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Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
Câu 1 :
a)
\(P = a + b - ab = 2 + \sqrt{3} + 2-\sqrt{3} - (2 + \sqrt{3})(2-\sqrt{3})\\ =4 - (2^2 - (\sqrt{3})^2) = 4 - (4 - 3) = 3\)
b)
\(\left\{{}\begin{matrix}3x+y=5\\x-2y=-3\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}3x+y=5\\3x-6y=-9\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}y-\left(-6y\right)=5-\left(-9\right)\\x=\dfrac{5-y}{3}\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}y=2\\x=\dfrac{5-2}{3}=1\end{matrix}\right.\)
Vậy nghiệm của hệ phương trình (x ; y) = (1 ; 2)
Câu 1:
a)
\(P=a+b-ab\\ =2+\sqrt{3}+2-\sqrt{3}-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)\\ =4-\left(4-2\sqrt{3}+2\sqrt{3}-3\right)\\ =4-1=3\)
Vậy \(P=3\)
b)
\(\left\{{}\begin{matrix}3x+y=5\\x-2y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x+2y=10\\x-2y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}7x=7\\x-2y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\1-2y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\2y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy pht có nghiệm là \(\left(x;y\right)=\left(1;2\right)\)
3a)\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\) (ĐK: x≠2;y≠\(\dfrac{1}{2}\))
Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{2y-1}=b\) (ĐK: a>0; b>0)
Hệ phương trình đã cho trở thành
\(\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\2\left(2-b\right)-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\4-2b-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\b=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\left(TM\text{Đ}K\right)\\b=\dfrac{3}{5}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Khi đó \(\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{2y-1}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\left(x-2\right)=5\\3\left(2y-1\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\6y-3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\left(TM\text{Đ}K\right)\\y=\dfrac{4}{3}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Vậy hệ phương trình đã cho có nghiệm duy nhất (x;y)=\(\left(\dfrac{19}{7};\dfrac{4}{3}\right)\)
b) Bạn làm tương tự như câu a kết quả là (x;y)=\(\left(\dfrac{12}{5};\dfrac{-14}{5}\right)\)
c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)(ĐK: x≥1;y≥0)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+4\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49\left(x-1\right)=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49x-49=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{218}{49}\\y=\dfrac{4}{49}\end{matrix}\right.\left(TM\text{Đ}K\right)\)
Bài 4:
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}3\left(3a-2\right)-2\left(2b+1\right)=30\\3\left(a+2\right)+2\left(3b-1\right)=-20\end{matrix}\right.\)
=>9a-6-4b-2=30 và 3a+6+6b-2=-20
=>9a-4b=38 và 3a+6b=-20+2-6=-24
=>a=2; b=-5
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
a, Với \(x>0;x\ne1\)
\(P=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)
\(=\left(\frac{x-1}{2\sqrt{x}}\right)^2\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\)
\(=\frac{x^2-2x+1}{4x}.\frac{-4\sqrt{x}}{x-1}=\frac{1-x}{\sqrt{x}}\)
Thay x = 4 => \(\sqrt{x}=2\)vào P ta được :
\(\frac{1-4}{2}=-\frac{3}{2}\)
c, Ta có : \(P< 0\Rightarrow\frac{1-x}{\sqrt{x}}< 0\Rightarrow1-x< 0\)vì \(\sqrt{x}>0\)
\(\Rightarrow-x< -1\Leftrightarrow x>1\)
Câu 5 :
Ta chứng minh bđt phụ: \(x^5+y^5\ge xy\left(x^3+y^3\right)\forall x\in N\Leftrightarrow x^5+y^5-x^4y-xy^4\ge0\Leftrightarrow\left(x-y\right)x^4-y^4\left(x-y\right)\ge0\Leftrightarrow\left(x-y\right)\left(x^4-y^4\right)\ge0\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\)
\(\Rightarrow x^5+y^5\ge xy\left(x^3+y^3\right)\) (1)
\(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\Rightarrow x^3+y^3\ge xy\left(x+y\right)\left(2\right)\)
Áp dụng bđt (1) và (2): \(\Rightarrow\dfrac{ab}{a^5+b^5+ab}\le\dfrac{ab}{ab\left(a^3+b^3\right)+ab}\le\dfrac{ab}{a^2b^2\left(a+b\right)+ab}=\dfrac{1}{ab\left(a+b\right)+1}=\dfrac{abc}{ab\left(a+b+c\right)}=\dfrac{c}{a+b+c}\) Tương tự:
\(\dfrac{bc}{b^5+c^5+bc}\le\dfrac{a}{a+b+c};\dfrac{ca}{c^5+a^5+ca}\le\dfrac{b}{a+b+c}\)
\(\Rightarrow\sum\dfrac{ab}{a^5+b^5+ab}\le\sum\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
Dấu = xảy ra \(\Leftrightarrow a=b=c\)=1
Câu 1:
a) Ta có: \(\left(x+3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy: S={1;-7}
b) Ta có: \(\left\{{}\begin{matrix}2x+y-3=0\\\dfrac{x}{4}=\dfrac{y}{3}-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=3\\\dfrac{1}{4}x-\dfrac{1}{3}y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+2y=6\\4x-\dfrac{16}{3}y=-16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{22}{3}y=22\\2x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\2x=3-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là (x,y)=(0;3)