giải hệ 3x+2y=-1 và 2x+3y=-4
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nhâ vế 1 vs 2
nhân vế 2 vs 3 là ra thôi bn
trừ 2 vế cho nhau nữa
\(\left\{{}\begin{matrix}3x+2y=4\\2x-3y=7\end{matrix}\right.< =>\left\{{}\begin{matrix}6x+4y=8\\6x-9y=21\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}13y=-13\\3x+2y=4\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-1\\3x=4+2=6\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)
\(a,\left\{{}\begin{matrix}2x-y=1\\3x+2y=5\end{matrix}\right.\\ =>\left\{{}\begin{matrix}4x-2y=2\\3x+2y=5\end{matrix}\right.\\ =>\left\{{}\begin{matrix}7x=7\\2x-y=1\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=1\\2.1-y=1\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;1\right)\)
\(b,\left\{{}\begin{matrix}4x+3y=-1\\3x-2y=2\end{matrix}\right.\\ =>\left\{{}\begin{matrix}4.2x+3.2y=-1.2\\3.3x-2.3y=2.3\end{matrix}\right.\\ =>\left\{{}\begin{matrix}8x+6y=-2\\9x-6y=6\end{matrix}\right.\\ =>\left\{{}\begin{matrix}17x=4\\3x-2y=2\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=\dfrac{4}{17}\\y=-\dfrac{11}{17}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\dfrac{4}{17};-\dfrac{11}{17}\right)\)
A)\(5xyz.4x^2y^2.\left(-2x^3y\right)=\left(5.4.\left(-2\right)\right).\left(xx^2x^3\right).\left(yy^2y\right)=\left(-40\right)x^6y^4\)
- BẬC : 10
- HỆ SỐ: -40
B) \(-xy.\left(\frac{1}{2}x^3y^4\right).\left(\frac{-4}{7}x^2y^5\right)=\left(\frac{1}{2}.\frac{-4}{7}.\left(-1\right)\right).\left(xx^3x^2\right).\left(y^4y^5y\right)=\frac{2}{7}x^6y^{10}\)
- BẬC : 16
- HỆ SỐ: 2/7
C) \(\frac{5}{3}x^2y^4.\left(\frac{-6}{5}xy^3\right).\left(-xy\right)=\left(\frac{5}{3}.\frac{-6}{5}.\left(-1\right)\right).\left(x^2xx\right).\left(y^4y^3y\right)=2x^4y^8\)
- BẬC : 12
- HỆ SỐ : 2
D) \(\left(\frac{-1}{3}x^2y^5\right).\left(\frac{3}{4}xy\right).5x=\left(\frac{-1}{3}.\frac{3}{4}.5\right).\left(x^2xx\right).\left(y^5y\right)=\frac{-5}{4}x^4y^6\)
- BẬC : 10
- HỆ SỐ : -5 /4
CHÚC BN HỌC TỐT!!
a: \(\left\{{}\begin{matrix}2x-2y+z=3\\2x+y-2z=-3\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-4y+2z=6\\8x+4y-8z=-3\\3x-4y-z=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}12x-6z=3\\11x-9z=1\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\z=\dfrac{1}{2}\\4y=3x-z-4=\dfrac{3}{2}-\dfrac{1}{2}-4=1-4=-3\end{matrix}\right.\)
=>x=1/2;z=1/2;y=-3/4
1) \(-2x^2+x+1-2\sqrt[]{x^2+x+1}=0\)
\(\Leftrightarrow2\sqrt[]{x^2+x+1}=-2x^2+x+1\left(1\right)\)
Ta có :
\(2\sqrt[]{x^2+x+1}=2\sqrt[]{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge\sqrt[]{3}\)
Dấu "=" xảy ra khi và chỉ khi \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow-2x^2+x+1=\sqrt[]{3}\)
\(\Leftrightarrow2x^2-x+\sqrt[]{3}-1=0\)
\(\Delta=1-8\left(\sqrt[]{3}-1\right)=9-8\sqrt[]{3}\)
\(pt\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\\x=\dfrac{1-\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\end{matrix}\right.\) \(\left(vì.x=-\dfrac{1}{2}\right)\)
Vậy phương trình cho vô nghiệm
a,Ta có hệ phương trình\(\left\{{}\begin{matrix}7x-2y=1\left(1\right)\\2x+3y=11\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}21x-6y=3\\4x+6y=22\end{matrix}\right.\)
=> \(21x-6y+4x+6y=25\)
=> \(25x=25\)
=> \(x=1\)
- Thay x = 1 vào phương trình 1 ta được :
\(7-2y=1\)
=> \(y=3\)
Vậy hệ phương trình có duy nhất 1 nghiệm là ( x, y ) = ( 1, 3 )
b, Ta có hệ phương trình\(\left\{{}\begin{matrix}3x+2y=16\\2x-y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+2y=16\\y=2x+1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3x+2\left(2x+1\right)=16\\y=2x+1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3x+4x+2=16\\y=2x+1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2\\y=2x+1\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}x=2\\y=2.2+1=5\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là ( x, y ) = ( 2, 5 )
c, Ta có hệ phương trình \(\left\{{}\begin{matrix}x+2y=5\\3x-2y=-1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=5-2y\\3x-2y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=5-2y\\3\left(5-2y\right)-2y=-1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=5-2y\\15-6y-2y=-1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=5-2y\\y=2\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}x=5-2.2=1\\y=2\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là ( x, y ) = ( 1, 2 )
\(\left\{{}\begin{matrix}3x+2y=-1\\2x+3y=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+4y=-2\\6x+9y=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-5y=10\\6x+4y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\6x+4.\left(-2\right)=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\6x-8=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=1\end{matrix}\right.\)