a) (x + 1 )$^{2}$ – 5 = $x^{2}$ + 11
b)3x ( 2x – 3 ) – 3 ( 3 + 2$x^{2}$ ) = 0
c)$\frac{x}{2}$- $\frac{2x}{3}$+ $\frac{1}{4}$= $\frac{2}{3}$
d)$\frac{x-2}{3}$ -$\frac{2x-3}{4}$ =x -1
e)$\frac{5-2x}{3}$+ $\frac{1}{2}$ = 2(x-1)
giúp mình với ạ mình cảm ơn trước><
`a)` `(x+1)^2-5=x^2+11`
`⇔ x^2+2x+1-5-x^2-11=0`
`⇔ 2x-15`=0`
`⇔ x=\frac{15}{2}`
Vậy `S={\frac{15}{2}}`
`b)` `3x(2x-3)-3(3+2x^2)=0`
`⇔ 6x^2-9x-9-6x^2=0`
`⇔ -9x-9=0`
`⇔ x=-1`
Vậy `S={-1}`
`c)` `\frac{x}{2}-\frac{2x}{3}+\frac{1}{4}=\frac{2}{3}`
`⇔ “\frac{x}{2}-\frac{2x}{3}+\frac{1}{4}-\frac{2}{3}=0`
`⇔ 6x-4.2x+3-2.4=0`
`⇔ -2x-5=0`
`⇔ x=-\frac{5}{2}`
Vậy `S={-\frac{5}{2}}`
`d)` `\frac{x-2}{3}-\frac{2x-3}{4}=x-1`
`⇔ \frac{x-2}{3}-\frac{2x-3}{4}-(x-1)=0`
`⇔ 4(x-2)-3(2x-3)-12(x-1)=0`
`⇔ 4x-8-6x+9-12x+12=0`
`⇔ -14x+13=0`
`⇔ x=\frac{13}{14}`
Vậy `S={\frac{13}{14}}`
`e)` `\frac{5-2x}{3}+\frac{1}{2}=2(x-1)`
`⇔ 2(5-2x)+3-6.2(x-1)=0`
`⇔ 10-4x+3-12x+12=0`
`⇔ -16x+25=0`
`⇔ x=\frac{25}{16}`
Vậy `S={\frac{25}{26}}`
Đáp án:
a) $x=\dfrac{15}{2}$
b) $x=-1$
c) $x=\dfrac{-5}{2}$
d) $x=\dfrac{-13}{10}$
e) $x=\dfrac{25}{16}$
Giải thích các bước giải:
a)
$(x+1)^2-5=x^2+11\\⇔x^2+2x+1-5=x^2+1\\⇔2x-4=11\\⇔2x=15\\⇔x=\dfrac{15}{2}$
b)
$3x(2x-3)-3(3+2x^2)\\⇔6x^2-9x-(9+6x^2)=0\\⇔6x^2-9x-9-6x^2=0\\⇔9x=-9\\⇔x=-1$
c)
$\dfrac{x}{2}-\dfrac{2x}{3}+\dfrac{1}{4}=\dfrac{2}{3}\\⇔\dfrac{x}{2}-\dfrac{2x}{3}+\dfrac{1}{4}-\dfrac{2}{3}\\⇔\dfrac{6x-4.2x+3-2.4}{12}=0\\⇔6x-8x+3-8=0\\⇔2x=-5\\⇔x=\dfrac{-5}{2}$
d)
$\dfrac{x-2}{3}-\dfrac{2x-3}{4}=x-1\\⇔\dfrac{x-2}{3}-\dfrac{2x-3}{4}-(x-1)=0\\⇔\dfrac{4(x-2)-3(2x-3)-12(x-1)}{12}=0\\⇔4(x-2)-3(2x-3)-12(x-1)=0\\⇔4x-8-(6x-9)-(12x-12)=0\\⇔4x-8-6x+9-12x+12=0\\⇔14x=13\\⇔x=\dfrac{13}{14}$
e)
$\dfrac{5-2x}{3}+\dfrac{1}{2}=2(x-1)\\⇔\dfrac{5-2x}{3}+\dfrac{1}{2}-2(x-1)=0\\⇔\dfrac{2(5-2x)+3-12(x-1)}{6}=0\\⇔2(5-2x)+3-12(x-1)=0\\⇔10-4x+3-12x+12=0\\⇔16x=25\\⇔x=\dfrac{25}{16}$