\(\dfrac{3}{5}-\dfrac{1}{4}+\dfrac{7}{20}=\dfrac{12}{20}-\dfrac{5}{20}+\dfrac{7}{20}=\dfrac{14}{20}=\dfrac{7}{10}\)

\(\dfrac{3}{5}-\dfrac{1}{4}+\dfrac{7}{20}=\dfrac{12}{20}-\dfrac{5}{20}+\dfrac{7}{20}=\dfrac{7}{20}+\dfrac{7}{20}=\dfrac{14}{20}\)

a)\(\frac{7}{3}.\left( { - 2,5} \right).\frac{6}{7} = \frac{7}{3}.\frac{6}{7}.\left( { - 2,5} \right) = 2.\left( { - 2,5} \right) =  - 5\)

b)

\(\begin{array}{l}0,8.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9} - 0,2\\ = \frac{4}{5}.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9}-\frac{2}{10}\\ = \frac{4}{5}.\left( {\frac{{ - 2}}{9} - \frac{7}{9}} \right) -\frac{1}{5}\\ = \frac{4}{5}.\left( { - 1} \right)-\frac{1}{5} \\= \frac{{ - 4}}{5}-\frac{1}{5}\\=\frac{-5}{5}\\=-1.\end{array}\)

= (12-12) + (11+10) - (9+8) - (7+5) - (4+3) + (2-1)

= 0 + 21 - 17 - 12 - 7 + 1

= 21- 17 - 12 - 7 +1

= 4 - 12 - 7 +1

= -8 - 7 + 1

= -15 + 1

= -14

hết

= 12 - 12 + 11 + ( 10 - 9 ) + ( 8 - 7 ) + ( 5 - 4 ) + 3 + ( 2 -1 ) 

= 0 + 11 + 1 + 1 + 1 + 3 + 1

= 11 +1 + 1 + 1 + 3 + 1

= 12 + 1 + 1 + 3 + 1 

= 13 + 1 + 3 + 1

= 14 + 3 + 1

= 17 + 1 

= 18

Đáp án đây nha bạn !!!

Chúc bạn học tốt !!!

Ta có 0,2+0,5+0,8+6,5=8 

Em điền vaod một số  sao cho 8 cộng với số đó lớn hơn 38,7

\(\frac{0,2\times317\times7+0,14+3520+331\times14}{0,2+0,5+0,8+31,7+6,5-38,7}\)

0,2 x 4 x 44 + 0,8 x 20 + 2,4 x 12

= 0,8 x 44 + 0,8 x 20 + 0,8 x 3 x 12

= 0,8 x 44 + 0,8 x 20 + 0,8 x 36

= 0,8 x (44 + 20 + 36)

= 0,8 x 100

= 80

Bài 6: M(x;y); A(1;3); B(4;2)

\(\overrightarrow{MA}=\left(1-x;3-y\right);\overrightarrow{MB}=\left(4-x;2-y\right)\)

=>\(MA^2=\left(1-x\right)^2+\left(3-y\right)^2;MB^2=\left(4-x\right)^2+\left(2-y\right)^2\)

ΔMAB cân tại M

=>MA=MB

=>\(MA^2=MB^2\)

=>\(\left(1-x\right)^2+\left(3-y\right)^2=\left(4-x\right)^2+\left(2-y\right)^2\)

=>\(x^2-2x+1+y^2-6y+9=x^2-8x+16+y^2-4y+4\)

=>-2x-6y+10=-8x-4y+20

=>-2x-6y+8x+4y=20-10

=>6x-2y=10

=>3x-y=5

=>y=3x-5

ΔMAB vuông tại M

=>\(\overrightarrow{MA}\cdot\overrightarrow{MB}=\overrightarrow{0}\)

=>(1-x)(4-x)+(3-y)(2-y)=0

=>(x-1)(x-4)+(y-3)(y-2)=0

=>(x-1)(x-4)+(3x-5-3)(3x-5-2)=0

=>(x-1)(x-4)+(3x-8)(3x-7)=0

=>\(x^2-5x+4+9x^2-45x+56=0\)

=>\(10x^2-50x+60=0\)

=>\(x^2-5x+6=0\)

=>(x-2)(x-3)=0

=>x=2 hoặc x=3

TH1: x=2

=>y=3x-5=3*2-5=6-5=1

=>M(2;1)

TH2: x=3

=>y=3x-5=3*3-5=9-5=4

=>M(3;4)

Bài 5:

a: A(1;3); B(4;2); C(1;0)

\(AB=\sqrt{\left(4-1\right)^2+\left(2-3\right)^2}=\sqrt{3^2+\left(-1\right)^2}=\sqrt{10}\)

\(AC=\sqrt{\left(1-1\right)^2+\left(0-3\right)^2}=3\)

\(BC=\sqrt{\left(1-4\right)^2+\left(0-2\right)^2}=\sqrt{\left(-3\right)^2+\left(-2\right)^2}=\sqrt{13}\)

Chu vi của tam giác ABC là:

AB+AC+BC

\(=3+\sqrt{10}+\sqrt{13}\)

b: A(1;3); B(4;2); C(1;0); H(x;y)

H là trực tâm của ΔABC

=>AH⊥BC và BH⊥AC

\(\overrightarrow{AH}=\left(x-1;y-3\right);\overrightarrow{BC}=\left(1-4;0-2\right)=\left(-3;-2\right)\)

\(\overrightarrow{BH}=\left(x-4;y-2\right);\overrightarrow{AC}=\left(1-1;0-3\right)=\left(0;-3\right)\)

BH⊥AC
=>\(\overrightarrow{BH}\cdot\overrightarrow{AC}=0\)

=>0(x-4)+(-3)(y-2)=0

=>-3(y-2)=0

=>y-2=0

=>y=2

AH⊥BC

=>\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0\)

=>-3(x-1)+(-2)(y-3)=0

=>3(x-1)+2(y-3)=0

=>3(x-1)+2(2-3)=0

=>3(x-1)+2*(-1)=0

=>3x-3-2=0

=>3x=5

=>x=5/3

=>H(5/3;2)

A;

1+2+3+4+5+6+7+8+9

=(1+9)+(2+8)+(3+7)+(4+6)+5

=10+10+10+10+5

=40+5

=45

B;0,1+0,2+0,3+0,4+0,5+0,6+0,7+0,8+0,9

= (0,1+0,9)+(0,2+0,8)+(0,3+0,7)+(0,4+0,6)+0,5

=1+1+1+1+0,5

=4+0,5

=4,5

Tick nhé mất công lắm đó

+) (1+9)+(2+8)+(3+7)+(4+6)+5

=10+10+10+10+5

=45

+) tương tự bài trên

a: \(\dfrac{x}{0.9}=\dfrac{5}{6}\)

\(\Leftrightarrow x=\dfrac{3}{4}\)

b: \(\dfrac{-6}{x}=\dfrac{9}{-15}\)

\(\Leftrightarrow x=10\)

c: \(\dfrac{\dfrac{14}{15}}{\dfrac{9}{10}}=\dfrac{x}{\dfrac{3}{7}}\)

\(\Leftrightarrow x=\dfrac{3}{7}\cdot\dfrac{14}{15}:\dfrac{9}{10}=\dfrac{2}{5}\cdot\dfrac{10}{9}=\dfrac{20}{45}=\dfrac{4}{9}\)