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1: a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37\) (Vì \(x-y=7\))
\(=100\)
Vậy \(A=100\)
b) Ta có: \(B=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
\(=25\)
Vậy \(B=25\)
c) Ta có : \(C=\left(x-y\right)^2\)
\(=x^2-2xy+y^2\)
\(=\left(x^2+y^2\right)-2xy\)
\(=26-2.5\) (Vì \(x^2+y^2=26\) ; \(xy=5\))
\(=16\)
Vậy \(C=16\)
2: a) \(\left(x+y\right)^2-y^2=x^2+2xy+y^2-y^2\)
\(=x^2+2xy\)
\(=x\left(x+2y\right)\) \(\left(dpcm\right)\)
b) \(\left(x^2+y^2\right)^2-2xy^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)^2\left(x+y\right)^2\) \(\left(dpcm\right)\)
c) \(\left(x+y\right)^2=x^2+2xy+y^2\)
\(=\left(x^2-2xy+y^2\right)+4xy\)
\(=\left(x-y\right)^2+4xy\) \(\left(dpcm\right)\)
Chúc bn học tốt ✔✔✔
a, \(A_{\left(x\right)}=2x^2+2xy+y^2-2x+2y+2\)
\(=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(x^2-4x+4\right)-3\)
\(=\left(x+y+1\right)^2+\left(x-2\right)^2-3\ge-3\) hay \(A_{\left(x\right)}\ge-3\)
Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
Vậy \(minA_{\left(x\right)}=-3\) khi x=-3; y=2
b, \(B_{\left(x\right)}=x^2-4xy+5y^2+10x-22y+28\)
\(=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)
\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\Leftrightarrow B_{\left(x\right)}\ge2\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
Vậy \(minB_{\left(x\right)}=2\Leftrightarrow x=-3;y=1\)
c, \(C_{\left(x\right)}=x^2-10xy+26y^2+14x-76y+59\)
\(=\left(x^2+25y^2+49-10xy+14x-70y\right)+\left(y^2-6y+9\right)+1\)
\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\Leftrightarrow C_{\left(x\right)}\ge1\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-5y+7\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-5y+7=0\\y-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)
Vậy \(minC_{\left(x\right)}=1\Leftrightarrow x=8;y=3\)
d, \(D_{\left(x\right)}=4x^2-4xy+2y^2-20x-4y+174\)
\(=\left(4x^2+y^2+25-4xy-20x+10y\right)+\left(y-14y+49\right)+74\)
\(=\left(2x-y-5\right)^2+\left(y-7\right)^2+74\ge74\Leftrightarrow D_{\left(x\right)}\ge74\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(2x-y-5\right)^2=0\\\left(y-7\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-y-5=0\\y-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\y=7\end{matrix}\right.\)
Vậy \(minD_{\left(x\right)}=74\Leftrightarrow x=6;y=7\)
e, \(E_{\left(x\right)}=x^2-2x+y^2+4y+5\)
\(=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(minE_{\left(x\right)}=0\Leftrightarrow x=1;y=-2\)
bạn ơi! Sao cái chỗ A(x) =(x+y+1)2+(x-2)2-3 mà chuyển sang lại là -3 v
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)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
Bác google được sinh ra để làm gì, đăng nhiều vc, google có hết mà ;v
Bài 1,2,3,4 đơn giản, tự làm :v
7) \(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=\dfrac{abc}{c^3}+\dfrac{abc}{a^3}+\dfrac{abc}{b^3}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc.\dfrac{1}{3abc}=\dfrac{1}{3}\)
P/S: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
5) ĐK: a>b>0
\(3a^2+3b^2=10ab\Leftrightarrow\left(a-3b\right)\left(3a-b\right)=0\)
Tự phân tích
Mà a>b>0=> Chọn a=3b
Thay vào
Bài 6 tương tự bài 5
Có bất mãn chỗ nào thì ib nha bạn :))
1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
b/
\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)
\(=16+8+20=44\)
\(\Rightarrow B\ge11\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
Bài 2:
a, \(x^2-6x+10=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\ge1>0\)
\(\Rightarrowđpcm\)
b, \(x^2-4xy+4y^2+1=\left(x-2y\right)^2+1>0\)
\(\Rightarrowđpcm\)
c, \(x^2-4x+7=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\ge3\)
\(\Rightarrowđpcm\)
d, \(x^2+y^2-2x+4y+5\)
\(=x^2-2x+1+y^2+4y+4\)
\(=\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
\(\Rightarrowđpcm\)
Ép người quá đáng >.<
Bài 1:
a, \(-\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)+\left(2x^2+1\right)\)
\(=-\left(4x^4-4x^3+2x^2+4x^3-4x^2+2x+2x^2-2x+1\right)+2x^2+1\)
\(=-\left(4x^4+1\right)+2x^2+1=-4x^4+2x^2\)
b, \(\left(x^2+x+2\right)^2+\left(x-1\right)^2-2\left(x^2+x+2\right)\left(x-1\right)\)
\(=\left(x^2+x+2-x+1\right)^2=\left(x^2+3\right)^2\)
d, \(-125x^3+225x^2-135x+27\)
\(=-\left(125x^3-225x^2+135x-27\right)\)
\(=-\left(125x^3-75x^2-150x^2+90x+45x-27\right)\)
\(=-\left[25x^2\left(5x-3\right)-30x\left(5x-3\right)+9\left(5x-3\right)\right]\)
\(=-\left[\left(5x-3\right)\left(25x^2-15x-15x+9\right)\right]\)
\(=-\left(5x-3\right)^3\)
a) \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(A=x^2+2x+y^2-2y-2xy+37\)
\(A=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(A=\left(x-y\right)^2+2\left(x-y\right)+37\)
Mà \(x-y=7\)
\(\Rightarrow A=7^2+2.7+37\)
\(A=100\)
b) \(B=x^2+4y^2-2x+10+4xy-4y\)
\(B=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)
\(B=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
Mà \(x+2y=5\)
\(\Rightarrow B=5^2-2.5+10\)
\(B=25\)