a, Khai triển: \(\left(x+2y\right)\times\left(x^2-2xy+4y^2\right)\)
b, So sánh:
1. \(2014\times2016vs2015^2\)
2. \(2012\times2016vs2014^2\)
3. \(2011\times2019vs2015^2\)
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5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
Sửa đề: \(x\left(x-3\right)+2y\left(2y-3\right)+4xy+19\)
a: \(x\left(x-3\right)+2y\left(2y-3\right)+4xy+19\)
\(=x^2-3x+4y^2-6y+4xy+19\)
\(=\left(x^2+4xy+4y^2\right)-3\left(x+2y\right)+19\)
\(=\left(x+2y\right)^2-3\left(x+2y\right)+19\)
\(=\left(-5\right)^2-3\cdot\left(-5\right)+19\)
=25+15+19=59
b: \(=x^3+x^2+8y^3+4y^2+2xy\left[3\left(x+2y\right)+2\right]+70\)
\(=x^3+8y^3+x^2+4y^2+2xy\cdot\left[3\cdot\left(-5\right)+2\right]+70\)
\(=\left(x+2y\right)^3-3\cdot x\cdot2y\left(x+2y\right)+\left(x+2y\right)^2-4xy+2xy\cdot\left(-13\right)+70\)
\(=\left(-5\right)^3+\left(-5\right)^2-6xy\cdot\left(-5\right)-4xy-26xy\)+70
\(=-125+25+70=-30\)
\(a,\left(2x+y+3\right)^2=4x^2+y^2+9+4xy+12x+6y\)
\(b,\left(x-2y+1\right)^2=x^2+4y^2+1-4xy+2x-4y\)
\(c,\left(x^2-2xy^2-3\right)^2=x^4+2x^2y^4+9-4x^3y^2-6x^2+12xy^2\)
Giải:
a) \(\left(2x+y+3\right)^2\)
\(=\left(2x+y\right)^2+2.3\left(2x+y\right)+3^2\)
\(=\left(2x\right)^2+2.2x.y+y^2+2.3\left(2x+y\right)+3^2\)
\(=4x^2+4xy+y^2+12x+6y+9\)
Vậy ...
b) \(\left(x-2y+1\right)^2\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1^2\)
\(=x^2-2.x.2y+\left(2y\right)^2+2x-4y+1^2\)
\(=x^2-4xy+4y^2+2x-4y+1\)
Vậy ...
c) \(\left(x^2-2xy^2-3\right)^2\)
\(=\left(x^2-2xy^2\right)^2+2.3.\left(x^2-2xy^2\right)-3^2\)
\(=\left(x^2\right)^2-2.x^2.2xy^2+\left(2xy^2\right)^2+2.3.\left(x^2-2xy^2\right)-3^2\)
\(=x^4-4x^3y^2+4x^2y^4+6x^2-12xy^2-9\)
Vậy ...
\(a.\left(2xy-3\right)^2=4x^2y^2-12xy+9\)
\(b.\left(\dfrac{1}{2}x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}x+\dfrac{1}{9}\)
\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)
\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)
\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)
\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
a. (x + 2y) . (x 2 - 2xy + 4y2)
= x3 + (2y)3
= x3 + 8y3
(Áp dụng HĐT x3 + y3)
b. 1. 2014.2016
= (2015 - 1).(2015 + 1)
= 20152 - 1 < 20152
Vậy 2014.2016 < 20152.
2. 2012.2016
= (2014 - 2).(2014 + 2)
= 20142 - 22
= 20142 - 4 < 20142
Vậy...
3. 2011.2019
= (2015 - 4).(2015 + 4)
= 20152 - 42
= 20152 - 16 < 20152
Vậy...
a) (x+2y)(x2-2xy+4y2)=x3+8y3
b)
1) Ta có: 2014.2016=(2015-1)(2015+1)=20152-1 <20152
Vậy 2014.2016<20152
câu 2 và 3 bạn làm tương tự câu 1 nhé