a: \(=\sqrt{3}-1-\sqrt{3}=-1\)

b: \(=2\sqrt{3}-10\sqrt{3}+4\sqrt{3}=-4\sqrt{3}\)

a) ( x - 3 ) ( x + 7 ) - ( x + 5 ) ( x - 1 )

= ( x² +7x - 3x - 21 ) - ( x² -x + 5x - 5 )

= x² + 4x -21 - x² - 4x +5

= -16

a: Ta có: \(\left(x-3\right)\left(x+7\right)-\left(x+5\right)\left(x-1\right)\)

\(=x^2+4x-21-x^2-4x+5\)

=-16

a: \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)

\(=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)

\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)

\(=x^3-16x^2+25x\)

a: \(\left(2x+3y\right)\left(x-2y\right)-\dfrac{\left(4x^3y-6x^2y^2-3xy^3\right)}{2xy}\)

\(=2x^2-4xy+3xy-6y^2-\dfrac{2xy\cdot\left(2x^2-3xy-1,5y^2\right)}{2xy}\)
\(=2x^2-xy-6y^2-2x^2+3xy+1,5y^2\)

\(=2xy-4,5y^2\)

b: \(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)-\left(3x-1\right)\left(3x-2\right)\)

\(=x^3-6x^2+12x-8-x\left(x^2-1\right)-\left(9x^2-6x-3x+2\right)\)

\(=x^3-6x^2+12x-8-x^3+x-9x^2+9x-2\)

\(=-15x^2+22x-10\)

a) \(P=\dfrac{\sqrt{3}+\sqrt{6}}{1+\sqrt{2}}=\dfrac{\left(\sqrt{3}+\sqrt{6}\right)\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}\)

                             \(=\dfrac{\sqrt{3}-\sqrt{6}+\sqrt{6}-\sqrt{12}}{1-2}=\sqrt{12}-\sqrt{3}\)

b) \(Q=\left(\sqrt{75}-\dfrac{3}{2}:\sqrt{3}-\sqrt{48}\right)\cdot\sqrt{\dfrac{16}{3}}\)

        \(=\left(5\sqrt{3}-\dfrac{3}{2}\cdot\dfrac{1}{\sqrt{3}}-4\sqrt{3}\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\sqrt{3}\left(5-\dfrac{1}{2}-4\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\left(1-\dfrac{1}{2}\right)\cdot4=2\)

a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012

2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013

3M=2^0+2^2013

M=(2^0+2^2013)÷3

Vậy.......

b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012

3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013

4N=3-3^2013

N=(3-3^2013)÷4

Vậy........

K tao nhé ko lên lớp tao đánh m😈😈😈

Bt dễ thế mà ko làm dc😂😂😂😂😂

phần c đâu

a: \(\left(2x-1\right)^2-3\left(x-1\right)\left(x+2\right)-\left(x-3\right)^2\)

\(=4x^2-4x+1-x^2+6x-9-3\left(x^2+x-2\right)\)

\(=3x^2+2x-8-3x^2-3x+6\)

=-x+2

b: \(\left(x-2\right)\left(2x-1\right)-3\left(x+1\right)^2-4x\left(x+2\right)\)

\(=2x^2-x-4x+2-3x^2-6x-3-4x^2-8x\)

\(=-5x^2-19x-1\)

\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\\ \Rightarrow3A=3+3^2+3^3+...+3^{100}+3^{101}\\ \Rightarrow3A-A=3^{101}-1\\ \Rightarrow2A=3^{101}-1\\ \Rightarrow A=\left(3^{101}-1\right).\dfrac{1}{2}\\ \Rightarrow\dfrac{3^{101}}{2}-\dfrac{1}{2}.\)

\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)

Ta có: \(3A=3+3^2+3^3+...+3^{99}+3^{100}\)

Khi đó: \(3A-A=3+3^2+3^3+...+3^{99}+3^{100}+3^{101}-\left(1+3+3^2+3^3+...+3^{99}+3^{100}\right)\)

\(=3^{101}-1\)

\(\Leftrightarrow2A=3^{101}-1\)

Vậy \(A=\left(3^{101}-1\right):2\)

em cảm ơn mọi nguoif ạ

 b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{12-3\sqrt{7}}-\sqrt{2}\cdot\sqrt{12+3\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{21}\right)^2-2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}\right)^2+2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{21}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}+\sqrt{3}\right)^2}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}\)

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

\(=-\sqrt{6}\)  

c) \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)

\(=\sqrt[3]{\dfrac{3\cdot9}{4\cdot16}}\)

\(=\sqrt[3]{\left(\dfrac{3}{4}\right)^3}\)

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

d) \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}\)

\(=\sqrt[3]{\dfrac{54}{-2}}\)

\(=\sqrt[3]{-27}\)

\(=\sqrt[3]{\left(-3\right)^3}\)

\(=-3\) 

a: Sửa đề: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}\cdot\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{\sqrt{6}+1}{3\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{2\sqrt{2}\left(\sqrt{6}+1\right)+\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)

e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{2\sqrt{2}+3\sqrt{2}+6+1}-\sqrt[3]{2\sqrt{2}-3\sqrt{2}+6-1}\)

\(=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}\)

\(=\sqrt{2}+1-\left(\sqrt{2}-1\right)\)

\(=\sqrt{2}+1-\sqrt{2}+1=2\)