\(\left\{{}\begin{matrix}x^3+3x=448y^3+6y\\385x^2-16y^2=96\end{matrix}\right.\)
K
Khách
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NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 7 2021
a.
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-448y^3=-3x+6y\\96=385x^2-16y^2\end{matrix}\right.\)
\(\Rightarrow96\left(x^3-448y^3\right)=\left(-3x+6y\right)\left(385x^2-16y^2\right)\)
\(\Leftrightarrow\left(x-4y\right)\left(417x^2+898xy+3576y^2\right)=0\)
\(\Leftrightarrow x-4y=0\)
\(\Leftrightarrow x=4y\)
Thế vào \(385x^2-16y^2=96\)
\(\Rightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 7 2021
b.
ĐKXĐ: \(x+y\ne0\)
\(\left\{{}\begin{matrix}\left(3x^3-y^3\right)\left(x+y\right)=1\\1=x^2+y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x^3-y^3\right)\left(x+y\right)=1\\1=\left(x^2+y^2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left(3x^3-y^3\right)\left(x+y\right)=\left(x^2+y^2\right)^2\)
\(\Leftrightarrow\left(x-y\right)\left(x+2y\right)\left(2x^2+xy+y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-2y\end{matrix}\right.\)
Thế vào \(x^2+y^2=1\)...
TC
0
HP
19 tháng 2 2021
2.
Ta cần tìm \(cosABC=\dfrac{AB^2+BC^2-AC^2}{2AB.BC}=\dfrac{3\left(AB^2+BC^2-AC^2\right)}{2AC^2}\)
Gọi H, K là trung điểm của AB, BC.
Theo giả thiết \(\overrightarrow{OM}\perp\overrightarrow{BI}\)
\(\Rightarrow\overrightarrow{OM}.\overrightarrow{BI}=0\)
\(\Leftrightarrow\left(2\overrightarrow{OA}+\overrightarrow{OB}+2\overrightarrow{OC}\right)\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=0\)
\(\Leftrightarrow\left(2\overrightarrow{OB}+2\overrightarrow{BA}+\overrightarrow{OB}+2\overrightarrow{OB}+2\overrightarrow{BC}\right)\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=0\)
\(\Leftrightarrow\left(5\overrightarrow{OB}+2\overrightarrow{BA}+2\overrightarrow{BC}\right)\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=0\)
\(\Leftrightarrow2\left(\overrightarrow{BA}+\overrightarrow{BC}\right)^2+5\overrightarrow{OB}.\overrightarrow{BA}+5\overrightarrow{OB}.\overrightarrow{BC}=0\)
\(\Leftrightarrow2\left(\overrightarrow{BA}+\overrightarrow{BC}\right)^2+5\left(\overrightarrow{OH}+\overrightarrow{HB}\right).\overrightarrow{BA}+5\left(\overrightarrow{OK}+\overrightarrow{KB}\right).\overrightarrow{BC}=0\)
\(\Leftrightarrow2\left(\overrightarrow{BA}+\overrightarrow{BC}\right)^2+5\overrightarrow{OH}.\overrightarrow{BA}+5\overrightarrow{HB}.\overrightarrow{BA}+5\overrightarrow{OK}.\overrightarrow{BC}+5\overrightarrow{KB}.\overrightarrow{BC}=0\)
\(\Leftrightarrow2\left(\overrightarrow{BA}+\overrightarrow{BC}\right)^2+0+\dfrac{5}{2}\overrightarrow{AB}.\overrightarrow{BA}+0+\dfrac{5}{2}\overrightarrow{CB}.\overrightarrow{BC}=0\) (Vì \(OH\perp AB,OK\perp BC\))
\(\Leftrightarrow-\dfrac{1}{2}\left(AB^2+BC^2\right)+4\overrightarrow{BA}.\overrightarrow{BC}=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(AB^2+BC^2\right)=2\left(AB^2+BC^2-AC^2\right)\)
\(\Leftrightarrow AB^2+BC^2=\dfrac{4}{3}AC^2\)
Khi đó \(cosABC=\dfrac{3\left(\dfrac{4}{3}AC^2-AC^2\right)}{2AC^2}=\dfrac{1}{2}\Rightarrow\widehat{ABC}=60^o\)
8 tháng 1 2021
1) \(\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}3x-2y=4\\7x=14\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}2x+3y=5\\4x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x=6y=10\end{matrix}\right.\)
=> Hệ có vô số nghiệm.
3)\(\left\{{}\begin{matrix}3x-4y=-2\\10x+4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}3x-4y=-2\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}6x+15y=9\\6x-4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}6x+15y=9\\19y=19\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)
HP
9 tháng 10 2021
6. \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}4x-6y=2\\x-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\\dfrac{2+6y}{4}-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-2\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\left(1-\dfrac{y}{2}\right).3\\6\left(1-\dfrac{y}{2}\right)+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(1-\dfrac{y}{2}\right)\\y=\left(VNghiệm\right)\end{matrix}\right.\Leftrightarrow\) không tồn tại x, y
(Các câu khác tương tự nhé.)
Còn lại tương tự
N
3 tháng 2 2019
a)\(\Leftrightarrow\left\{{}\begin{matrix}12x+16y=-1\\3x+4y=-2\end{matrix}\right.\)(vô nghiệm)
Vậy hpt vô nghiệm.
b)\(\left\{{}\begin{matrix}\dfrac{5x-1}{5y-1}=\dfrac{1}{2}\\5x-7y=-9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}10x-2=10y-1\\5x-7y=-9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}10x-10y=1\\5x-7y=-9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{97}{20}\\y=\dfrac{19}{4}\end{matrix}\right.\)
Vậy hpt có tập nghiệm là \(\left(\dfrac{97}{20};\dfrac{19}{4}\right)\).
21 tháng 3 2020
1) \(\left\{{}\begin{matrix}4x+y=2\\8x+3y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2-4x\\8x+3\left(2-4x\right)=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{4}\\y=1\end{matrix}\right.\)
2) 2 pt 3 ẩn không giải được.
3) \(\left\{{}\begin{matrix}3x+2y=6\\x-y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\3x+2\left(x-2\right)=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}2x-3y=1\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+1}{2}\\-4\cdot\frac{3y+1}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
5) \(\left\{{}\begin{matrix}2x+3y=5\\5x-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-3y+5}{2}\\5\cdot\frac{-3y+5}{2}-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
6) \(\left\{{}\begin{matrix}3x-y=7\\x+2y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x+2\left(3x-7\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
7) \(\left\{{}\begin{matrix}x+4y=2\\3x+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2-4y\\3\left(2-4y\right)+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{1}{5}\\x=\frac{6}{5}\end{matrix}\right.\)
8) \(\left\{{}\begin{matrix}-x-y=2\\-2x-3y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-x-2\\-2x-3\left(-x-2\right)=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
9) \(\left\{{}\begin{matrix}2x-3y=2\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+2}{2}\\-4\cdot\frac{3y+2}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 10 2020
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{3x-1}=a\ge0\\\sqrt{8y+3}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+2\left(a^2+1\right)=b+2\left(b^2-3\right)+8\)
\(\Leftrightarrow2a^2-2b^2+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a+2b+1\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow3x-1=8y+3\) (1)
Lại xét pt đầu:
\(\left(x+4y\right)\left(x^2+16y^2+8xy\right)=8xy\left(x+4y\right)+32xy\left(x+4y-3\sqrt{xy}\right)\)
\(\Leftrightarrow\left(x+4y\right)^3-40xy\left(x+4y\right)+96xy\sqrt{xy}=0\)
Đặt \(\left\{{}\begin{matrix}x+4y=m\\\sqrt{xy}=n\ge0\end{matrix}\right.\)
\(\Rightarrow m^3-40mn^2+96n^3=0\)
\(\Leftrightarrow\left(m-4n\right)\left(m^2+4mn-24n^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4y=4\sqrt{xy}\\\left(x+4y\right)^2+4\left(x+4y\right)\sqrt{xy}-24xy=0\end{matrix}\right.\) (2)
Rút x hoặc y từ (1) và thế vào (2) để giải
Dài quá làm biếng.