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a) \(\left\{{}\begin{matrix}x+3y=11\\2x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=11\\6x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=14\\x+3y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất (x; y) = (2; 3)
b) \(3x^4+9x^2-12=0\) (1)
Đặt \(x^2=t\left(t\ge0\right)\)
Phương trình (1) trở thành: \(3t^2+9t-12=0\)
Ta có: \(a+b+c=3+9+\left(-12\right)=0\)
\(\Rightarrow t_1=1\) (Thỏa mãn \(t\ge0\) )
\(t_2=-4\) (Không thỏa mãn \(t\ge0\) )
+) Với \(t=1\) , ta có: \(x^2=1\Leftrightarrow x_1=1;x_2=-1\)
Vậy phương trình có hai nghệm: \(x_1=1;x_2=-1\)
1/ \(\left\{{}\begin{matrix}7x-3y=5\\\dfrac{x}{2}+\dfrac{y}{3}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x-3y=5\\\dfrac{1}{2}x+\dfrac{1}{3}y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{3}x-y=\dfrac{5}{3}\\\dfrac{3}{2}x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{23}{6}x=\dfrac{23}{3}\\7x-3y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(2;3\right)\)
2/ \(\left\{{}\begin{matrix}3x+y=3\\2x-y=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x=10\\3x+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(2;-3\right)\)
3/ \(\left\{{}\begin{matrix}2x+y=5\\3x-2y=11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=10\\3x-2y=11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=21\\2x+y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(3;-1\right)\)
1. Giải phương trình, hệ phương trình:
a) 2x2 - 5x + 3 = 0
\(\Leftrightarrow2x^2-2x-3x+3=0\)
\(\Leftrightarrow2x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
b) x2 - 3x = 0
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}2\left(x+1\right)-5\left(y+1\right)=5\\3\left(x+1\right)-2\left(y+1\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6\left(x+1\right)-15\left(y+1\right)=15\\6\left(x+1\right)-4\left(y+1\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-11\left(y+1\right)=13\\3\left(x+1\right)-2\left(y+1\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=\dfrac{-13}{11}\\3\left(x+1\right)-2.\left(-\dfrac{13}{11}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{24}{11}\\3\left(x+1\right)=-\dfrac{15}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{24}{11}\\x=-\dfrac{16}{11}\end{matrix}\right.\)
Hix ,mệt quá.
\(d,\left\{{}\begin{matrix}\dfrac{15}{x}-\dfrac{7}{y}=9\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{60}{x}-\dfrac{28}{y}=36\\\dfrac{60}{x}+\dfrac{135}{y}=525\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{163}{y}=-489\\\dfrac{60}{x}+\dfrac{135}{y}=525\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\\dfrac{60}{x}+405=525\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)
\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
mình giải tắt nhé vì mình không giỏi dùng công thức. Thông cảm nha.
1.
\(\left\{{}\begin{matrix}3x-y=2m+3\\x+y=3m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m}{4}+1\\y=\dfrac{-5m}{4}\end{matrix}\right.\)
vậy phương trình có nghiệm duy nhất là \(\left(\dfrac{m}{4}+1;\dfrac{-5m}{4}\right)\)
Thay vào đẳng thức ta được:
\(\left(\dfrac{m}{4}+1\right)^2+\left(\dfrac{-5m}{4}\right)^2=5\\ \Leftrightarrow x=\)
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) \(\left\{{}\begin{matrix}2x-3y=2\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+2}{2}\\-4\cdot\frac{3y+2}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
a) Đặt t = \(x^2\)
ta có p/t : \(3x^2-4x+1=0\)
Bn lập Δ rồi giải
b)⇔ \(\left\{{}\begin{matrix}6x+3y=15\\6x-4y=22\end{matrix}\right.\)
tự giải
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