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Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
a: \(=xy^2+xy+x-y^3-y^2-y+\dfrac{2}{3}x^3y+\dfrac{1}{3}x^2y^3-2xy-y^3\)
\(=xy^2-xy+x-2y^3-y^2-y+\dfrac{2}{3}x^3y+\dfrac{1}{3}x^2y^3\)
b: \(=2x^3-4x^2+3x^3-3x^2-6x-15+5x^2\)
\(=5x^3-2x^2-6x-15\)
c: \(=x^2-4x+3+\left(x-4\right)\left(2x-1\right)-3x^3+2x-5\)
\(=-3x^3+x^2-2x-2+2x^2-x-8x+4\)
\(=-3x^3+3x^2-11x+2\)
a: \(=4x^2-25-4x^2+12x-9-12x=-34\)
b: \(=8y^3-12y^2+6y-1-2y\left(4y^2-12y+9\right)-12y^2+12y\)
\(=8y^3-24y^2+18y-1-8y^3+24y^2-18y=-1\)
c: \(=x^3+27-x^3-20=7\)
d: \(=3y\left(9y^2+12y+4\right)-27y^3+1-36y^2-12y-1\)
\(=27y^3+36y^2+12y-27y^3-36y^2-12y\)
=0
a: \(=2x^2-x+5\)
b: \(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)
c: \(=-x^3+\dfrac{3}{2}-2x\)
d: \(=-2x^2+4xy-6y^2\)
e: \(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
Xin câu a :3
a) (x + y + 1)2 = 3(x2 + y2) + 1
<=> x2 + y2 + 1 + 2xy + 2x + 2y = 3x2 + 3y2 + 1
<=> 2x2 + 2y2 - 2xy - 2x - 2y = 0
<=> (x2 - 2xy + y2) + (x2 - 2x + 1) + (y2 - 2y + 1) = 2
<=> (x - y)2 + (x - 1)2 + (y - 1)2 = 2
Vì 2 = 02 + 12 + 12 nên ta có các TH sau:
TH1:
\(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-1\right)^2=1\\\left(y-1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=2\\x=y=0\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}\left(x-y\right)^2=1\\\left(x-1\right)^2=0\\\left(y-1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1;y=0\\x=1;y=2\end{matrix}\right.\)
TH3:
\(\left\{{}\begin{matrix}\left(x-y\right)^2=1\\\left(x-1\right)^2=1\\\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2;y=1\\x=0;y=1\end{matrix}\right.\)
Vậy ...
a) ta có : \(\left(x+y+1\right)^2=3\left(x^2+y^2\right)+1\)
\(\Leftrightarrow x^2+y^2+1+2xy+2y+2x=3x^2+3y^2+1\)
\(\Leftrightarrow-\left(x-1\right)^2-\left(y-1\right)^2=\left(x-y\right)^2-2\le0\)
\(\Leftrightarrow-\sqrt{2}\le x-y\le\sqrt{2}\) --> ...
b) \(\left(2x-y-2\right)^2=7\left(x-2y-y^2-1\right)\)
\(\Leftrightarrow4x^2+y^2+4-4xy+4y-4x=7x-14y-7y^2-7\)
\(\Leftrightarrow2x^2-4xy+2y^2+2x^2-11x+\dfrac{121}{16}+6y^2+18y+\dfrac{9}{4}=\dfrac{-19}{16}\left(vl\right)\)
câu c tương tự .
a. \(\left(20x^4y-25x^2y^2-3x^2y\right):5x^2y\)
\(=4x^2-5y-\frac{3}{5}\)
b. \(\left(15xy^2+17xy^3+18y^2\right):6y^2\)
\(=\frac{5}{2}x+\frac{17}{6}xy+3\)
c. \(\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(y-x\right)^2\)
\(=\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(x-y\right)^2\)
\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)
d. \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left(y-x\right)\)
\(=\left(y-x\right)^2:\left(y-x\right)\)
\(=y-x\)
a/ \(x^3+2x^2+3x+2=y^3\)
Với \(\orbr{\begin{cases}x>1\\x< -1\end{cases}}\)thì
\(x^3< x^3+2x^2+3x+2=y^3< \left(x+1\right)^3\)
Nên không tồn tại số nguyên x, y thỏa mãn đề bài.
Từ đây ta suy ra \(-1\le x\le1\)
Với \(x=-1\Rightarrow y=0\)
\(x=0\Rightarrow y=\sqrt[3]{2}\left(l\right)\)
\(x=1\Rightarrow y=2\)
b/ \(y^2+2\left(x^2+1\right)=2y\left(x+1\right)\)
\(\Leftrightarrow2y^2+4\left(x^2+1\right)=4y\left(x+1\right)\)
\(\Leftrightarrow\left(y^2-4xy+4x^2\right)+\left(y^2-4y+4\right)=0\)
\(\Leftrightarrow\left(y-2x\right)^2+\left(y-2\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}y=2x\\y=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)