\(\dfrac{x}{3}=\dfrac{y}{4}\)và x2+y2=25
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H9
HT.Phong (9A5)
CTVHS
29 tháng 10 2023
a) \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)
b) \(\left(x+\dfrac{1}{4}\right)^2=x^2+2\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)
c) \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4}{25}y^2\)
d) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y^2+3\cdot2x\cdot\left(y^2\right)^2+\left(y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
e) \(\left(3x^2-2y\right)^2=\left(3x^2\right)^2-2\cdot3x^2\cdot2y+\left(2y\right)^2=9x^4-12x^2y+4y^2\)
f) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)
g) \(\left(x^2-\dfrac{1}{3}\right)\cdot\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)
23 tháng 6 2023
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
AH
Akai Haruma
Giáo viên
9 tháng 12 2021
Bạn tham khảo bài này:
https://hoc24.vn/cau-hoi/cho-biet-y-ti-le-thuan-voi-x1-x2-la-cac-gia-tri-cua-x-y1y2-la-cac-gia-tri-tuong-uong-cua-y-a-biet-xy-ti-le-thuan-va-x1-2-x2-3-y1-12-tim-y2-b-biet-xy-ti-le-nghich-v.3536605510330
5 tháng 10 2021
a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)
b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)
3 tháng 9 2021
a) Ta có: \(\dfrac{x}{y}=\dfrac{10}{9}\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}\)
\(\dfrac{y}{z}=\dfrac{3}{4}\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{9}=\dfrac{z}{12}\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}=\dfrac{x-y+z}{10-9+12}=\dfrac{78}{13}=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.10=60\\y=6.9=54\\z=6.12=72\end{matrix}\right.\)
b)Ta có: \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)
c) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{9+16+9}=\dfrac{200}{34}=\dfrac{100}{17}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{900}{17}\\y^2=\dfrac{1600}{17}\\z^2=\dfrac{900}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{30\sqrt{17}}{17}\\y=\pm\dfrac{40\sqrt{17}}{17}\\z=\pm\dfrac{30\sqrt{17}}{17}\end{matrix}\right.\)
Vậy\(\left(x;y;z\right)\in\left\{\left(\dfrac{30\sqrt{17}}{17};\dfrac{40\sqrt{17}}{17};\dfrac{30\sqrt{17}}{17}\right),\left(-\dfrac{30\sqrt{17}}{17};-\dfrac{40\sqrt{17}}{17};-\dfrac{30\sqrt{17}}{17}\right)\right\}\)
T
27 tháng 10 2023
a, \(8^3yz+12^2yz+6xyz+yz\)
\(=512yz+144yz+6xyz+yz\)
\(=yz\left(512+14+6x+1\right)\)
\(=yz\left(527+6x\right)\)
$---$
b, \(81x^4\left(z^2-y^2\right)-z^2+y^2\)
\(=81x^4\left(z^2-y^2\right)-\left(z^2-y^2\right)\)
\(=\left(z^2-y^2\right)\left(81x^4-1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(9x^2\right)^2-1^2\right]\)
\(=\left(z-y\right)\left(z+y\right)\left(9x^2-1\right)\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(3x\right)^2-1^2\right]\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left(3x-1\right)\left(3x+1\right)\left(9x^2+1\right)\)
$---$
c, \(\dfrac{x^3}{8}-\dfrac{y^3}{27}+\dfrac{x}{2}-\dfrac{y}{3}\)
\(=\left[\left(\dfrac{x}{2}\right)^3-\left(\dfrac{y}{3}\right)^3\right]+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}\right)+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}+1\right)\)
$---$
d, \(x^6+x^4+x^2y^2+y^4-y^6\)
\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^4+x^2y^2+y^4\right)\left(x^2-y^2+1\right)\)
$Toru$
BT
1
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 3 2023
\(P=\dfrac{x^3+y^3}{x^3y^3}=\dfrac{\left(x+y\right)\left(x^2+y^2-xy\right)}{x^3y^3}=\dfrac{x^2y^2\left(x+y\right)}{x^3y^3}=\dfrac{x+y}{xy}=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{x^2+y^2-xy}=\dfrac{4\left(x^2+y^2-xy\right)-3\left(x^2+y^2-2xy\right)}{x^2+y^2-xy}\)
\(=4-\dfrac{3\left(x-y\right)^2}{x^2+y^2-xy}\le4\)
\(P_{max}=4\) khi \(x=y=\dfrac{1}{2}\)
Theo t/c của dãy tỉ số bằng nhau ta có:
x/3=y/4=(x^2+y^2)/(3^2+4^2)=1
=>x=1.3=3
=>y=1.4=4
Mình viết bằng đt nên hơi khó hiểu thông cảm nhé
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)
Ta có: \(x^2+y^2=25\)
\(\Leftrightarrow k^2=1\)
Trường hợp 1: k=1
\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=3\\y=4k=4\end{matrix}\right.\)
Trường hợp 2: k=-1
\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=-3\\y=4k=-4\end{matrix}\right.\)