X=18 div 18=1

10.213

11. 2 giờ 55 phút

12. 4 ngày 21 giờ

13.

a: 6,008

b: 4500

c: 5,628

d: 4,009

14.

a,8579 cm2

b,32,04 m2

c,187200m2

d,5,017 tấn

14:

a: \(85.79km^2=857900000000cm^2\)

b: \(32m^24dm^2=32.04m^2\)

c: \(1872dm^2=18.72m^2\)

d: 5 tấn 17kg=5,017 tấn

13:

a: \(6dm^38cm^3=6.008dm^3\)

b: \(4.5m^3=4500dm^3\)

c: \(5628dm^3=5.628m^3\)

d: \(4m^39dm^3=4.009m^3\)

12: =108h:3+80h

=36h+80h=116h

10:

3h18p+7h21p:3

=198p:441p:3

=198p+147p=345p

=(4x0.25)x(2x0.5)

=1x1=1

a) ĐKXĐ : \(x\ne0;x\ne\pm2;x\ne3\)

\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

Đặt \(B=\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\)

\(B=\frac{\left(x+2\right)\left(x+2\right)}{-\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(2-x\right)\left(x-2\right)}{\left(2+x\right)\left(x-2\right)}\)

\(B=\frac{-\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{-\left(x+2\right)^2-4x^2--\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{-4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{-4x}{x-2}\)

\(\Rightarrow A=\frac{-4x}{x-2}:\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

\(\Leftrightarrow A=\frac{-4x\cdot x^2\cdot\left(2-x\right)}{\left(x-2\right)\cdot x\cdot\left(x-3\right)}\)

\(\Leftrightarrow A=\frac{4x^2\cdot x\cdot\left(x-2\right)}{\left(x-3\right)\cdot x\cdot\left(x-2\right)}\)

\(\Leftrightarrow A=\frac{4x^2}{x-3}\)

b) \(\left|x-7\right|=4\)

\(\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\end{cases}}}\)

Mà ĐKXĐ x khác 3 => x = 11

\(\Leftrightarrow A=\frac{4\cdot11^2}{11-3}=\frac{121}{2}\)

c) \(A=\frac{4x^2}{x-3}\)

Để A dương thì hoặc cả tử và mẫu âm hoặc cả tử và mẫu dương

Dễ thấy \(4x^2\ge0\forall x\)

=> Để A dương thì x - 3 dương

hay x - 3 > 0

<=> x > 3

Vậy x > 3 thì A > 0

Ta có: \(x=\sqrt{\dfrac{2}{3}}:\sqrt{\dfrac{3}{2}}\)

\(=\sqrt{\dfrac{2}{3}:\dfrac{3}{2}}=\sqrt{\dfrac{2}{3}\cdot\dfrac{2}{3}}=\dfrac{2}{3}\)

Ta có: \(M=\sqrt{6x+5}\)

\(=\sqrt{6\cdot\dfrac{2}{3}+5}=\sqrt{9}=3\)

Ta có  \(x=\sqrt{\dfrac{2}{3}}:\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{2}{3}.\dfrac{2}{3}}=\dfrac{2}{3}\)

Thay \(x=\dfrac{2}{3}\) vào biếu thức \(M\), ta được :

\(M=\sqrt{6.\dfrac{2}{3}+5}=\sqrt{4+5}=\sqrt{9}=3\)

\(x=\sqrt{\dfrac{2}{3}}:\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{2}{3}\cdot\dfrac{2}{3}}=\dfrac{2}{3}\)

\(M=\sqrt{6\cdot\dfrac{2}{3}+5}=\sqrt{9}=3\)

\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7..3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{2^{11}.3^5}=\frac{3}{2^4}=\frac{3}{16}\)

a: \(M=\dfrac{18+5x+15+3x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{8x+24}{\left(x+3\right)\left(x-3\right)}=\dfrac{8}{x-3}\)

b: Thay x=11 vào M, ta được:

\(M=\dfrac{8}{11-3}=1\)

a) \(M=\dfrac{18}{x^2-9}+\dfrac{5}{x-3}+\dfrac{3}{x+3}.\left(x\ne\pm3\right).\)

\(M=\dfrac{18}{\left(x-3\right)\left(x+3\right)}+\dfrac{5}{x-3}+\dfrac{3}{x+3}=\dfrac{18+5\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{18+5x+15+3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{24+8x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{8\left(3+x\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{8}{x-3}.\)

b) Thay \(x=11\left(TM\right)\) vào biểu thức M: 

\(\dfrac{8}{11-3}=\dfrac{8}{8}=1.\)