\(< =>3\sqrt{2x}-5\sqrt{2^2.2x}+7\sqrt{3^2.2x}=28\)

\(< =>3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

\(< =>14\sqrt{2x}=28\)

\(< =>\sqrt{2x}=\dfrac{28}{14}=2=\sqrt{4}\)

\(< =>\sqrt{2x}=\sqrt{2.2}=>x=2\)

b) 2³³²<2³³³=2^3.111=8¹¹¹

3²²³>3²²²=3^2.111=9¹¹¹

Ta có 8¹¹¹<9¹¹¹=>2³³³<3²²²=>2³³²<3²²²(mà 3²²²<3²²³)=>2³³²<3²²³

a) 2²⁴=2^3.8=8⁸

3³⁶=3^3.12=9¹²

8⁸<9⁸ mà 9⁸<9¹² =>8⁸<9¹²=>2²⁴<3³⁶

3³.5³=(3.5)³=15³

a)(x+3y)(x²-3xy+9y²)=x³+27y³

b)(2x-3y)(4x²+6xy+9y²)=8x³-27y³

a) 4x²-y²+4x+1=(4x²+4x+1)-y²=(2x+1)²-y² = (2x+1-y)(2x+1+y)

b) x³-x+y³-y=(x³+y³)-(x+y)=(x+y)(x²-xy+y²)-(x+y) = (x+y)(x²-xy+y²-1)

a) 2012/|x|+2013 ta có |x| luôn lớn hơn hoặc bằng 0

muốn 2012/|x|+2013 lớn nhất thì |x|+2013(mẫu số) phải nhỏ nhất =>|x|=0=>x=0

Giá trị nhỏ nhất của 2012/|x|+2013 là 2012/0+2013=2012/2013

a)(x+1)(x²+2x)=(x+1)x(x+2)=0

\(=>\left\{{}\begin{matrix}x+1=0=>x=-1\\x=0\\x+2=0=>x=-2\end{matrix}\right.\)

b)x(3x-2)-5(2-3x)=x(3x-2)+5(3x-2)=(3x-2)(x+5)=0

\(=>\left\{{}\begin{matrix}3x-2=0=>x=\dfrac{2}{3}\\x+5=0=>x=-5\end{matrix}\right.\)

c)\(\dfrac{4}{9}-25x^2=\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)\)

=0

\(=>\left\{{}\begin{matrix}\dfrac{2}{3}-5x=0=>x=\dfrac{2}{15}\\\dfrac{2}{3}+5x=0=>x=\dfrac{-2}{15}\end{matrix}\right.\)

d)\(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2=0\)

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

<=>x²+3x+2-x²-2x+15=0

<=>x+17=0

=>x=-17

Bài 1

a) x:(3/4)³=(3/4)²=>x=(3/4)².(3/4)³=(3/4)²⁺³=(3/4)⁵

b) (2/5)⁵.x=(2/5)⁸=>x=(2/5)⁸:(2/5)⁵=(2/5)^8-5=(2/5)³

Bài 2

(0,36)⁸=[ (0,6)² ]⁸=(0,6)^2.8=(0,6)¹⁶

(0,216)⁴=[ (0,6)³ ]⁴=(0,6)^3.4=(0,6)¹²

Bài 3

a) (1/3)^m=1/81=(1/3)⁴ => m=4

b) (3/5)ⁿ=(9/25)⁵=[ (3/5)² ]⁵=(3/5)^2.5=(3/5)¹⁰=>n=10

c) (-0,25)^p=1/256=(1/4)⁴=(0,25)⁴=(-0,25)⁴=>p=4