Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\Leftrightarrow\left\{{}\begin{matrix}9x-3y=15\\2x+3y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=3x-5=4\end{matrix}\right.\)
mk cảm ơn bạn đã giúp mk nhưng mà bạn làm chi tiết giùm mk nhé
Câu 2b
\(\left\{{}\begin{matrix}2x+y=5\\2x-6y=14m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=5-14m+2\\x=\dfrac{5-y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-2m\\x=\dfrac{5-1+2m}{2}=2+m\end{matrix}\right.\)
Ta có \(2\left(m+2\right)^2-\left(2m-1\right)^2=17\)
\(\Leftrightarrow2m^2+8m+8-4m^2+4m-1=17\Leftrightarrow-2m^2+12m-10=0\)
Ta có a + b + c = -2 + 12 - 10 = 0
vậy pt có 2 nghiệm m = 1 ; m = 5
b: \(=\dfrac{x\left(\sqrt{x-1}-1+\sqrt{x-1}+1\right)}{1+x-2}=\dfrac{x\cdot2\sqrt{x-1}}{x-1}=\dfrac{2x}{\sqrt{x-1}}\)
c:
Sửa đề: 1/căn x
\(=\dfrac{3\sqrt{x}-x+x+9}{9-x}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\sqrt{x}}{\sqrt{x}+4}\)
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ =\dfrac{x-1}{\sqrt{x}}\)
Bài 2:
a: Thay x=1 vào B, ta được:
\(B=\dfrac{2\cdot\left(1-1\right)}{1+1}=0\)
b: \(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}=2\)
Bài 2.
a.Thế \(x=1\) vào B ta có:
\(B=\dfrac{2\left(\sqrt{1}-1\right)}{\sqrt{1}+1}=\dfrac{2.0}{2}=\dfrac{0}{2}=0\)
b.
\(A=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\)
\(A=\dfrac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x^2+x\sqrt{x}-\sqrt{x}-1-x^2+x\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{2x\sqrt{x}-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{2\sqrt{x}\left(x-1\right)}{\sqrt{x}\left(x-1\right)}\)
\(A=2\)
c.\(P=1:\left(A:B\right)=1:\dfrac{2\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)}=1:\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}+1}-\dfrac{2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Đê P lớn nhất thì \(\sqrt{x}+1\) nhỏ nhất, mà \(\sqrt{x}+1\ge1\) => Min =1
\(\Rightarrow P\le1-\dfrac{2}{1}=1-2=-1\)
giúp mk vs mk cần gấp
giúp mk vs mk cần gấp
