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23 tháng 7 2018
a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
AT
17 tháng 6 2017
\(a\)) \(-2,5ab\left(-2a^2+3b^2\right)=5a^3b-7,5ab^3\)
b) \(-2x^3\left(3x+0,5x^2-7x^3-2\right)\)
\(=-6x^4-1x^5+14x^6+4x^3\)
c/ \(\left(x^3-2x^2+3x-5\right)\left(-xy\right)\)
\(=-x^4y+2x^3y-3x^2y+5xy\)
d/ \(\left(-\dfrac{1}{2}x^2y\right)\left(3x^3-\dfrac{2}{7}x^2-\dfrac{4}{5}x+8\right)\)
\(=-\dfrac{3}{2}x^5y+\dfrac{1}{7}x^4y+\dfrac{2}{5}x^3y-4x^2y\)
17 tháng 6 2017
\(-2,5ab\left(-2a^2+3b^2\right)=5a^3b-7,5ab^3\)
\(-2x^3\left(3x+0,5x^2-7x^3-2\right)=-6x^4-x^5+14x^6+4x^3\)
\(\left(x^3-2x^2+3x-5\right)\left(-xy\right)=-x^4y+2x^3y-3x^2y+5xy\)
\(\left(\dfrac{1}{2}x^2y\right)\left(3x^3-\dfrac{2}{7}x^2-\dfrac{4}{5}x+8\right)\)
\(=\dfrac{-3}{2}x^5y+\dfrac{1}{7}x^4y+\dfrac{2}{5}x^3y-4x^2y\)
29 tháng 1 2019
a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
2 tháng 7 2022
a: \(A=x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)
Dấu '=' xảy ra khi x=3/2
c: \(x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\)
=>\(\dfrac{3}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}< =3:\dfrac{7}{4}=\dfrac{12}{7}\)
=>C>=-12/7
Dấu '=' xảy ra khi x=1/2
28 tháng 5 2022
Câu 1:
Sửa đề: \(\left(2x+1\right)^3+\left(x-5\right)^3+\left(-3x+4\right)^3=0\)
\(\Leftrightarrow\left(2x+1\right)^3+\left(x-5\right)^3-\left(3x-4\right)^3=0\)
Đặt a=2x+1; b=x-5
Phương trình sẽ là \(a^3+b^3-\left(a+b\right)^3=0\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-5\right)\left(3x-4\right)=0\)
hay \(x\in\left\{-\dfrac{1}{2};5;\dfrac{4}{3}\right\}\)
27 tháng 5 2022
a: \(\left(3x-1\right)^2-\left(x+3\right)^3=\left(2-x\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow9x^2-6x+1-x^3-9x^2-27x-27=8-x^3\)
\(\Leftrightarrow-x^3-33x-26-8+x^3=0\)
=>-33x=34
hay x=-34/33
b: \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow x^4-1-x^4+2x^2-1=2\)
\(\Leftrightarrow2x^2=4\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: \(x^2-2\sqrt{3}x+3=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)^2=0\)
hay \(x=\sqrt{3}\)
d: \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-x+\sqrt{2}\right)=0\)
\(\Leftrightarrow x-\sqrt{2}=0\)
hay \(x=\sqrt{2}\)
11 tháng 12 2022
Bài 2:
\(=\dfrac{-3x-1}{3\left(x-1\right)\left(x+1\right)}+\dfrac{5}{3\left(x-1\right)}+\dfrac{1}{3\left(x+1\right)}\)
\(=\dfrac{-3x-1+5x+5+x-1}{3\left(x-1\right)\left(x+1\right)}=\dfrac{3x+3}{3\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
TN
7 tháng 8 2018
a) \(\dfrac{1}{8}x^3y^3-27=\left(\dfrac{1}{2}xy\right)^3-3^3=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}x^2y^2+\dfrac{1}{6}xy+9\right)\)
b)\(\dfrac{8}{125}x^3+27y^3=\left(\dfrac{2}{5}x\right)^3+\left(3y\right)^3=\left(\dfrac{2}{5}x+3y\right)\left(\dfrac{4}{25}x^2-\dfrac{6}{5}xy+9y^2\right)\)
c) \(0.008x^6-27y^3=\left(0.2x^2\right)^3-\left(3y\right)^3=\left(0.2x^2-3y\right)\left(0.04x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d)\(\left(2x+y\right)^3-\left(x-y\right)^3=\left(2x+y-x+y\right)[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2]\\ =\left(x+2y\right)\left(4x^2+4xy+y^2+2x^2-2xy+xy-y^2+x^2-2xy+y^2\right)\\ =\left(x+2y\right)\left(6x^2+xy+y^2\right)\)
Y
7 tháng 8 2018
Bài 1:
a) \(\dfrac{1}{8}x^3y^3-27\)
\(=\left(\dfrac{1}{2}xy\right)^3-3^3\)
\(=\left(\dfrac{1}{2}xy-3\right)\left[\left(\dfrac{1}{2}xy\right)^2+\dfrac{1}{2}xy.3+3^2\right]\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}xy+\dfrac{3}{2}xy+9\right)\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{7}{4}xy+9\right)\)
b) \(\dfrac{8}{125}x^3+\dfrac{1}{8}y^3\)
\(=\left(\dfrac{2}{5}x\right)^3+\left(\dfrac{1}{2}y\right)^3\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left[\left(\dfrac{2}{5}x\right)^2-\dfrac{2}{5}x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\right]\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left(\dfrac{4}{25}x-\dfrac{1}{5}xy+\dfrac{1}{4}y\right)\)
c) \(0.008x^6-27y^3\)
\(=\left(\dfrac{1}{5}x^2\right)^3-\left(3y\right)^3\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left[\left(\dfrac{1}{5}x^2\right)^2+\dfrac{1}{5}x^2.3y+\left(3y\right)^2\right]\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left(\dfrac{1}{25}x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d) \(\left(2x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(2x+y\right)-\left(x-y\right)\right]\left[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(2x+y-x+y\right)\left(4x^2+4xy+y^2+2x^3-2xy+xy-y^2+x^2-2xy+y^2\right)\)
\(=\left(x-2y\right)\left(4x^2+2x^3+xy\right)\)
\(\left(3x^2-2y\right)^3=27x^6-54x^4y+36x^2y^2-8y^3\)
\(\left(2x+\dfrac{1}{2}\right)^3=8x^3+6x^2+\dfrac{3}{2}x+\dfrac{1}{8}\)