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cái này mk làm ở câu dưới của bạn r` đó -_-" nèCâu hỏi của Phạm Hoa - Toán lớp 8 - Học toán với OnlineMath
a, =(x+2)*(y+2*x)
= (88+2)(y+2.-76)
= 90*y-6660
b, = (x-7)*(y+x)
\(\left(7\frac{3}{5}-7\right)\left(2\frac{2}{5}+7\frac{3}{5}\right)\)
= 3/5 . 10
=6
k cho tớ nha :))))))
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\(Q=x^2+y^2+xy+x+y+10\)
\(=\left(x^2+xy+x\right)+y^2+y+10\)
\(=x^2+x\left(y+1\right)+y^2+y+10\)
\(=x^2+2.x.\frac{y+1}{2}+\left(\frac{y+1}{2}\right)^2+y^2+y-\left(\frac{y+1}{2}\right)^2+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{\left(y+1\right)^2}{4}+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{y^2+2y+1}{4}+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{1}{4}y^2-\frac{1}{2}y-\frac{1}{4}+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}y^2+\frac{1}{2}y+\frac{39}{4}\)
\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left(y^2+\frac{2}{3}y+13\right)=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left(y^2+2.y.\frac{2}{6}+\frac{4}{36}-\frac{4}{36}+13\right)\)
\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left[\left(y+\frac{2}{6}\right)^2+\frac{116}{9}\right]=\left(\frac{2x+y+1}{2}\right)^2+\frac{3}{4}\left(y+\frac{2}{6}\right)^2+\frac{29}{3}\)
Vì \(\left(\frac{2x+y+1}{2}\right)^2\ge0;\frac{3}{4}\left(y+\frac{2}{6}\right)^2\ge0=>\left(\frac{2x+y+1}{2}\right)^2+\frac{3}{4}\left(y+\frac{2}{6}\right)^2+\frac{29}{3}\ge\frac{29}{3}>0\) (với mọi x;y)
Vậy biểu thức Q luôn dương với mọi giá trị của biến
=>4Q=4x2+4xy+4y2+4x+4y+40
=4x2+4x(y+1)+(y+1)2+4y2-y2+4y-2y+40-1
=(2x+y+1)2+3y2+2y+39
\(=\left(2x+y+1\right)^2+\left(\sqrt{3}y+\frac{\sqrt{3}}{3}\right)^2+\frac{116}{3}\)
\(\Rightarrow Q=\left(\frac{2x+y+1}{2}\right)^2+\left(\frac{\sqrt{3}y+\frac{\sqrt{3}}{3}}{2}\right)^2+\frac{29}{3}>0\)
=>đpcm
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\(D=-x^2-y^2+xy+2x+2y\)
\(\Rightarrow D=-\dfrac{x^2}{2}+xy-\dfrac{y^2}{2}-\dfrac{x^2}{2}+2x-\dfrac{y^2}{2}+2y\)
\(\Rightarrow D=-\left(\dfrac{x^2}{2}-xy+\dfrac{y^2}{2}\right)-\left(\dfrac{x^2}{2}-2x\right)-\left(\dfrac{y^2}{2}-2y\right)\)
\(\Rightarrow D=-\left(\dfrac{x^2}{2}-2.\dfrac{x}{\sqrt[]{2}}.\dfrac{y}{\sqrt[]{2}}+\dfrac{y^2}{2}\right)-\left(\dfrac{x^2}{2}-2.\dfrac{x}{\sqrt[]{2}}.\sqrt[]{2}+2\right)-\left(\dfrac{y^2}{2}-2.\dfrac{y}{\sqrt[]{2}}.\sqrt[]{2}+2\right)+2+2\)
\(\Rightarrow D=-\left(\dfrac{x}{\sqrt[]{2}}-\dfrac{y}{\sqrt[]{2}}\right)^2-\left(\dfrac{x}{\sqrt[]{2}}-\sqrt[]{2}\right)^2-\left(\dfrac{y}{\sqrt[]{2}}-\sqrt[]{2}\right)^2+4\)
mà \(\left\{{}\begin{matrix}-\left(\dfrac{x}{\sqrt[]{2}}-\dfrac{y}{\sqrt[]{2}}\right)^2\le0,\forall x;y\\-\left(\dfrac{x}{\sqrt[]{2}}-\sqrt[]{2}\right)^2\le0,\forall x\\-\left(\dfrac{y}{\sqrt[]{2}}-\sqrt[]{2}\right)^2\le0,\forall y\end{matrix}\right.\)
\(\Rightarrow D=-\left(\dfrac{x}{\sqrt[]{2}}-\dfrac{y}{\sqrt[]{2}}\right)^2-\left(\dfrac{x}{\sqrt[]{2}}-\sqrt[]{2}\right)^2-\left(\dfrac{y}{\sqrt[]{2}}-\sqrt[]{2}\right)^2+4\le4\)
\(\Rightarrow GTLN\left(D\right)=4\left(tạix=y=2\right)\)
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A= 2x^2 + 4x + xy + 2y
=(xy+2x2)+(2y+4x)
=x(y+2x)+2(y+2x)
=(x+2)(y+2x)
Thay x=88,y=-76 ta được:
A=(88+2)*(-76+2*88)
=90*100
=9 000
B= x^2 +xy - 7x - 7y
=(xy-7y)+(x2-7x)
=y(x-7)+x(x-7)
=(x-7)(y+x).Thay vào tính bình thường
![](https://rs.olm.vn/images/avt/0.png?1311)
đổi x= 38/5 ; y = 12/5
B= x(x+y) -7(x+y) = (x+y)(x-7)
B= (38/5 + 12/5)( 38/5-7)= 10.3/5 = 6
mới mở máy thấy làm liền đó
áp dụng bất đẳng thức Cosi ta có 2xy<=(x+y)2
thay x+y=10 vào là ra
\(x+y=10\Rightarrow x=10-y\)
\(\Rightarrow xy=\left(10-y\right)y=10y-y^2\)
\(=-\left(y^2-10y+25\right)+25\)
\(=-\left(y-5\right)^2+25\le25\)
Vậy \(P_{max}=25\Leftrightarrow y-5=0\Leftrightarrow y=5\)
\(\Rightarrow x=5\)