rút gọn biểu thức sau:
2√3-3√2/√6-v2-2/1-√2+3-√3/1+√3
rút gọn các biểu thức sau:
a)rút gọn biểu thức:N=3-3^2+3^3-3^4+3^5-2^6+...+3^2011-3^2012
b)rút gọn biểu thức:e=2^100-2^99-2^98-2^97-...-2^3-2^2-2-1
GIÚP MÌNH NHA
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😈😈😈
Rút gọn biểu thức sau: \(\sqrt{9-3\sqrt{8}}\) - \(\dfrac{\sqrt{3}-1}{\sqrt{2}}\) + \(\sqrt{5-2\sqrt{6}}\) - \(\sqrt{2-\sqrt{3}}\)
\(\sqrt{9-3\sqrt{8}}-\dfrac{\sqrt{3}-1}{\sqrt{2}}+\sqrt{5-2\sqrt{6}}-\sqrt{2-\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2.\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\dfrac{\sqrt{6}-\sqrt{2}}{2}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
\(=\sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}-\sqrt{6}+\sqrt{2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{6}-\sqrt{3}\right|-\sqrt{6}+\sqrt{2}+\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\sqrt{6}-\sqrt{3}-\sqrt{6}+\sqrt{2}+\sqrt{3}-\sqrt{2}\) (do \(\sqrt{6}-\sqrt{3}>0;\sqrt{3}-\sqrt{2}>0\))
\(=0\)
\(=\sqrt{9-6\sqrt{2}}-\dfrac{\sqrt{6}-\sqrt{2}}{2}+\sqrt{3}-\sqrt{2}-\dfrac{1}{\sqrt{2}}\left(\sqrt{3}-1\right)\)
\(=\sqrt{6}-\sqrt{3}-\dfrac{1}{2}\sqrt{6}+\dfrac{1}{2}\sqrt{2}+\sqrt{3}-\sqrt{2}-\dfrac{1}{2}\sqrt{6}+\dfrac{1}{2}\sqrt{2}\)
\(=0\)
Rút gọn biểu thức
(2√3-3√2)/√6-(3-√2)/(1-√2)+3/√3
Rút gọn biểu thức sau: (x-2)^3+6(x-1)^2-(x+1)(x^2-x+1)
rút gọn các biểu thức sau
\(\dfrac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}\)+\(\dfrac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\)
\(\dfrac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}\)+\(\dfrac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)
a: \(=\dfrac{6+4\sqrt{2}}{\sqrt{2}+2+\sqrt{2}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}-2+\sqrt{2}}\)
\(=\dfrac{6+4\sqrt{2}}{2+2\sqrt{2}}+\dfrac{6-4\sqrt{2}}{2\sqrt{2}-2}\)
\(=\dfrac{3+2\sqrt{2}}{\sqrt{2}+1}+\dfrac{3-2\sqrt{2}}{\sqrt{2}-1}\)
=căn 2+1+căn 2-1=2căn 2
b: \(=\dfrac{\sqrt{3}+\sqrt{3+\sqrt{3}}+\sqrt{3}-\sqrt{3+\sqrt{3}}}{1-\sqrt{3}-1}=\dfrac{-2\sqrt{3}}{\sqrt{3}}=-2\)
a) \(A=1+3+3^2+...+3^{100}\)
\(3A=3+3^2+3^3+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...-2^3+2^2-2+1\)
\(2B=2^{101}-2^{100}+2^{99}-2^{98}+...-2^4+2^3-2^2+2\)
\(B+2B=\left(2^{100}-2^{99}+...-2+1\right)+\left(2^{101}-2^{100}+...-2^2+2\right)\)
\(3B=2^{101}+1\)
\(B=\frac{2^{101}+1}{3}\)
Rút gọn biểu thức sau :
A = (3x-x^2/9-x^2 - 1) : (9-x^2/x^2+x-6 + x-3/2-x - x+2/x+3)
\(A=\left(\dfrac{3x-x^2}{9-x^2}-1\right):\left(\dfrac{9-x^2}{x^2+x-6}+\dfrac{x-3}{2-x}-\dfrac{x+2}{x+3}\right)\left(dk:x\ne\pm3,x\ne2\right)\)
\(=\dfrac{3x-x^2-9+x^2}{9-x^2}:\left(\dfrac{9-x^2}{\left(x-2\right)\left(x+3\right)}-\dfrac{x-3}{x-2}-\dfrac{x+2}{x+3}\right)\)
\(=\dfrac{3x-9}{9-x^2}:\dfrac{9-x^2-\left(x-3\right)\left(x+3\right)-\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=-\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2-\left(x^2-9\right)-\left(x^2-4\right)}\)
\(=-\dfrac{3}{x+3}.\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2-x^2+9-x^2+4}\)
\(=\dfrac{-3\left(x-2\right)}{22-3x^2}\)
\(=\dfrac{-3x+6}{22-3x^2}\)
Vậy \(A=\dfrac{-3x+6}{22-3x^2}\) với \(x\ne\pm3,x\ne2\)
Kết quả sau khi rút gọn biểu thức A = (x – y – 1)3 – (x – y + 1)3 + 6(x – y)2 là: …
Ta có: \(A=\left(x-y-1\right)^3-\left(x-y+1\right)^3+6\left(x-y\right)^2\)
\(=\left(x-y-1-x+y-1\right)\left[\left(x-y-1\right)^2+\left(x-y-1\right)\left(x-y+1\right)+\left(x-y+1\right)^2\right]+6\left(x-y\right)^2\)
\(=-2\cdot\left[3\left(x-y\right)^2+1\right]+6\left(x-y\right)^2\)
\(=-6\left(x-y\right)^2+6\left(x-y\right)^2-2\)
=-2
Câu 1: Rút gọn biểu thức: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{2}}+\dfrac{6}{x+3\sqrt{x}}\right)\) với x > 0
Câu 2: Rút gọn biểu thức:
\(P=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\) với x > 0; x \(\ne\) 2
Câu 3: Rút gọn biểu thức:
\(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\) với a > 0; a \(\ne\) 4
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
rút gọn các biểu thức sau: (giả thiết các biểu thức chữ đều có nghĩa)
a) \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)
b) \(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\)
c) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}\)
\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)