a) với x = 0 (TM) thay vào biểu thức ta được

A =\(\dfrac{8x+13}{2x+5}\)= \(\dfrac{8.0+13}{2.0+5}\)= \(\dfrac{13}{5}\)

với x =1 (TM) thay vào biểu thức ta được

A= \(\dfrac{8x+13}{2x+5}\)= \(\dfrac{8.1+13}{2.1+5}\)= \(\dfrac{21}{7}\)= 3

với x =-2 (TM) thay vào biểu thức ta được

A =\(\dfrac{8x+13}{2x+5}\)= \(\dfrac{8.\left(-2\right)+13}{2.\left(-2\right)+5}\)= \(\dfrac{-16+13}{-4+5}\)= -3

vậy khi x= 0 thì A= \(\dfrac{13}{5}\)

khi x= 1 thì A = 3

khi x=-2 thì A= -3

b) ta có A= \(\dfrac{1}{2}\) \(\Rightarrow\)\(\dfrac{8x+13}{2x+5}\)= \(\dfrac{1}{2}\)

\(\Rightarrow\) 2( 8x+13) =1(2x+5)

\(\Rightarrow\) 16x+26=2x+5

\(\Rightarrow\)16x-2x=5-26

\(\Rightarrow\) 14x = -21

\(\Rightarrow\)x = \(\dfrac{-21}{14}\)=\(\dfrac{-3}{2}\)

vậy x = \(\dfrac{-3}{2}\)để A=\(\dfrac{1}{2}\)

c) ta có A= \(\dfrac{8x+13}{2x+5}\)= 4 - \(\dfrac{7}{2x+5}\)

để x\(\in\)z thì 2x+5 \(\in\)Ư(7)

Ư(7) = \(\pm\)1 , \(\pm\)7

lập bảng

ta có 2x+5=1\(\Rightarrow\)2x=-4 \(\Rightarrow\)x = \(\dfrac{-4}{2}\)=-2 (KTM)

2x+5 =-1 \(\Rightarrow\)2x= -6 \(\Rightarrow\)x = \(\dfrac{-6}{2}\)=-3(KTM)

2x+5 = 7\(\Rightarrow\)2x=2\(\Rightarrow\)x= \(\dfrac{2}{2}\)=1

2x+5 = -7 \(\Rightarrow\)2x= -12 \(\Rightarrow\)x = \(\dfrac{-12}{2}\)=-6(KTM)

vậy1 là giá trị cần tìm để x \(\in\)z

a) 2|x-3|+|2x+5|=11

2x-6+2x+5=11

2x+2x = 11-5+6

4x = 12

x= \(\dfrac{12}{4}\)

x= 3

vậy x =3

b) |x-3| +|5-x| +2|x-4|=2

x-3+5-x+2x-8=2

x-x+2x = 2+3-5+8

2x = 8

x = \(\dfrac{8}{2}\)

x=4

vậy x = 4

3712,5 kg 

Tick nha !

x⁴+x³+x²+x+1

= ( x⁴+x³) +( x²+x+1)

= x³(x+1)+(x+1)²

= ( x+1) (x³+x+1)

bài 2

a) | 3x+2| =\(\dfrac{4}{3}\)

3x+2 = \(\dfrac{4}{3}\)

3x = \(\dfrac{4}{3}\)-2

3x = \(\dfrac{-2}{3}\)

x = \(\dfrac{-2}{3}\):3

x= \(\dfrac{-2}{3}\). \(\dfrac{1}{3}\)

x = \(\dfrac{-2}{9}\)

vậy x = \(\dfrac{-2}{9}\)

b) | 2x- \(\dfrac{1}{3}\)| - \(\dfrac{1}{3}\)= \(\dfrac{-1}{5}\)

2x -\(\dfrac{1}{3}\)-\(\dfrac{1}{3}\)= \(\dfrac{-1}{5}\)

2x =\(\dfrac{-1}{5}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)

2x = \(\dfrac{7}{15}\)

x = \(\dfrac{7}{15}\): 2

x = \(\dfrac{7}{15}\).\(\dfrac{1}{2}\)

x= \(\dfrac{7}{30}\)

vậy x = \(\dfrac{7}{30}\)

c) | 4x +\(\dfrac{2}{7}\)| = | \(\dfrac{-5}{7}\)|

4x +\(\dfrac{2}{7}\)= \(\dfrac{-5}{7}\)

4x = \(\dfrac{-5}{7}\)-\(\dfrac{2}{7}\)

4x = \(\dfrac{-7}{7}\)

4x = -1

x= \(\dfrac{-1}{4}\)

x= -0,25

vậy x = - 0,25

| x+2|-2=x

x+2-2=x

x+x = -2+2

2x = 0

x = \(\dfrac{0}{2}\)

x = 0

vậy x =0

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

=2x-1=3

=2x=3+1

=2x=4

x=\(\dfrac{4}{2}\)

x=2

vậy x =2

sữa dê 1 chút nhé \(\sqrt{7x}\)-\(\sqrt{2x-3}\)=0

= 7x-2x-3=0

= 5x -3=0

= 5x =3

x= \(\dfrac{3}{5}\)

vậy x = \(\dfrac{3}{5}\)

\(\dfrac{3}{5}\)

a) ta có 345² = 119025

342.348= 119016

vì 119025>119016

\(\Rightarrow\) 345² > 342.348