Viết dưới dạng tích.
1)\(\sqrt{x^3}-\sqrt{x}\)
2)\(\sqrt{a^3}+a\)
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HT
1
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TL
1
31 tháng 10 2016
Bài 1:
a,Viết số 650 dưới dạng tích 2 số tự nhiên liên tiếp
b,Viết số 35904 dưới dạng tích 3 số tự nhiên liên tiếp
giải
a) 650 = 25 x 26
b) 35904 = 32 x 33 x 34
TL
1
1 tháng 8 2015
b, 8x^3 - y^3 = (2x)^3 - y^3 = ( 2x - y)(4x^2 + 2xy + y^2)
c, ( x + 2 )( x^2 - 2x + 4 ) = x^3 + 2^3 = x^3 + 8
Tick đúng nha
9
2
PT
23 tháng 9 2018
\(A=9^{25}.27^4.81^3=\left(3^2\right)^{25}.\left(3^3\right)^4.\left(3^4\right)^3=3^{50}.3^{12}.3^{12}=3^{74}\)
k mk nhé
CC
11 tháng 10 2019
a,\(a\sqrt{b}-b\sqrt{a}\)= \(\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\)
b,\(x\sqrt{x}+\sqrt{x}-x-1\)
=\(\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)-\(\sqrt{x}\left(\sqrt{x}-1\right)\)
=\(\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1-\sqrt{x}\right)\)
=\(\left(\sqrt{x}-1\right)\left(x+1\right)\)
c,\(\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)=\(\sqrt{a}\left(\sqrt{b}+2\right)\)+\(3\left(\sqrt{b}+2\right)\)
=\(\left(\sqrt{b}+2\right)\left(3+\sqrt{a}\right)\)
d,\(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)
=\(\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)\)
=\(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}+\sqrt{b}\right)\)
28 tháng 8 2021
a: Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Ta có: \(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}-1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}+1+1\)
\(=a-\sqrt{a}+2\)
28 tháng 8 2021
a,ĐKXĐ: tự tìm :v
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(x+2\sqrt{x}+1\right)-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+1\right)^2-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6+2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9\sqrt{x}-x-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(9\sqrt{x}-9\right)-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(10-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\dfrac{10-\sqrt{x}}{\sqrt{x}+3}\)
\(\sqrt{x^3}-\sqrt{x}=\sqrt{x}\left(\sqrt{x}^2-1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)đk : \(x>0\)
\(\sqrt{a^3}+a=a\left(\sqrt{a}+1\right)\)đk: \(a>0\)
có sai xót mong mn bỏ qa cho