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a,\(\dfrac{x^2-9}{2x+6}:\dfrac{3-x}{2}=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}.\dfrac{2}{3-x}=\dfrac{x-3}{3-x}=\dfrac{-\left(3-x\right)}{3-x}=-1\)
b, \(\dfrac{2x}{x-y}-\dfrac{2y}{x-y}=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{x-y}=2\)
\(a,=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\dfrac{2}{-\left(x-3\right)}=\dfrac{x-3}{2}\cdot\dfrac{2}{-\left(x-3\right)}=-1\\ b,=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{\left(x-y\right)}=2\)
a: \(\dfrac{x^2}{3x+6}+\dfrac{4x+4}{3x+6}=\dfrac{x^2+4x+4}{3x+6}=\dfrac{x+2}{3}\)
b: \(\dfrac{x+3}{x}+\dfrac{x}{3-x}-\dfrac{9}{3x-x^2}\)
\(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}\)
=0
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
\(=\dfrac{3}{2\left(x+3\right)}+\dfrac{6-x}{2x\left(x+3\right)}=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+6\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{1}{x}\)
\(=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)
\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=4x\left(x^2-9\right)-x^3+27\)
\(=4x^3-36x-x^3+27\)
\(=3x^3-36x+27\)
\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)
\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)
\(=\left(x+6\right).0\)
\(=0\)
2^16x(-9)^3/(-6)^5x16=98304
nếu ko đúng thì cho mình xin lỗi nhé!
\(a.\dfrac{3^{27}}{9^6.3^{16}}=\dfrac{3^{27}}{3^{12}.3^{16}}=\dfrac{3^{27}}{3^{28}}=\dfrac{1}{3}\)
\(\left(x-\dfrac{5}{2}\right)^2=\dfrac{9}{4}\\ \Rightarrow x-\dfrac{5}{2}=\pm\dfrac{3}{2}\)
\(TH1:x-\dfrac{5}{2}=\dfrac{3}{2}\Rightarrow x=\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{8}{2}=4\)
\(TH2:x-\dfrac{5}{2}=-\dfrac{3}{2}\Rightarrow x=-\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{2}{2}=1\)
a: \(=\dfrac{3^{27}}{3^{12}\cdot3^{16}}=\dfrac{1}{3}\)
a) Rút gọn thu được kết quả: 3;
b) Ta có MC = 3x (x - 3)
Thực hiện tính toán thu được kết quả: x 2 − 6 x + 9 3 x ( x − 3 ) = x − 3 3 x
c) Trước tiên biến đổi: 3 + 3 x = 3 ( x + 1 ) x ; 3 3 ( x + 1 ) x = x x + 1
Thay vào A và thu gọn ta được A = 4 x + 3 x