1.\(\frac{43141}{3}\)

2.18

ko dung r

=14380,333333333333

=18

(*<>*)

\(A=\dfrac{\sqrt{4x^2-4x+1}}{4x-2}=\dfrac{\sqrt{\left(2x-1\right)^2}}{2\left(2x-1\right)}=\dfrac{\left|2x-1\right|}{2\left(2x-1\right)}\)

\(\Rightarrow\left|A\right|=\left|\dfrac{\left|2x-1\right|}{2\left(2x-1\right)}\right|=\dfrac{\left|2x-1\right|}{2\left|2x-1\right|}=\dfrac{1}{2}\)

Ta có: \(A=\dfrac{\sqrt{4x^2-4x+1}}{4x-2}\)

\(=\dfrac{\left|2x-1\right|}{2\left(2x-1\right)}\)

\(=\left[{}\begin{matrix}-\dfrac{\left(2x-1\right)}{2\left(2x-1\right)}=-\dfrac{1}{2}\left(x< \dfrac{1}{2}\right)\\\dfrac{2x-1}{2\left(2x-1\right)}=\dfrac{1}{2}\left(x\ge\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left|A\right|=0.5\)

a) \(=x^3-8-x^3+x=x-8\)

b) \(=x^2+x-3x-3-x^2+16=13-2x\)

Bài 3:

Số tiền phải trả khi mua 10 chai là:

12000*10-10000=110000(đồng)

a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)

\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)

\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)

\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)

\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)

\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)

a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)

do \(x\ge0;y\ge0\)

\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)

b) Ta có :

\(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2.2a+2^2.a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left[1^2-2.1.2a+\left(2a\right)^2\right]}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}.\sqrt{5}.\sqrt{a^2}.\sqrt{\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|\)

Vì a > 0,5 <=> a > 0 <=> | a | = a

Vì a > 0,5 <=> 2a > 2 . 0,5 <=> 2a > 1 hay 1 < 2a

<=> 1 - 2a < 0 <=> | 1 - 2a | = - ( 1 - 2a )

= -1 + 2a = 2a - 1

Thay vào trên ta được : 

\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|=\frac{2}{2a-1}\sqrt{5}.a.\left(2a-1\right)=2a\sqrt{5}\)

Vậy : \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=2a\sqrt{5}\)

a, <=> 2,5 : 4x = 2,5

<=> 4x = 2,5 : 2,5 = 1

<=> x=1 : 4 = 1/4

b, <=> 1/5.x:3 = 8/3

<=> 1/5.x = 8/3 . 3 = 8

<=> x = 8 : 1/5 = 40

cai lon me may

a)2,5:4x=0,5:0,2

2,5:4x=2,25

4x=2,5:2,25

4x=\(\frac{10}{9}\)

x=\(\frac{10}{9}\):4

x=\(\frac{10}{9}\).\(\frac{1}{4}\)

x=\(\frac{5}{18}\)

Vậy x=\(\frac{5}{18}\)

b)\(\frac{1}{5}\)x:3=\(\frac{2}{3}\):0,25

\(\frac{1}{5}\)x:3=\(\frac{2}{3}\):\(\frac{1}{4}\)

\(\frac{1}{5}\)x:3=\(\frac{8}{3}\)

\(\frac{1}{5}\)x=\(\frac{8}{3}\).3

0,2x=8

x=8:0,2

x=40

Vậy x=40

Vui lòng viết yêu cầu bài :>

a, (x-2)(3x+5)=(2x-4)(x+1)
<=> (x-2)(3x+5)-2(x-2)(x+1)=0
<=>(x-2)(3x+5-2x-2)=0
<=>(x-2)(x+3)=0
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

b, (2x+5)(x-4)=(x-5)(4-x)
<=>(2x+5)(x-4)-(x-5)(4-x)=0
<=>(x-4)(2x+5+x-5)=0
<=>(x-4)3x=0
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}}\)