how to ko hĩu
how to ko hĩu
\(\sqrt{\sqrt[]{}\frac{ }{ }\hept{\begin{cases}\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}^{ }^2_{ }|^{ }_{ }\tanh\forall\perp}\)
\(\hept{\sqrt[1]{3}}\sqrt{2\sqrt[3]{2}2}3332133333\hept{\begin{cases}3\\3\\3\orbr{\begin{cases}6\\9\end{cases}}9\end{cases}}\)
Giải các hệ phương trình sau :
a) \(\hept{\begin{cases}\sqrt{2x}-\sqrt{3y}=1\\x+\sqrt{3y}=\sqrt{2}\end{cases}}\) b) \(\hept{\begin{cases}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\x+\left(\sqrt{2}+1\right)y=1\end{cases}}\) c) \(\hept{\begin{cases}x-2\sqrt{2y}=\sqrt{5}\\\sqrt{2x}+y=1-\sqrt{10}\end{cases}}\) d) \(\hept{\begin{cases}\sqrt{3x}-\sqrt{2y}=1\\\sqrt{2x}+\sqrt{3y}=\sqrt{3}\end{cases}}\)
Giải Phương Trình
\(\sqrt{4x^2-9}=2\sqrt{2x+3}\)\(2\left(\sqrt{\frac{x-1}{4}}-3\right)=2\sqrt{\frac{4x-4}{9}}-\frac{1}{3}\)Áp dụng \(\sqrt{a^2}=\left|a\right|=\orbr{\begin{cases}a\left(a\ge0\right)\\-a\left(a< 0\right)\end{cases}}\)và \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\)
moi nguoi oi giup em may cau nay voi
1) Cho \(\hept{\begin{cases}a,b,c,d\ge0\\a+b+c+d\le3\end{cases}}\)tim max \(P=2a+3b^2+4b^3+5b^4\)
2) Cho \(\hept{\begin{cases}a,b,c\ge0\\a+b+c=3\end{cases}}\)tim min \(P=\left(a-1\right)^3+\left(b-1\right)^3+\left(c-1\right)^3\)
3) Cho \(\hept{\begin{cases}a,b\ge0;0\le c\le1\\a^2+b^2+c^2=3\end{cases}}\) tim max,min \(P=ab+bc+ca+3\left(a+b+c\right)\)
4) Cho \(\hept{\begin{cases}a,b,c\ge0\\a+b+c=3\end{cases}}\)tim max \(P=a\sqrt{b}+b\sqrt{c}+c\sqrt{a}-\sqrt{abc}\)
5) Cho \(\hept{\begin{cases}a,b\ge0;0\le c\le1\\a+b+c=3\end{cases}}\)tim max, min \(P=a^2+b^2+c^2+abc\)
em cam on nhieu
Giải các HPT sau:
a) \(\hept{\begin{cases}\sqrt{xy}+\sqrt{1-y}=\sqrt{y}\\2\sqrt{xy-y}-\sqrt{y}=-1\end{cases}}\)
b) \(\hept{\begin{cases}\sqrt{\frac{2x}{y}}+\sqrt{\frac{2y}{x}}=3\\x-y+xy=3\end{cases}}\)
c) \(\hept{\begin{cases}2x+2y-\sqrt{xy}=3\\\sqrt{3x+1}+\sqrt{3y+1}=4\end{cases}}\)
d) \(\hept{\begin{cases}x^3\left(2+3y\right)=8\\x\left(y^3-2\right)=6\end{cases}}\)
p/s: m.n giúp mk nha, ko cần phải làm hết đâu :)
a, \(\hept{\begin{cases}\left(x+y+z\right)^2=3\left(xy+yz+xz\right)\\x^{2017}+y^{2017}+z^{2017}=3^{2018}\end{cases}}\)
b,\(\hept{\begin{cases}x^3=y^3+9\\x-x^2=2y^2+4y\end{cases}}\)
c,\(\hept{\begin{cases}\sqrt{x}+\sqrt{2017-y}=\sqrt{2017}\\\sqrt{y}+\sqrt{2017-x}=\sqrt{2017}\end{cases}}\)
d,\(\hept{\begin{cases}x+y=z\\x^3+y^3=2z^2\end{cases}}\)với x,y,z là các số nguyên
Giải hpt
a/\(\hept{\begin{cases}\left|x\right|+4\left|y\right|=18\\3\left|x\right|+\left|y\right|=10\end{cases}}\)
b/ \(\hept{\begin{cases}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{cases}}\)
a) \(\hept{\begin{cases}\sqrt{x+y-1}=1\\\sqrt{x-y+2}=2\left(y-1\right)\end{cases}}\)
b)\(\hept{\begin{cases}\sqrt{x+3y+1}=2\\\sqrt{2x-y+2}=7y-6\end{cases}}\)