\(\sqrt{A}\ge0\) ; \(\sqrt{B}\ge0\) Mà \(\sqrt{A}+\sqrt{B}=0\) => \(A=0;B=0\)
Ta có \(\sqrt{A}\ge0,\sqrt{B}\ge0\Rightarrow\sqrt{A}+\sqrt{B}\ge0\)
Vậy muốn xảy ra dấu '=' thì \(\sqrt{A}=\sqrt{B}=0\Leftrightarrow A=B=0\)(đpcm)
\(\sqrt{A}\ge0\) ; \(\sqrt{B}\ge0\) Mà \(\sqrt{A}+\sqrt{B}=0\) => \(A=0;B=0\)
Ta có \(\sqrt{A}\ge0,\sqrt{B}\ge0\Rightarrow\sqrt{A}+\sqrt{B}\ge0\)
Vậy muốn xảy ra dấu '=' thì \(\sqrt{A}=\sqrt{B}=0\Leftrightarrow A=B=0\)(đpcm)
Tính: a)
\(\dfrac{x^2-8x-5\sqrt{x^2-8x+10}+14}{\left(x+1\right)\left(\left(4+\sqrt{22}\right)—x\right)}\)= 0
b) \(\left\{{}\begin{matrix}x+108y=200\\100x-87y=113\end{matrix}\right.\). Tính \(\left(x^2-3y^2\right)^{2018}\).
c) \(\left\{{}\begin{matrix}x^2-y^2=0\\2018x+y=2019\end{matrix}\right.\)
Cho x,y,z>0 tm : \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+\sqrt{z}=2\\x+y+z=2\end{matrix}\right.\) .Tính:
P= \(\sqrt{\left(x+1\right).\left(y+1\right).\left(z+1\right)}.\left(\frac{\sqrt{x}}{x+1}+\frac{\sqrt{y}}{y+1}+\frac{\sqrt{z}}{z+1}\right)\)
\(\left\{{}\begin{matrix}x\left(x+y\right)+y^2-4y+1=0\\y\left(x+y\right)^2-2x^2-7y=2\end{matrix}\right.\)
Chứng minh rằng:
\(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\le\sqrt{ab}\) với a>0, b>0, c>0
Ai giỏi toán giúp mình giải mấy câu này với:
Câu a) \(\left\{{}\begin{matrix}7x+5y=19\\3x+5y=31\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+5y=-6\\3x+2y=11\end{matrix}\right.\) c) \(\left\{{}\begin{matrix}2x+3y=-2\\3x-2y=-3\end{matrix}\right.\)
1. Cho
A= \(\sqrt{\frac{x\sqrt{x}+1}{\sqrt{x}+1}+\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}}với\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
a) Rút gọn A
b) Tính giá trị của biểu thức X= M + \(\frac{2012}{2013}\) biết x= \(1+2012^2+\frac{2012^2}{2013^2}\)
cho a,b,c >0 chứng minh rằng
\(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}>=2\left(\sqrt{\dfrac{c}{a+b}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{a}{b+c}}\right)\)
Rút gọn:
a) \(\frac{a-b}{\sqrt{a}-\sqrt{b}}\)-\(\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)(\(a\ge0\),\(b\ge0\),\(a\ne b\))
b)\(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)\(\left(a>0,b>0,a\ne b\right)\)
C)\(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)\(\left(a>0,a\ne1,a\ne4\right)\)
d)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\)\(\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)\(\left(a>0,b>0,a\ne b\right)\)
e)\(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right)\):\(\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)\(\left(x>0,x\ne9\right)\)
\(P=\left(\frac{1}{a-1}+\frac{3\sqrt{a}+5}{a\sqrt{a}-a-\sqrt{a}+1}\right).\frac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}\left(a>0,a\ne1\right)\)
a)Rút gọn
b)Đặt Q= \(\left(a-\sqrt{a}+1\right)P\).Chứng minh Q>1