Bài 44: (SBT/12):

a. (7.3⁵ - 3⁴ + 3⁶) : 3⁴

= (7.3⁵ : 3⁴) + (-3⁴ : 3⁴) + (3⁶ : 3⁴)

= 7 . 3 - 1 + 3²

= 21 - 1 + 9

= 29

b. (16³ - 64²) : 8³

= [(2.8)³ - (8²)² ] : 8³

= (2³ . 8³ - 8⁴) : 8³

= ( 2³ . 8³ : 8³) + (-8⁴ : 8³)

= 2³ - 8

= 8 - 8

= 0

a) \(\left(7.3^5-3^4+3^6\right):3^4\)

\(=7.3^5:3^4-3^4:3^4+3^6:3^4\)

\(=7.3^{5-4}-3^{4-4}+3^{6-4}\)

\(=7.3^1-3^0+3^2\)

\(=7.3-1+9\)

\(=21-1+9\)

\(=20+9\)

\(=29\)

b) \(\left(16^3-64^2\right):8^3\)

\(=\left[\left(2^4\right)^3-\left(2^6\right)^2\right]:\left(2^3\right)^3\)

\(=\left(2^{4.3}-2^{6.3}\right):2^{3.3}\)

\(=\left(2^{12}-2^{12}\right):2^9\)

\(=2^{12-9}-2^{12-9}\)

\(=2^3-2^3\)

\(=8-8\)

\(=0\)

Giải:

1) \(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)

\(=\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)

\(=\dfrac{-1}{12}-\dfrac{55}{24}\)

\(=\dfrac{-19}{8}\)

2) \(-1,75-\left(\dfrac{-1}{9}-2\dfrac{1}{18}\right)\)

\(=-\dfrac{7}{4}+\dfrac{1}{9}+2\dfrac{1}{18}\)

\(=-\dfrac{7}{4}+\dfrac{1}{9}+\dfrac{37}{18}\)

\(=\dfrac{5}{12}\)

3) \(-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{10}\right)\)

\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)

\(=-\dfrac{67}{120}\)

4) \(\dfrac{2}{5}+\left(-\dfrac{4}{3}\right)+\left(-\dfrac{1}{2}\right)\)

\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{1}{2}\)

\(=-\dfrac{43}{30}\)

5) \(\dfrac{3}{12}-\left(\dfrac{6}{15}-\dfrac{3}{10}\right)\)

\(=\dfrac{3}{12}-\dfrac{6}{15}+\dfrac{3}{10}\)

\(=\dfrac{3}{20}\)

6) \(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\)

\(=8\dfrac{5}{11}+3\dfrac{5}{8}-3\dfrac{5}{11}\)

\(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}\)

\(=8+\dfrac{5}{8}\)

\(=\dfrac{69}{8}\)

7) \(-\dfrac{1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)

\(=-\dfrac{1}{4}.13\dfrac{9}{11}-\dfrac{1}{4}.6\dfrac{2}{11}\)

\(=-\dfrac{1}{4}\left(13\dfrac{9}{11}+6\dfrac{2}{11}\right)\)

\(=-\dfrac{1}{4}\left(13+\dfrac{9}{11}+6+\dfrac{2}{11}\right)\)

\(=-\dfrac{1}{4}\left(13+6+1\right)\)

\(=-\dfrac{1}{4}.20=-5\)

8) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)

\(=\dfrac{4}{9}\left(-7\right)+6\dfrac{5}{9}\left(-7\right)\)

\(=-7\left(\dfrac{4}{9}+6\dfrac{5}{9}\right)\)

\(=-7\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)\)

\(=-7\left(6+1\right)\)

\(=-7.7=-49\)

Vậy ...

P/s : Phá ngoặc ra là ok : 

a ) 

\(\left[4x-2\left(x-3\right)\right].\left(-3x\right)\)

\(=\left[4x-2x+6\right]\left(-3x\right)\)

\(=-12x^2+6x^2-18x\)

b ) 

\(3\left[x-3\left(4-2x\right)+8\right]\)

\(=3\left[x-12+6x+8\right]\)

\(=3\left[7x-4\right]\)

\(=21x-12\)

c ) 

\(5\left(3x^2-4y^3\right)+9\left(2x^2-y^3\right)\)

\(=15x^2-20y^3+18x^2-9y^3\)

\(=33x^2-29y^3\)

d ) 

\(3x^2\left(2y-1\right)-2x^2\left(5y-3\right)\)

\(=6x^2y-3x^2-10x^2y+6x^2\)

\(=-4x^2y+3x^2\)

\(\frac{-1}{2}+\frac{1}{3}+\frac{2}{4}=\frac{-6}{12}+\frac{4}{12}+\frac{6}{12}\)

= \(\frac{4}{12}\)

a)(7.3^5-3^4+3^6):3^4 (7.(3^5=243)-(3^4=81)+(3^6=729)):(3^4=81)=29

b)(16^3-64^2):8^2 ((16^3=4096)-(64^2=4096)):(8^2=64)=0

c)(3x^2y^2+6y^2):3y lấy 3x^2y^2:3y=x^2y rồi lấy 6y^2:3y=2y cộng 2 kết quả lạ

a)(7.243-81+729):81=29

b)(4096-4096):64=0

c)(9x^2.3x^2y+36y^2):3y=3x^2y+12y

Cái này mình chịu nha

1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)

=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)

=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)

2, \(\frac{1}{1-x}-\frac{2x}{1-x^2}\)=\(\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{2x}{\left(1-x\right)\left(1+x\right)}\)=\(\frac{1+x+2x}{\left(1-x\right)\left(1+x\right)}=\frac{3x+1}{\left(1-x\right)\left(1+x\right)}\)

a) \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}=\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x\left(x-3\right)}\)

\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x.x}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)

\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)

b) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)

\(=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{1\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(\frac{3x+2-12x+2+10x-8}{\left(3x-2\right)\left(3x+2\right)}=\frac{x-4}{\left(3x-2\right)\left(3+2\right)}\)

c) \(\frac{4a^2-3a+5}{a^3-1}-\frac{1-2a}{a^2+a+1}-\frac{6}{a-1}\)

\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{2a-1}{a^2+a+1}-\frac{6}{a-1}\)

\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\frac{6\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{4a^2-3a+5+2a^2-2a-a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{-12}{\left(a-1\right)\left(a^2+a+1\right)}\)

d) \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)

\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x^2-6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{\left(x-3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x-3y}{x\left(x+3y\right)}\)

e) \(\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)

\(=\frac{3x-2}{\left(x-1\right)^2}-\frac{6}{\left(x-1\right)\left(x+1\right)}-\frac{3x-2}{\left(x+1\right)^2}\)

\(=\frac{\left(3x+2\right)\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\frac{6\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)\left(x+1\right)}-\frac{\left(3x-2\right)\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}\)

\(=\frac{3x^3+6x^2+3x+2x^2+4x+2-6x^2+6-3x^3+6x^2-3x+2x^2-4x+2}{\left(x-1\right)^2\left(x+1\right)^2}\)

\(=\frac{8x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\) 

f) \(\frac{5}{a+1}-\frac{10}{a-\left(a^2+1\right)}-\frac{15}{a^3+1}=\frac{5a^2}{a^3+1}+\frac{10}{a^3+1}-\frac{15}{a^3+1}\)

\(=\frac{5a^2+10-15}{a^3+1}=\frac{5a^2-5}{a^3+1}\)

a) Đk: x \(\ne\)-2

Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)

<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

<=> 2x² - 4x + 8 - 2x² - 16 = 5x + 10

<=> -4x - 8 = 5x + 10

<=> -4x - 5x = 10 + 8

<=> -9x = 18

<=> x = -2 (ktm)

=> pt vô nghiệm

b) Đk: x \(\ne\)2; x \(\ne\)-3

Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)

<=> \(\frac{x+3}{\left(x-2\right)\left(x+3\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{5}{\left(x-2\right)\left(x+3\right)}\)

<=> x + 3 - 6x + 12 = -5

<=> -5x = -5 - 15

<=> -5x = -20

<=> x = 4 

vậy S = {4}

c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11

Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)

<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)

<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)

<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)

<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)

Vậy S = {0}

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)