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a) \(\sqrt{\dfrac{9x^2}{25}}+\dfrac{1}{5}x\) (x<0)
=\(\dfrac{-3x}{5}+\dfrac{x}{5}\) (vì x<0)
=\(\dfrac{-2x}{5}\)
b)2xy\(\sqrt{\dfrac{9x^2}{y^6}}-\sqrt{\dfrac{49x^2}{y^2}}\) (x<0 , y>0)
=2xy\(\dfrac{-3x}{y^3}+\dfrac{7x}{y}\)(vì x<y<0)
=\(\dfrac{-6x}{y^2}+\dfrac{7xy}{y^2}\)
=\(\dfrac{7xy-6x}{y^2}\)
c) \(\dfrac{1}{ab}\sqrt{a^6\left(a-b\right)^2}\) (a<b<0)
=\(\dfrac{1}{ab}\sqrt{a^6}\sqrt{\left(a-b\right)^2}\)
=\(\dfrac{1}{ab}\left(-a^3\right)\left(b-a\right)\) (vì a<b<0)
=\(\dfrac{\left(a-b\right)a^3}{a-b}\)
=a3
Cảm ơn bạn Thu Trang nhiều nhé, sau này có gì giúp đỡ nhau nha. 
# Bài 1
* Ta cm BĐT sau \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\) (1) bằng cách biến đổi tương đương
* Với \(x,y>0\) áp dụng (1) ta có
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{\left(\sqrt{x}\right)^2}+\dfrac{1}{\left(\sqrt{y}\right)^2}\ge\dfrac{1}{2}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\)
Mà \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\)
\(\Rightarrow\) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\le1\) \(\Leftrightarrow\) \(0< \dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\le1\) (I)
* Ta cm BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) với \(a,b>0\) (2)
Áp dụng (2) với x , y > 0 ta có
\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\ge\dfrac{4}{\sqrt{x}+\sqrt{y}}\) (II)
* Từ (I) và (II) \(\Rightarrow\) \(\dfrac{4}{\sqrt{x}+\sqrt{y}}\le1\)
\(\Leftrightarrow\) \(\sqrt{x}+\sqrt{y}\ge4\)
Dấu "=" xra khi \(x=y=4\)
Vậy min \(\sqrt{x}+\sqrt{y}=4\) khi \(x=y=4\)
Áp dụng bất đẳng thức Cauchy-Schwarz ta có:
\(B=\dfrac{1}{2-x}+\dfrac{1}{x}\ge\dfrac{\left(1+1\right)^2}{2-x+x}=\dfrac{4}{2}=2\)
Dấu "=" xảy ra khi: \(x=1\)
p/s Mình nghĩ đề phải là \(0< x\le1\) nhé
áp dụng bunhia
\(\left[\left(\sqrt{\dfrac{2}{1-x}}\right)^2+\left(\sqrt{\dfrac{1}{x}}\right)^2\right]\left[\left(\sqrt{1-x}\right)^2+\left(\sqrt{x}\right)^2\right]\)
\(\ge\left(\sqrt{\dfrac{2}{1-x}}.\sqrt{1-x}+\sqrt{\dfrac{1}{x}}.\sqrt{x}\right)^2\)
\(\Leftrightarrow\left(\dfrac{2}{1-x}+\dfrac{1}{x}\right)\left(1\right)\ge\left(\sqrt{2}+\sqrt{1}\right)^2\)
\(\Rightarrow B\ge\left(\sqrt{2}+1\right)^2\)
dấu = xảy ra khi \(\dfrac{\dfrac{2}{1-x}}{1-x}=\dfrac{\dfrac{1}{x}}{x}\Leftrightarrow x=\sqrt{2-1}\)
cau a) =\((\dfrac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{\sqrt{x}+2}{(\sqrt{x}+1)^{2}})\)x\(\dfrac{(\sqrt{x}-1)^{2}}{2} \)
=\(\dfrac{(\sqrt{x}-2)(\sqrt{x}+1)-(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)^{2}}\)x\(\dfrac{(\sqrt{x}-1)^{2}}{2} \)
=\(\dfrac{-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)^{2}}\)x\(\dfrac{(\sqrt{x}-1)^{2}}{2} \)
=\(\dfrac{-(\sqrt{x})(\sqrt{x}-1)}{(\sqrt{x}+1)^{2}}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky với $x>0; 1-x> 0$ ta có:
\(\left(\frac{2}{1-x}+\frac{1}{x}\right)[(1-x)+x]\geq (\sqrt{2}+1)^2\)
\(\Rightarrow \frac{2}{1-x}+\frac{1}{x}\geq \frac{(\sqrt{2}+1)^2}{1-x+x}=(\sqrt{2}+1)^2\)
Vậy \(y_{\min}=(\sqrt{2}+1)^2\)
Dấu bằng xảy ra khi \(\frac{\sqrt{2}}{1-x}=\frac{1}{x}\Rightarrow x=\sqrt{2}-1\)
\(A=\dfrac{9x}{2-x}+\dfrac{2}{x}=\dfrac{9x}{2-x}+\dfrac{2-x+x}{x}=\dfrac{9x}{2-x}+\dfrac{2-x}{x}+1\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{9x}{2-x}+\dfrac{2-x}{x}\) ≥ \(2\sqrt{\dfrac{9x}{2-x}.\dfrac{2-x}{x}}=2.3=6\)
⇔ \(\dfrac{9x}{2-x}+\dfrac{2-x}{x}+1\text{≥}6+1=7\)
⇒ \(A_{Min}=7."="\text{⇔}x=\dfrac{1}{2}\)
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
\(\dfrac{9x}{2-x}+\dfrac{2}{x}=\dfrac{9x}{2-x}+\dfrac{2-x}{x}+1\ge2\sqrt{\dfrac{9x}{2-x}.\dfrac{2-x}{x}}+1=2.3+1=7\)
GTNN của A là 7 khi x=0,5
a: \(P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{1}\cdot\dfrac{\sqrt{x}-1}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b: \(P=-x+\sqrt{x}-\dfrac{1}{4}+\dfrac{1}{4}=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)
Dấu = xảy ra khi x=1/4
c: Để P>0 thì -(căn x-1)>0
=>căn x-1<0
=>0<x<1
`A=(9(x-2)+18)/(2-x)+2/x`
`=-9+18/(2-x)+2/x`
`=-9+2(9/(2-x)+1/x)`
Áp dụng bđt cosi-schwarts ta có:
`9/(2-x)+1/x>=(3+1)^2/(2-x+x)=8`
`=>A>=16-9=7`
Dấu "=" xảy ra khi `3/(2-x)=1/x`
`<=>3x=2-x`
`<=>4x=2<=>x=1/2(tm)`
b
`y=x/(1-x)+5/x`
`=(x-1+1)/(1-x)+5/x`
`=1/(1-x)+5/x-1`
Áp dụng cosi-schwarts ta có:
`1/(1-x)+5/x>=(1+sqrt5)^2/(1-x+x)=(1+sqrt5)^2=6+2sqrt5`
`=>y>=5+2sqrt5`
Dấu "=" xảy ra khi `1/(1-x)=sqrt5/x`
`<=>x=sqrt5-sqrt5x`
`<=>x(1+sqrt5)=sqrt5`
`<=>x=sqrt5/(sqrt5+1)=(sqrt5(sqrt5-1))/(5-1)=(5-sqrt5)/4`
`c)C=2/(1-x)+1/x`
Áp dụng bđt cosi schwarts ta có:
`C>=(sqrt2+1)^2/(1-x+x)=3+2sqrt2`
Dấu "=" xảy ra khi `sqrt2/(1-x)=1/x`
`<=>sqrt2x=1-x`
`<=>x(sqrt2+1)=1`
`<=>x=1/(sqrt2+1)=(sqrt2-1)/(2-1)=sqrt2-1`
cho hỏi là câu a sao lại thế ở mấy dòng đầu ạ