bài 5 Tìm x biết
a 5 x( 1/5 x – 2) + 3(6 – 1/2 x bình )+ = 12
b 7 x (x – 2 )- 5 (x – 1) = 7 xbình + 3
c 2 (5x -8) – 3(4 x- 5) =4(3x-4)+11
d 5x-3 {4x-2[4x-3(5x-2)]}=182
bài 6 chứng minh đẳng thức
a)a(b-c)-b(a+c)+c(a-b)=-2bc
b)a(1-b)+a(a²-1)=a(a²-b)
B5,
b, 7x(x – 2 )-5(x – 1)=7x^2 + 3
<=>7x^2-14x-5x+5=7x^2+3
<=>7x^2-19x+5=7x^2+3
<=>-19x+5=3
<=>-19x=-2
<=>x=2/19
c, 2(5x -8)-3(4x-5)=4(3x-4)+11
<=>10x-16-12x+15=12x-16+11
<=>-2x-1=12x-5
<=>14x-5=-1
<=>14x=4
<=>x=4/14=2/7
B6,
a, a(b-c)-b(a+c)+c(a-b)=-2bc
VT=a(b-c)-b(a+c)+c(a-b)=ab-ac-ab-bc+ac-bc=-2bc=VP (đpcm)
b, a(b-c)-b(a+c)+c(a-b)=-2bc
VT=a(1-b)+a(a^2-1)=a-ab+a^3-a=a(a^2-b)=VP (đpcm)
Bài 5:
`a) 5x(1/5 x -2 )+3(6-1/2 x^2)=12`
`⇔` x^2-10x+18-3/2x^2-12=0`
`⇔-1/2x^2-10x +6= 0`
`⇔-x^2-20x+12=0`
`⇔x^2+20x-12=0`
`⇔x^2+20x+100-112=0`
`⇔(x+10)^2-(\sqrt[112])^2=0`
`⇔(x+10-\sqrt[112])(x+10+\sqrt[112])=0`
`⇒` \(\left[ \begin{array}{l}x+10-\sqrt{112}=0\\x+10+\sqrt{112}=0\end{array} \right.\)
`⇒ x=-10±\sqrt{112}.`
`⇒ x=-10±4\sqrt{7}.`
Vậy ` x=-10±4\sqrt{7}.`
`b)7x(x-2)-5(x-1)=7x^2+3`
`⇔7x^2-14x-5x+5-7x^2-3=0`
`⇔(7x^2-7x^2)+(-14x-5x)+(-3+5)=0`
`⇔-19x+2=0`
`⇔-19x=0-2`
`⇔-19x=-2`
`⇔x= (-2) : (-19)`
`⇔x=2/19.`
Vậy ` x=2/19.`
`c)2(5x-8)-3(4x-5)=4(3x-4)+11`
`⇔10x-16-12x+15=12x-16+11`
`⇔-2x-1-12x+16-11=0`
`⇔-14x+4=0`
`⇔-14x=-4`
`⇔x=2/7.`
Vậy ` x=2/7.`
`d)5x-3{4x-2[4x-3(5x-2)]}=182`
`⇔5x-3[4x-2(4x-15x+6)]=182`
`⇔5x-3(4x-8x+30x-12)=182`
`⇔5x-12x+24x-90x+36-182=0`
`⇔-73x-146=0`
`⇔-73x=146`
`⇔x=146 : (-73)`
`⇔x=-2.`
Vậy ` x=-2.`
Bài 6:
`a)` Có: `a(b-c)-b(a+c)+c(a-b)=ab-ac-ab-bc+ac-bc=(ab-ab)+(-ac+ac)-(bc+bc)=-2bc` $(đpcm).$
Vậy `a(b-c)-b(a+c)+c(a-b)=-2bc.`
`b) ` Có `a(1-b)+a(a^2-1)=a(1-b+a^2-1)=a(a^2-b)` $(đpcm).$
Vậy `a(1-b)+a(a^2-1)=a(a^2-b).`