b: 1/2x-4=0

=>1/2x=4

hay x=8

a: x+7=0

=>x=-7

e: 4x²-81=0

=>(2x-9)(2x+9)=0

=>x=9/2 hoặc x=-9/2

g: x²-9x=0

=>x(x-9)=0

=>x=0 hoặc x=9

a)\(x+7=0=>x=-7\)

b)\(\dfrac{1}{2}x-4=0=>\dfrac{1}{2}x=4=>x=8\)

c)\(-8x+20=0=>-8x=-20=>x=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100=>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

e)\(4x^2-81=0=>4x^2=81=>x^2=\dfrac{81}{4}=>\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\)

f)\(x^2-7=0=>x^2=7=>x=\sqrt{7}\)

g)\(x^2-9x=0=>x\left(x-9\right)=0=>\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

a: x+7=0

nên x=-7

b: x-4=0

nên x=4

c: -8x+20=0

=>-8x=-20

hay x=5/2

d: x²-100=0

=>(x-10)(x+10)=0

=>x=10 hoặc x=-10

a) x +7 =0

=>x = -7

b) x - 4 =0=>x = 4

c) -8x + 20 = 0 =>-8x =-20 =>\(x=-\dfrac{20}{-8}=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

e)\(4x^2-81=0=>4x^2=81=>x^2=\dfrac{81}{4}=>\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\)

f)\(x^2-7=0=>x^2=7=>x=\sqrt{7}\)

g)\(x^2-9x=0=>x\left(x-9\right)=0=>\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

H)\(x^3+3x=0=>x\left(x^2 +3\right)=0=>\left[{}\begin{matrix}x=0\\x^2=-3\left(vl\right)\end{matrix}\right.\)

Đáp án là A

Đáp án A

(Lưu các giá trị này vào các biến A, B để thuận tiện tính toán).

\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)

\(\Leftrightarrow-5x-18=0\)

\(\Leftrightarrow x=-\dfrac{18}{5}\)

Vậy ...

\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)

\(\Leftrightarrow12x+6=0\)

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

Vậy ...

\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)

\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)

Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)

\(\Rightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy ...

\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)

\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)

Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ...

( 3x - 6 ) . 3 = 81

⇒ 3x - 6 = 27

⇒ 3x = 33

⇒ x = 11.

Vậy x = 11

\(\left(3x-6\right)\cdot3=81\)

\(\Rightarrow9x-18=81\)

\(\Rightarrow9x=81+18\)

\(\Rightarrow9x=99\)

\(\Rightarrow x=\dfrac{99}{9}\)

\(\Rightarrow x=11\)

\(\left(3x-6\right).3=81\)

\(3x-6=81:3\)

\(3x-6=27\)

\(3x=27+6\)

\(3x=33\)

\(x=33:3\)

\(x=11\)

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.