Bài 1 : NĂNG KHIẾU 2016-2017
A) Tính S=a+b biết a;b>0, a \(\ne\)b và \(\left(\dfrac{a\left(a-4b\right)+b\left(b+2a\right)}{a+b}\right):\left[\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\right]=2016\)
B) Giải: \(x\sqrt{x+5}=2x^2-5x\left(1\right)và\left\{{}\begin{matrix}\left(\sqrt{y}+x-3\right)\left(y+\sqrt{x}\right)=0\\x^2+y=5\end{matrix}\right.\)
Giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}4x^2-4xy-14x-3y^2+y+10=0\\5\sqrt{xy}+2x+2y=6\sqrt{y}-8\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x^4+3x^2y+4x^2-2y^2+3y+2=0\\\sqrt{x\left(y-1\right)}+2y+2\sqrt{y-1}=3x+2\sqrt{x}+2\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^6+3x^2-y^3-6y^2-15y-14=0\\\sqrt{xy+2x-y-2}+6x-2y=10\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
Cho \(\left\{{}\begin{matrix}\left|a\right|\le1\\\left|b\right|\ge1\\\left|a+b\right|=\sqrt{3}\end{matrix}\right.\)
Tìm GTLN of \(A=\sqrt{1-a^2}+\sqrt{1-b^2}\)
giải hệ:
a) \(\left\{{}\begin{matrix}\sqrt{x+3y}+\sqrt{x+y}=2\\\sqrt{x+y}+y-x=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\left(x-\frac{1}{y}\right)\left(y+\frac{1}{x}\right)=2\\2x^2y+xy^2-4xy=2x-y\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}2x^2+xy=y^2-3y+2\\x^2-y^2=3\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}x^2+y^2+z^2+2xy-xz-zy=3\\x^2+y^2-2xy-xz+zy=-1\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}x^2-y^2+5x-y+6=0\\x^2+\left(x-y\right)^2=2+\sqrt{6x+7}+2\sqrt{x+y+1}\end{matrix}\right.\)
Tim cac so nguyen duong a, b, c thoa man: \(\left\{{}\begin{matrix}\sqrt{a-b+c}=\sqrt{a}-\sqrt{b}+\sqrt{c}\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\end{matrix}\right.\)
giải pt: \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
làm thế này mà chả hiểu sao lại bị gạch, ai biết chỉ với, cảm ơn nak:
+ ĐK:\(\left\{{}\begin{matrix}x\ge1\\x+3-4\sqrt{x-1}\ge0\\x+8-6\sqrt{x-1}\ge0\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)
+ pt đã cho \(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\) (*)
Th1: \(\left\{{}\begin{matrix}\sqrt{x-1}-2< 0\\\sqrt{x-1}-3< 0\end{matrix}\right.\)
(*) \(\Leftrightarrow2-\sqrt{x-1}+3-\sqrt{x-1}=1\Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\Leftrightarrow x=5\left(N\right)\)
Th2: \(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\ge0\end{matrix}\right.\)
(*) \(\Leftrightarrow\sqrt{x-1}-2+\sqrt{x-1}-3=1\Leftrightarrow2\sqrt{x-1}=6\Leftrightarrow\sqrt{x-1}=3\Leftrightarrow x=10\left(N\right)\)
Th3: \(\sqrt{x-1}-3< 0\le\sqrt{x-1}-2\)
(*) \(\Leftrightarrow\sqrt{x-1}-2+3-\sqrt{x-1}=1\Leftrightarrow1=1\left(đúng\right)\)
Kl: \(x\ge1\)
\(\left\{{}\begin{matrix}x\sqrt{2}-y\sqrt{3}=1\\x+y\sqrt{3}=\sqrt{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\x+\left(\sqrt{2}+1\right)y=1\end{matrix}\right.\)
rút gọn giá trị biểu thức sau
a)\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
b) \(\sqrt{2x+2\sqrt{2x^2-1}}\) với \(\left\{{}\begin{matrix}x\ge1\\\sqrt{x-1}+\sqrt{x+1}=\sqrt{7}\end{matrix}\right.\)
1.Giải hpt
\(\left\{{}\begin{matrix}\frac{7}{\sqrt{x+7}}-\frac{4}{\sqrt{y-6}}=\frac{-1}{4}\\\frac{5}{\sqrt{x+7}}+\frac{3}{\sqrt{y-6}}=\frac{11}{4}\end{matrix}\right.\)
