 =0`
`=> 12 . (x^2 – 4/3x + 4/9) – 19/3 =0`
`=> 12. [ x^2 – 2 . 2/3 . x + (2/3)^2 ] = 19/3`
`=> 12 . (x – 2/3)^2 = 19/3`
`=> (x-2/3)^2 = 19/36`
`=> x – 2/3 = (\sqrt{19})/6` hoặc `x – 2/3 = (-\sqrt{19})/6`
`+) x – 2/3 = (\sqrt{19})/6`
`=> x = (4 + \sqrt{19})/6`
`+) x – 2/3 = (-\sqrt{19})/6`
`=> x = (4 – \sqrt{19})/6`
Vậy `x\in{ (4 + \sqrt{19})/6 ; (4 – \sqrt{19})/6}`
`f) (x-3)(x^2 + 3x+9) + x (5-x^2) = 6x`
`=> x^3 – 27 + 5x – x^3 = 6x`
`=> 5x – 27 =6x`
`=> 5x – 6x =27`
`=> -x =27`
`=> x = -27`
Vậy `x = -27`
`g) 8x (x-1) – 8(x-1) (x+1) = 20`
`=> 8x^2 – 8x – 8 .(x^2 -1) = 20`
`=> 8x^2 – 8x – 8x^2 + 8 = 20`
`=> -8x + 8 =20`
`=> -8x = 12`
`=> x = -3/2`
Vậy `x=-3/2`
Đáp án:
`a)x=1`
`b)x=2/5`
`c)x=10/11`
`d)x=13/5`
`e)x∈{`$\dfrac{\sqrt[]{19}+4}{6};\dfrac{4-\sqrt[]{19}}{6}$`}`
`f)x=-27`
`g)x=-3/2`
Giải thích các bước giải:
`a)(12x-5)(4x-1)+(3x-7)(1-16x)=81`
`⇔48x²-12x-20x+5+3x-48x²-7+112x=81`
`⇔(48x²-48x²)+(-12x-20x+3x+112x)+(5-7)=81`
`⇔83x-2=81`
`⇔83x=81+2`
`⇔83x=83`
`⇔x=83:83`
`⇔x=1`
Vậy `x=1`
`b)5x(1/5x-2)+3(6-1/3x²)=14`
`⇔x²-10x+18-x²=14`
`⇒(x²-x²)-10x+18=14`
`⇔-10x+18=14`
`⇔-10x=14-18`
`⇔-10x=-4`
`⇔x=4/10`
`⇔x=2/5`
Vậy `x=2/5`
`c)3x(4/3x+1)-4x(x-2)=10`
`⇔4x²+3x-4x²+8x=10`
`⇔(4x²-4x²)+(3x+8x)=10`
`⇔11x=10`
`⇔x=10/11`
Vậy `x=10/11`
`d)(x-3)(x+7)-(x-5)(x-1)=0`
`⇔x²+7x-3x-21-(x²-x-5x+5)=0`
`⇔x²+7x-3x-21-x²+x+5x-5=0`
`⇔(x²-x²)+(7x-3x+x+5x)+(-21-5)=0`
`⇔10x-26=0`
`⇔10x=26`
`⇔x=26/10`
`⇔x=13/5`
Vậy `x=13/5`
`e)3x(12x-4)-9(4x-3)=30`
`⇔36x²-12x-36x+27=30`
`⇔36x²-48x+27=30`
`⇔36x²-48x+27-30=0`
`⇔36x²-48x-3=0`
`⇔3(12x²-16x-1)=0`
`⇔12x²-16x-1=0`
`⇔12(x²-4/3x-1/12)=0`
`⇔12(x²-4/3x+4/9-19/36)=0`
`⇔12[x²-2.x.2/3+(2/3)^2]-19/3=0`
`⇔12(x-2/3)^2=19/3`
`⇔(x-2/3)^2=19/3:12`
`⇔(x-2/3)^2=19/3 . 1/12`
`⇔(x-2/3)^2=19/36`
`⇔(x-2/3)^2=(`$\dfrac{\sqrt[]{19}}{6}$ `)^2`
`⇔`\(\left[ \begin{array}{l}x-\dfrac{2}{3}=\dfrac{\sqrt[]{19}}{6}\\x-\dfrac{2}{3}=-\dfrac{\sqrt[]{19}}{6}\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{\sqrt[]{19}}{6}+\dfrac{2}{3}\\x=-\dfrac{\sqrt[]{19}}{6}+\dfrac{2}{3}\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{\sqrt[]{19}+4}{6}\\x=\dfrac{4-\sqrt[]{19}}{6}\end{array} \right.\)
Vậy `x∈{`$\dfrac{\sqrt[]{19}+4}{6};\dfrac{4-\sqrt[]{19}}{6}$`}`
`f)(x-3)(x²+3x+9)+x(5-x²)=6x`
`⇔x³-27+5x-x³=6x`
`⇔x³+5x-x³-6x=27`
`⇔(x³-x³)+(5x-6x)=27`
`⇔-x=27`
`⇔x=-27`
Vậy `x=-27`
`g)8x(x-1)-8(x-1)(x+1)=20`
`⇔8x²-8x-8(x²-1)=20`
`⇔8x²-8x-8x²+8=20`
`⇔(8x²-8x²)-8x+8=20`
`⇔-8x+8=20`
`⇔-8x=20-8`
`⇔-8x=12`
`⇔x=-12/8`
`⇔x=-3/2`
Vậy `x=-3/2`