\(x+y=1\Rightarrow y=1-x\\ \Rightarrow A=x^2+y^2\\ =x^2+\left(1-x\right)^2\\ =x^2+1-2x+x^2\\ =2x^2-2x+1\\ =2x^2-2x+\dfrac{1}{2}+\dfrac{1}{2}\\ =\left(2x^2-2x+\dfrac{1}{2}\right)+\dfrac{1}{2}\\ =2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}\\ =2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)

Do \(2\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow A=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\)

Dấu "=" xảy ra khi:

\(2\left(x-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{1}{2}=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ \Leftrightarrow y=1-x=1-\dfrac{1}{2}=\dfrac{1}{2}\)

Vậy \(A_{Min}=\dfrac{1}{2}\) khi \(x=y=\dfrac{1}{2}\)

Bài này không cho các số x;y dương nên ko áp dụng BDT được.

x=\(\frac{-55}{7}\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}}-\dfrac{5\sqrt{x}+3}{x+\sqrt{x}}\\ ĐKXĐ:x>0;x\ne1\\ \Rightarrow A=\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}-\dfrac{5\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{x+3\sqrt{x}+3-5\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

Vậy \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) với \(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

ai ngang qua , kết bạn với mk nhé

tách ra rồi rút gọn

a) \(m_{NaOH}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{200\cdot15}{100}=30\left(g\right)\)

\(m_{d^2A\text{ sau khi pha thêm }100\left(g\right)nước}=100+200=300\left(g\right)\)

\(\Rightarrow C\%\left(NaOH\right)=\dfrac{m_{NaOH}\cdot100}{m_{d^2}}=\dfrac{30\cdot100}{300}=10\%\)

b) \(m_{NaOH\text{ sau khi cho thêm }5\left(g\right)NaOH}=30+5=35\left(g\right)\)

\(m_{d^2B}=5+200=205\left(g\right)\)

\(\Rightarrow C\%\left(NaOH\right)=\dfrac{m_{NaOH}\cdot100}{m_{d^2}}=\dfrac{35\cdot100}{205}=17,07\%\)

c) \(C\%\left(NaOH\right)=\dfrac{30\cdot100}{150}=20\%\)

bai toan @gmail.com

số nguyên tố lẻ nhỏ nhất là số 3 

coppy trên mạng

có thì trả lời nhé