9: \(A=\dfrac{\dfrac{1}{4}-5\cdot\left(\dfrac{3}{2}\right)^2}{10\dfrac{5}{9}+\left(-\dfrac{2}{3}\right)^2}=\dfrac{\dfrac{1}{4}-5\cdot\dfrac{9}{4}}{10+\dfrac{5}{9}+\dfrac{4}{9}}\)

\(=\dfrac{\dfrac{1}{4}-\dfrac{45}{4}}{10+1}=\dfrac{-44}{4}:11=-\dfrac{44}{44}=-1\)

\(B=\dfrac{5}{12}\cdot3,7-\dfrac{5}{12}\cdot6,7=\dfrac{5}{12}\cdot\left(3,7-6,7\right)\)

\(=\dfrac{5}{12}\cdot\left(-3\right)=-\dfrac{5}{4}\)

\(A-B=\left(-1\right)-\left(-\dfrac{5}{4}\right)=-1+\dfrac{5}{4}=\dfrac{1}{4}\)

10: \(P=\left(6,8;1,36-\dfrac{29}{3}:\dfrac{58}{9}\right):\dfrac{0.27^3}{0.09^3\cdot2}\)

\(=\left(5-\dfrac{29}{3}\cdot\dfrac{9}{58}\right):\dfrac{\left(0,3\right)^6\cdot3^3}{0,3^6\cdot2}\)

\(=\left(5-\dfrac{3}{2}\right):\dfrac{3^3}{2}=\dfrac{7}{2}\cdot\dfrac{2}{27}=\dfrac{7}{27}\)

\(P+\dfrac{1}{27}=\dfrac{7}{27}+\dfrac{1}{27}=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\)

=>\(P+\dfrac{1}{27}\) là bình phương của một số hữu tỉ

9: \(A=\dfrac{\dfrac{1}{4}-5\cdot\left(\dfrac{3}{2}\right)^2}{10\dfrac{5}{9}+\left(-\dfrac{2}{3}\right)^2}=\dfrac{\dfrac{1}{4}-5\cdot\dfrac{9}{4}}{10+\dfrac{5}{9}+\dfrac{4}{9}}\)

\(=\dfrac{\dfrac{1}{4}-\dfrac{45}{4}}{10+1}=\dfrac{-44}{4}:11=-\dfrac{44}{44}=-1\)

\(B=\dfrac{5}{12}\cdot3,7-\dfrac{5}{12}\cdot6,7=\dfrac{5}{12}\cdot\left(3,7-6,7\right)\)

\(=\dfrac{5}{12}\cdot\left(-3\right)=-\dfrac{5}{4}\)

\(A-B=\left(-1\right)-\left(-\dfrac{5}{4}\right)=-1+\dfrac{5}{4}=\dfrac{1}{4}\)

10: \(P=\left(6,8;1,36-\dfrac{29}{3}:\dfrac{58}{9}\right):\dfrac{0.27^3}{0.09^3\cdot2}\)

\(=\left(5-\dfrac{29}{3}\cdot\dfrac{9}{58}\right):\dfrac{\left(0,3\right)^6\cdot3^3}{0,3^6\cdot2}\)

\(=\left(5-\dfrac{3}{2}\right):\dfrac{3^3}{2}=\dfrac{7}{2}\cdot\dfrac{2}{27}=\dfrac{7}{27}\)

\(P+\dfrac{1}{27}=\dfrac{7}{27}+\dfrac{1}{27}=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\)

=>\(P+\dfrac{1}{27}\) là bình phương của một số hữu tỉ

\(\dfrac{x}{\left(x+1\right)\left(x+4\right)}+\dfrac{x}{\left(x+4\right)\left(x+7\right)}+\dfrac{x}{\left(x+7\right)\left(x+10\right)}=\dfrac{x}{\left(x+1\right)\left(x+10\right)}\left(x\notin\left\{-1;-4;-7;-10\right\}\right)\\ \Leftrightarrow x\left[\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}\right]=\dfrac{x}{\left(x+1\right)\left(x+10\right)}\\ \Leftrightarrow\dfrac{1}{3}x\left(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+10}\right)=\dfrac{x}{\left(x+7\right)\left(x+10\right)}\\ \Leftrightarrow\dfrac{1}{3}x\left(\dfrac{1}{x+1}-\dfrac{1}{x+10}\right)=\dfrac{x}{\left(x+1\right)\left(x+10\right)}\\ \Leftrightarrow\dfrac{1}{3}x\cdot\dfrac{9}{\left(x+1\right)\left(x+10\right)}-\dfrac{x}{\left(x+1\right)\left(x+10\right)}=0\\ \Leftrightarrow\dfrac{3x}{\left(x+1\right)\left(x+10\right)}-\dfrac{x}{\left(x+1\right)\left(x+10\right)}\\ =0\\ \Leftrightarrow\dfrac{2x}{\left(x+1\right)\left(x+10\right)}=0\\ \Leftrightarrow2x=0\\ x=0\left(tm\right)\)

a) AB < AC (3 <  6)

=> B nằm giữa A và C 

b) B nằm giữa A và C 

=> AC = AB + BC

=> BC = AC - AB

=> BC = 6 - 3 = 3 (cm)

Mà: AB = 3 (cm) => AB = BC

Câu 4:

\(-0,40=\dfrac{-4}{10}=\dfrac{-2}{5}\)

\(0,25=\dfrac{25}{100}=\dfrac{1}{4}\)

\(-3,125=\dfrac{-3125}{1000}=\dfrac{-25}{8}\)

\(-5,24=\dfrac{-524}{100}=\dfrac{-131}{25}\)

Câu 1: -9,02<-1,23<0

0<0,5<2<13,1

Do đó: -9,02<-1,23<0,5<2<13,1

Câu 2:

\(-\dfrac{347}{10}=-34,7\)

\(\dfrac{6741}{100}=67,41\)

\(-\dfrac{53}{1000}=-0,053\)

\(\dfrac{86}{100}=0,86\)

Câu 3:

Số đối của 8,4 là -8,4

Số đối của -34,24 là 34,24

Số đối của -0,9 là 0,9

Số đối của 11,1 là -11,1

Lời giải:

Gọi khối lượng đầu cá và thân cá lần lượt là $a$ và $b$ (gam). Theo bài ra ta có:

$a = \frac{1}{2}b+350$

$b=a+350$

Thay $b=a+350$ vào điều kiện ban đầu thì:

$a=\frac{1}{2}(a+350)+350$

$a=\frac{1}{2}a+525$

$\frac{1}{2}a=525$

$a=525.2=1050$

$b=a+350=1050+350=1400$ 

Khối lượng con cá: $a+b+350=1050+1400+350=2800$ (gam) hay $2,8$ kg.

\(15\cdot23+4\cdot3^2-5\cdot7\)

\(=15\cdot23+4\cdot9-35\)

=315+36-35

=315+1

=316

15 x 23 + 4x 3^2 - 5 x 7

= 15 x 23 + 4 x 9 - 5 x 7

= 345 + 36 - 35

= 381 - 35

= 346

35x16x2+4x65x8+65

=32x35+32x65+65

=32x(35+65)+65

=32x100+65

=3200+65=3265

35x16x2+4x65x8+65

=35x32+32x65+65

= 32x(35+65)+65

= 32x100+65

= 3200+65=3265

\(2^4-12=16-12=4\)

\(3^3-16=27-16=11\)

Do 11 > 4 

=> \(2^4-12< 3^3-16\)

Ta có:

+) \(2^4+\left(-12\right)=16-12=4\)

+) \(3^3+\left(-16\right)=27-16=11\)

Vì \(4< 11\) nên \(2^4+\left(-12\right)< 3^3+\left(-16\right)\)

Vậy...

a) \(15+2x=5^{10}:5^8\\ 15+2x=5^2\\ 2x=25-15\\ 2x=10\\ x=10:2\\ x=5\)

b) \(48:x+17=33\\ 48:x=33-17\\ 48:x=16\\ x=48:16\\ x=3\)

c) \(7^{2x-6}=49=7^2\\ 2x-6=2\\ 2x=2+6\\ 2x=8\\ x=8:2\\ x=4\)

d) Bạn xem lại đề nhé

a, \(15+2x=\dfrac{5^{10}}{5^8}=5^2\Leftrightarrow15+2x=25\Leftrightarrow2x=10\Leftrightarrow x=5\)

b, \(\dfrac{48}{x}+17=33\Leftrightarrow\dfrac{48}{x}=33-17=16\Leftrightarrow x=\dfrac{48}{16}=3\)

c, \(7^{2x-6}=49=7^2\Rightarrow2x-6=2\Leftrightarrow x=4\)

d, sửa \(\left(9x+2\right).5=28\Leftrightarrow9x+2=\dfrac{28}{5}\Leftrightarrow9x=\dfrac{18}{5}\Leftrightarrow x=\dfrac{2}{5}\)