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Lời giải
a) Thay a=2+√3a=2+3 và b=2−√3b=2−3 vào P, ta được:
P=a+b−abP=2+√3+2−√3−(2+√3)(2−√3)P=2+2−(22−√32)P=4−(4−3)P=4−4+3=3P=a+b−abP=2+3+2−3−(2+3)(2−3)P=2+2−(22−32)P=4−(4−3)P=4−4+3=3
b) {3x+y=5x−2y=−3⇔{6x+2y=10x−2y=−3⇔{7x=7x−2y=−3⇔{x=1y=2{3x+y=5x−2y=−3⇔{6x+2y=10x−2y=−3⇔{7x=7x−2y=−3⇔{x=1y=2
Vậy nghiệm hệ phương trình (1; 2)
Có gì bạn tham khảo nha//
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
a + b + c = 6
=> (a + b + c)2 = 36
<=> a2 + b2 + c2 + 2(ab + bc + ca) = 36
<=> a2 + b2 + c2 = 36 - 2.12 = 12
<=> a2 + b2 + c2 = ab + bc + ca
<=> 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
<=> (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0
<=> (a - b)2 + (b - c)2 + (c - a)2 = 0
<=> a = b = c
=> a = b = c = 2
P = (a - 3)2018 + (b - 3)2018 + (c - 3)2018 = (-1)2018 + (-1)2018 + (-1)2018 = 1 + 1 + 1 = 3
Lời giải:
Có: \(\left\{\begin{matrix} a+b+c=9\\ a^2+b^2+c^2=27\end{matrix}\right.\Rightarrow \left\{\begin{matrix} (a+b+c)^2=81\\ a^2+b^2+c^2=27\end{matrix}\right.\)
\(\Rightarrow (a+b+c)^2-(a^2+b^2+c^2)=54\)
\(\Leftrightarrow 2(ab+bc+ac)=54\Leftrightarrow ab+bc+ac=27\)
Do đó: \(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow \frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}=0(*)\)
Ta thấy: \((a-b)^2; (b-c)^2; (c-a)^2\geq 0\forall a,b,c\in\mathbb{R}\)
Suy ra \((*)\) xảy ra khi và chỉ khi
\((a-b)^2=(b-c)^2=(c-a)^2=0\Leftrightarrow a=b=c\)
Khi đó: \(a=b=c=\frac{9}{3}=3\) (thỏa mãn)
\(P=(a-2)^{2015}+(b-3)^{2016}+(c-4)^{2017}=1^{2015}+0^{2016}+(-1)^{2017}\)
\(P=1+0+(-1)=0\)
a)\(\left\{{}\begin{matrix}3ax-\left(b+1\right)y=93\\bx+4ay=-3\end{matrix}\right.\)
có nghiệm \(\left(x;y\right)=\left(1;-5\right)\) ta thay \(x=1;y=-5\) vào hệ pt trên, ta có:
\(\left\{{}\begin{matrix}3a.1-\left(b+1\right).\left(-5\right)=93\\b.1+4a.\left(-5\right)=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3a+5b+5=93\\b-20a=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3a+5b=93-5\\-\left(20a-b\right)=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3a+5b=88\\20a-b=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3a+5b=88\\100a-5b=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}103a=103\\3a+5b=88\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=1\\3.1+5b=88\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=1\\5b=88-3=85\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=1\\b=17\end{matrix}\right.\)
vậy để hệ pt trên có nghiệm (1;-5) thì a=1; b=17.
b) \(\left\{{}\begin{matrix}\left(a-2\right)x+5by=25\\2ax-\left(b-2\right)y=5\end{matrix}\right.\)
có nghiệm (x; y) =(3; -1), ta thay x =3; y = -1 vào pt, ta có:
\(\left\{{}\begin{matrix}\left(a-2\right).3+5b.\left(-1\right)=25\\2a.3-\left(b-2\right).\left(-1\right)=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3a-6-5b=25\\6a+b-2=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3a-5b=25+6\\6a+b=5+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3a-5b=31\\6a+b=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6a-10b=62\\6a+b=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-11b=55\\6a+b=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=-5\\6.a-5=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=-5\\6a=7+5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=-5\\6a=12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=-5\\a=2\end{matrix}\right.\)
Vậy hệ pt trên có nghiệm (3; -1) khi a=2, b=-5.
\(ab=\dfrac{\left(a+b\right)^2-a^2-b^2}{2}=\dfrac{13^2-89}{2}=\dfrac{80}{2}=40\)
\(P=\left(a+b\right)^3-3ab\left(a+b\right)=13^3-3\cdot40\cdot13=637\)
\(\left\{{}\begin{matrix}a+b=13\\a^2+b^2=89\end{matrix}\right.\)
\(\left(a+b\right)^2=169\)
\(a^2+2ab+b^2=169\)
\(ab=40\)
\(P=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=13^3-3.40.13=637\)