tính ( rút gọn)
(x-10)^2 - 2(x-10)(5x-3)+(5x-3)^2
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24 tháng 6 2021
a) \(2x\left(5-3x^2\right)-10\left(6+x\right)\)
\(=10x-6x^3-60-10x\)
\(=\) \(-6x^3-60\)
PT
24 tháng 6 2021
a) \(2x\left(5-3x^2\right)-10\left(6+x\right)\\ =2x.5-2x.3x^2-10.6-10.x\\ =10x-6x^3-60-10x\)
b) \(3\left(-x+2\right)-6\left(1-x+5x^{20}\right)\\ =-3.x+3.2-6.1+6.x-5.5x^{20}\\ =-3x+6-6+6x-25x^{20}=25x^{20}+3x\)
c) \(7x\left(2-5x^2+\dfrac{1}{2}x^3\right)-14x\left(1-2x^2\right)\\ =7x.2-7x.5x^2+7x.\dfrac{1}{2}x^3-14x.1+14x.2x^2\\ =14x-25x^3+\dfrac{7}{2}x^4-14x+28x^3=3x^2+\dfrac{7}{2}x^4\)
MT
0
H9
HT.Phong (9A5)
CTVHS
8 tháng 1
\(\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x^2-5x+6}-\dfrac{2x-4}{x-2}\left(ĐK:x\ne3;x\ne2\right)\)
\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x\left(x-2\right)-3\left(x-2\right)}-\dfrac{2x-4}{x-2}\)
\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{2x-4}{x-2}\)
\(=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x-2+3x^2-8x+10-\left(2x^2-6x-4x+12\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{3x^2-7x+8-2x^2+10x-12}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+3x-4}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+3x-4}{x^2-5x+6}\)
5 tháng 10 2021
\(A=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{5x\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\left(x>0;x\ne4\right)\\ A=\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{5x\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\\ A=\dfrac{10x\left(\sqrt{x}-2\right)}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}\)
5 tháng 10 2021
\(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}-4}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
22 tháng 6 2018
a. \(2x^2+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)\)
\(=2x^2+3\left(x^2-1\right)-5x^2-5x\)
\(=2x^2+3x^2-3-5x^2-5x\)
\(=\left(2x^2+3x^2-5x^2\right)-5x-3\)
\(=-5x-3\)
b,c mk ms học lớp 7
TT
0
LK
0
\(\left(x-10\right)^2-2\left(x-10\right)\left(5x-3\right)+\left(5x-3\right)^2\)
\(=\left(x-10-5x+3\right)^2=\left(-4x-7\right)^2=\left(4x+7\right)^2\)
\(=16x^2+56x+49\)
Ta có: \(\left(x-10\right)^2-2\left(x-10\right)\left(5x-3\right)+\left(5x-3\right)^2\)
\(=\left(x-10-5x+3\right)^2\)
\(=\left(-4x+6\right)^2\)
\(=16x^2-48x+36\)