Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tham khảo:
a) \((4{x^2} - 5):(x - 2) = \dfrac{{4{x^2} - 5}}{{x - 2}} = 4x + 8 + \dfrac{{11}}{{x - 2}}\)
Vậy \( (4{x^2} - 5):(x - 2)= 4x + 8 + \dfrac{{11}}{{x - 2}}\)
b) \((3{x^3} - 7x + 2):(2{x^2} - 3) = \dfrac{{3{x^3} - 7x + 2}}{{2{x^2} - 3}}\)
Vậy \( (3{x^3} - 7x + 2):(2{x^2} - 3)= \dfrac{3}{2}x + \dfrac{{\dfrac{-5}{2}x + 2}}{{2{x^2} - 3}}\)
a) 4x²(x² - 5x + 2)
= 4x².x² - 4x².5x + 4x².2
= 4x⁴ - 20x³ + 8x²
b) (2x² - 5x + 3) : (2x - 3)
= (2x² - 3x - 2x + 3) : (2x - 3)
= [(2x² - 3x) - (2x - 3)] : (2x - 3)
= [x(2x - 3) - (2x - 3)] : (2x - 3)
= (2x - 3)(x - 1) : (2x - 3)
= x - 1
a) \(\begin{array}{l}(4x - 3)(x + 2) = 4x(x + 2) - 3(x + 2)\\ = 4{x^2} + 8x - 3x - 6\end{array}\)
\( = 4{x^2} + 5x - 6\)
b) \((5x + 2)( - {x^2} + 3x + 1)\)
\( = 5x( - {x^2} + 3x + 1) + 2( - {x^2} + 3x + 1)\)
\( = - 5{x^3} + 15{x^2} + 5x - 2{x^2} + 6x + 2\)
\( = - 5{x^3} + 13{x^2} + 11x + 2\)
c) \((2{x^2} - 7x + 4)( - 3{x^2} + 6x + 5)\)
\( = 2{x^2}( - 3{x^2} + 6x + 5) - 7x( - 3{x^2} + 6x + 5) + 4( - 3{x^2} + 6x + 5)\)
\( = 2{x^2}( - 3{x^2}) + 2{x^2}.6x + 2{x^2}.5 + 7x.3{x^2} - 7x.6x - 7x.5 + 4( - 3{x^2}) + 4.6x + 4.5\)
\(= - 6{x^4} + 33{x^3} - 44{x^2} - 11x + 20\)
\(\dfrac{2}{67}-\left(\dfrac{3}{7}+\dfrac{2}{67}\right)\\ =\dfrac{2}{67}-\dfrac{215}{469}\\ =\dfrac{-3}{7}\)
\(a,2010:\left(-5\right)+400-1\\ =-402+400-1\\ =-3\\ b,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\\ =\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c,\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\\ =\dfrac{1}{12}.\left(\dfrac{1}{20}\right)^2\\ =\dfrac{1}{12}.\dfrac{1}{400}\\ =\dfrac{1}{4800}\)
a) \(2010:\left(-5\right)+400-1=-400+400-1=-1\)
b) \(\dfrac{2}{3}+\dfrac{3}{4}\cdot\dfrac{-4}{9}=\dfrac{2}{3}+\dfrac{-1}{3}=\dfrac{1}{3}\)
c) \(\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{1}{12}\cdot\dfrac{1}{400}=\dfrac{1}{4800}\)
\(a)(2{y^4} - 13{y^3} + 15{y^2} + 11y - 3):({y^2} - 4y - 3)=2y^2-5y+1\)
b) \((5{x^3} - 3{x^2} + 10):({x^2} + 1)=5x-3+\dfrac{-5x+13}{x^2+1}\)
A = 6 -2/3 + 1 /2 - 5 -5/3 +3/2 -3 + 7/3 - 5/2
= (6 - 5 - 3) - ( 2/3 -5/3 + 7/3 ) + ( 1/2 +3/2 - 5/2)
= -2 + 0 -1/2 = -5/2
a)
\(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{18}{24}-\dfrac{20}{24}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-12\times\dfrac{144}{1}\)
\(=-1728\)
b)
\(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\left(4\times\dfrac{3}{4}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\left(3-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\times\dfrac{6}{5}-17\)
\(=3-17\)
\(=-14\)
a)\(=-12:\left(-\dfrac{1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)\(=-12.144=-1728\)
b)\(=\left(8:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\left(6-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\dfrac{11}{2}.\dfrac{6}{5}-17=\dfrac{33}{5}-17=\dfrac{33}{5}-\dfrac{85}{5}=-\dfrac{2}{5}\)
\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\dfrac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(=\dfrac{2^{12}.3^4.2}{2^{12}.3^5.4}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)
\(=\dfrac{1}{6}-\dfrac{-10}{3}\)
\(=\dfrac{7}{2}\)