Xét x < -5 ta có /x-5/=- ( x-5)=x+5=5-x

Khi đó /x+5 = -( x+ 5)= -x-5

  B= 10 + ( 5-x) - (-x-5)

   = 10+5-x+x+5=20

Xét 5 bé hoặc bằng x<5 ta có /x-5/=-(x-5)= -x+5=5-x

                 và /x+5/=x+5

Khi đó B= 10+(5-x) - (x+5)=10+5-x=x-5=10-2-x

Xét x lớn hoặc bằng 5 ta có /x-5/ = x-5 và /x+5/=x+5

Khi đó B=10+/x-5/-/x+5/=10+x-5-x=0

Vậy khi x < 5 thì 

    B= 10-2-x

    Khi x>5 thì 

  B=0

A )  2/17 + 30/46 + 15/17 + 4/49 + 8/23=102/49

B ) 5/8 x ( 19/31 x 8/5 )=19/31

C ) 7/9 x 8/15 x 9/7 x 15/4=2

D ) 11/12 x 24/7 x 21/22=3

nhìu quá nên mk chỉ cho kết quả thôi nha ^-^!!!

1/ (a + a).(b + b) = 2a.2b = 4ab

2/ (a - a) . (b - b) = 0.0 = 0

lạ quá

1:

=ax2xbx2

=2x(a=b)

2:

=0x0

=0

1) - 7264 + ( 1543 + 7264 )

= - 7264 + 1543 + 7264 

= ( - 7264 + 7264 ) + 1543

= 0 + 1543 

= 1543

2) ( 144 - 97 ) - 144

= 144 - 97 - 144

= ( 144 - 144 ) - 97

= 0 - 97 

= - 97

3) ( - 145 ) - ( 18 - 145 )

= -145 - 18 + 145

= ( - 145 + 145 ) - 18

= 0 - 18 

=  - 18 

4) 111 + ( - 11 + 27 )

= 111 - 11 + 27

= 100 + 27 

= 127 

các câu còn lại bn tự làm nha

1. -7264+(1543+7264)                                                     

=-7264+1543+7264

=[-7264+7264]+1543

=0

1,-7264+(1543+7264)=-7264+1543+7264=(-7264+7264)+1543=0+1543=1543

2,(144-97)-144=144-97-144=(144-144)-97=0-97=-97

3,(-145)-(18-145)=-145-18+145=(-145+145)-18=0-18=-18

4,111+(-11+27)=111-11+27=100+27=127

5,(27+514)-(486-73)=27+514-486+73=(27+73)+(514-486)=100+28=128

6,(36+79)+(145-79-36)=36+79+145-79-36=(36-36)+(79-79)+145=0+0+145=145

7,10-[12-(-9-1)]=10-[12+9+1]=10-12-9-1=10-10-12=0-12=-12

8,(38-29+43)-(43+38)=38-29+43-43-38=(38-38)+(43-43)-29=0+0-29=-29

9,271-[(-43)+271-(-17)]=271-[-43+271+17]=271+43-271-17=(271-271)+(43-17)=0+26=26

10,-144-[29-(+144)-(+144)]=-144-[29-144-144]=-144-[29-144-144]=-144-29+144+144=(-144+144)+(144-29)=0+115=115

b: \(=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3x}{x\left(x+5\right)}=\dfrac{3}{x+5}\)

\(a,=-3x^3+x^2+9x^2-3x-12x+4=-3x^3+10x^2-15x+4\\ b,=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3x}{x\left(x+5\right)}=\dfrac{3}{x+5}\)

a: -3x+4+5x=-10-x

=>2x+4=-x-10

=>3x=-14

hay x=-14/3

b: \(-x+1=-3x-8\)

=>-x+3x=-8-1

=>2x=-9

hay x=-9/2

c: \(8-\left(x-1\right)=10+\left(x+5\right)\)

=>x+15=8-x+1

=>x+15=9-x

=>2x=-6

hay x=-3

d: \(100+\left(x+7\right)-\left(-x+3\right)=8+\left(x+100\right)\)

=>x+7+x-3=8+x

=>2x+4-x-8=0

=>x=4

Bài 1:

a)\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2\cdot4\right)^5}{\left(0,2\cdot2\right)^6}=\frac{\left(0,2\right)^5\cdot\left(2^2\right)^5}{\left(0,2\right)^6\cdot2^6}=\frac{\left(0,2\right)^5\cdot2^{10}}{\left(0,2\right)^6\cdot2^6}=\frac{2^4}{0,2}=\frac{16}{\frac{2}{10}}=80\)

b)\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=256\)

Bài 2:

a)\(2^{x-1}=16\)

\(\Rightarrow2^{x-1}=2^4\)

\(\Rightarrow x-1=4\Rightarrow x=5\)

b)\(\left(x-1\right)^2=25\)

\(\Rightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)

\(\Rightarrow x-1=5\) hoặc \(x-1=-5\)

\(\Rightarrow x=6\) hoặc \(x=-4\)

Vậy \(x=6\) hoặc \(x=-4\)

c)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\1=\left(x-1\right)^4\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\\left(x-1\right)^4=\left(-1\right)^4=1^4\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x-1=1\\x-1=-1\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=0\end{array}\right.\)

d)\(\left(x+20\right)^{100}+\left|y+4\right|=0\left(1\right)\)

Ta thấy: \(\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\left(2\right)\)

Từ (1) và (2) suy ra \(\begin{cases}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{cases}\)

\(\Rightarrow\begin{cases}x+20=0\\y+4=0\end{cases}\)\(\Rightarrow\begin{cases}x=-20\\y=-4\end{cases}\)

Bài 1.

a.\(\left(x-8\right)\left(x^3+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

b.\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)

\(\Leftrightarrow4x-3-x-5=30-3x\)

\(\Leftrightarrow4x-x+3x=30+5+3\)

\(\Leftrightarrow6x=38\)

\(\Leftrightarrow x=\dfrac{19}{3}\)

Bài 1:

a. $(x-8)(x^3+8)=0$

$\Rightarrow x-8=0$ hoặc $x^3+8=0$

$\Rightarrow x=8$ hoặc $x^3=-8=(-2)^3$

$\Rightarrow x=8$ hoặc $x=-2$

b.

$(4x-3)-(x+5)=3(10-x)$

$4x-3-x-5=30-3x$

$3x-8=30-3x$

$6x=38$
$x=\frac{19}{3}$

Bài 2:

$f(x)=(x-1)(x+2)=0$

$\Leftrightarrow x-1=0$ hoặc $x+2=0$

$\Leftrightarrow x=1$ hoặc $x=-2$

Vậy $g(x)$ cũng có nghiệm $x=1$ và $x=-2$

Tức là:

$g(1)=g(-2)=0$

$\Rightarrow 1+a+b+2=-8+4a-2b+2=0$

$\Rightarrow a=0; b=-3$

a. (13 - 135 + 49) - (13 + 49)

= 13 - 135 + 49 - 13 - 49

= (13 - 13) + (49 - 49) - 135

= 0 + 0 - 135

= -135

=-135 nha bạn

tik đi

a) (13 - 135 + 49) - (13 + 49)

= 13 - 135 + 49 -13 - 19

= [ 13 + (-13 ) ] + ( 49 - 49 ) - 135

=      0             +     0         - 135

=                  0                   - 135

=                   -135