\(\dfrac{3}{2x-1}=\dfrac{12}{20}\)

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

\(\Rightarrow2x-1=5\)

\(\Rightarrow2x=5+1\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=6:2=3\)

Bài 1

1.\(x\left(x+3\right)\)

\(=x^2+3x\)

2.\(3x\left(x+2\right)\)

\(=3x^2+6x\)

3,\(x^2\left(3x-1\right)\)

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

4.\(-5x^3\left(3x^2-7\right)\)

\(=-15x^5+35x^3\)

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

\(=15x^3-6x^2-3x\)

6.\(-x^2\left(5x^3-x-\dfrac{1}{2}\right)\)

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

7.\(\left(x^2+2x-3\right).\left(-x\right)\)

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

8.\(4x^3\left(-2x^2+4x^4-3\right)\)

\(=-8x^5+16x^7-12x^3\)

9.\(-5x^2\left(3x^2-2x+1\right)\)

\(=-15x^4+10x^3-5x^2\)

10.\(-4x^5\left(x^3-4x^2+7x-3\right)\)

\(=-4x^8+16x^7-28x^6+12x^5\)

11.\(\left(x+2\right)\left(x+3\right)\)

\(=x^2+3x+2x+6\)

12.\(\left(x-7\right)\left(x-5\right)\)

\(=x^2-5x-7x+35\)

13.\(\left(3x+5\right)\left(2x-7\right)\)

\(=6x^2-21x+10x-35\)

14.\(\left(x-3\right)\left(x^2-2x-1\right)\)

\(x^3-2x^2-x-3x^2+6x+3\)

15.\(\left(2x-1\right)\left(x^2-5x+3\right)\)

\(=2x^3-10x^2+6x-x^2+5x-3\)

16.\(\left(x-5\right)\left(-x^2+x-1\right)\)

\(=-x^3+x^2-x+5x^2-5x+5\)

17,\(\left(\dfrac{1}{2}x+3\right)\left(2x^2-4x-6\right)\)

\(=x^3-2x^2-3x+6x^2-12x-18\)

P/s:mình làm hơi tắt tại bài dài quá:))

Bài 2

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

\(=5x^2-3x^2+2\)

\(=2x^2+2\)

2.\(-2x^2+3\left(x^2+2\right)\)

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

\(=x^2+6\)

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

\(=-4x^2+2x-4x^2+20x\)

\(=22x\)

4.\(3x\left(x-5\right)-5x\left(x+7\right)\)

\(=3x^2-15x-5x^2-35x\)

\(=-2x^2-50x\)

5.\(-3x\left(3x-4\right)+2x\left(3x+1\right)\)

\(=-9x^2+12x+6x^2+2x\)

\(=-3x^2+14x\)

6.\(x\left(x^2+x+1\right)-x\left(2x^2+1\right)\)

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

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

7.\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

\(=2x^3-3x-5x^3-x^2+x^2\)

\(=-3x^3-3x\)

8.\(4x\left(x^2-x+1\right)-x\left(3x-2x-5\right)\)

\(=4x^3-4x^2+4x-3x^2+2x^2+5x\)

\(=4x^3-5x^2+9x\)

9.\(4x\left(x^2-x+1\right)+\left(x-1\right)\left(x^2-x\right)\)

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

\(=5x^3-6x^2+5x\)

10.\(\left(2x-3\right)\left(x+4\right)+\left(-x+1\right)\left(x-2\right)\)

\(=2x^2+8x-3x-12+-x^2+2x+x-2\)

\(=x^2+8x-14\)

11.\(\left(-x+5\right)\left(x+3\right)+\left(2x-1\right)\left(x+3\right)\)

\(=-x^2-3x+5x+15+2x^2+6x-x-3\)

\(=x^2+7x+12\)

12.\(\left(x+3\right)\left(x-1\right)-4x\left(x^2-2x\right)\)

\(=x^2-x+3x-3-4x^3+8x^2\)

\(=9x^2+2x-3-4x^3\)

Bài nhiều câu như thế này lần sau bạn lưu ý tách ra các post riêng nhé.

đề kiểu này thì tự làm đi bn

=5*[55+13+32]

=5*100                     [* là dấu nhân]

=500

5x55+13x5+32

=5x(55+13+32)

=5x100

=500

5 x 55 + 13 x 5 + 5 x 32 = 5 x ( 55 + 13 + 32 )

                                       = 5 x 100

                                       = 500

Chúc học tốt!

\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{24\cdot12}{24+12}=8\Omega\)

\(I=\dfrac{U}{R}=\dfrac{12}{8}=1,5A\)

\(P=\dfrac{U^2}{R}=\dfrac{12^2}{8}=18W\)

\(Q_{tỏa1}=A_1=U_1\cdot I_1\cdot t=12\cdot\dfrac{12}{24}\cdot1\cdot3600=21600J\)

\(Q_{tỏa2}=A_2=U_2\cdot I_2\cdot t=12\cdot\dfrac{12}{12}\cdot1\cdot3600=43200J\)

     2\(\sqrt{\dfrac{16}{3}}\)  - 3\(\sqrt{\dfrac{1}{27}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{8}{\sqrt{3}}\) - \(\dfrac{3}{3\sqrt{3}}\)  - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{8}{\sqrt{3}}\) - \(\dfrac{1}{\sqrt{3}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{16}{2\sqrt{3}}\) - \(\dfrac{2}{2\sqrt{3}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{11}{2\sqrt{3}}\)

= \(\dfrac{11\sqrt{3}}{6}\)

f, 2\(\sqrt{\dfrac{1}{2}}\)- \(\dfrac{2}{\sqrt{2}}\) + \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{2}{\sqrt{2}}\) - \(\dfrac{2}{\sqrt{2}}\) + \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{5\sqrt{2}}{4}\)

(1 + \(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)).(1- \(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\)) 

= \(\dfrac{\sqrt{3}-1+3-\sqrt{3}}{\sqrt{3}-1}\).\(\dfrac{\sqrt{3}+1-3+\sqrt{3}}{\sqrt{3}+1}\)

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

= \(\dfrac{-4}{3-1}\)

= \(\dfrac{-4}{2}\)

= -2

   \(\dfrac{2}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}+\dfrac{5}{\sqrt{6}}\)

= \(\dfrac{2.\left(\sqrt{6}+2\right)+2\left(\sqrt{6}-4\right)}{\left(\sqrt{6}-2\right)}\) + \(\dfrac{5}{\sqrt{6}}\)

= \(\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{6-4}\) + \(\dfrac{5\sqrt{6}}{6}\)

= \(\dfrac{4\sqrt{6}}{2}\) + \(\dfrac{5\sqrt{6}}{6}\)

= \(\dfrac{12\sqrt{6}+5\sqrt{6}}{6}\)

= \(\dfrac{17\sqrt{6}}{6}\)

CM đẳng thức hay tìm x,y vậy 

Mình sẽ làm theo đề bài của mình nếu đúng thì ... nha 

Biến đổi vế phải  ta có :

( x + y) [ ( x - y)^2 + xy ] = ( x + y)( x^2 - 2xy + y^2 + xy)

                                      = ( x+  y)( x^2 - xy+ y^2)

                                       = x^3 + y^3

VẬy VT  = VP đẳng thức được CM 

hảo thật đấy

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