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\(A=\dfrac{x^2+2x+17}{2\left(x+1\right)}=\dfrac{x^2-6x+9+8x+8}{2\left(x+1\right)}\\ =\dfrac{\left(x-3\right)^2+8\left(x+1\right)}{2\left(x+1\right)}=\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}+4\)
Vì \(\left(x-3\right)^2\ge0;2\left(x+1\right)>0\) (do \(x>-1\))
\(\Rightarrow\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}\ge0\Leftrightarrow A=\dfrac{\left(x-3\right)^2}{2\left(x+1\right)}+4\ge4\)
Dấu "=" xảy ra khi \(x=3\)
Vậy....
\(a,P=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
\(=\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)+1\)
\(=\left(x^2+5x+5\right)^2-1+1\)
\(=\left(x^2+5x+5\right)^2\ge0\forall x\)
Vậy \(P\ge0\forall x\)
\(b,P=\left(x^2+5x+5\right)^2\left(cmt\right)\)
Thay \(x=\frac{\sqrt{7}-5}{2}\)vào P ta được
\(P=\left(\left(\frac{\sqrt{7}-5}{2}\right)^2+5.\frac{\sqrt{7}-5}{2}+5\right)^2\)
\(=\left(\frac{7-10\sqrt{7}+25}{4}+\frac{10\sqrt{7}-50}{4}+\frac{20}{4}\right)^2\)
\(=\left(\frac{32-10\sqrt{7}+10\sqrt{7}-50+20}{4}\right)^2\)
\(=\left(\frac{2}{4}\right)^2\)
\(=\frac{1}{4}\)
a,
P=(x+1)(x+2)(x+3)(x+4)+1
P=[(x+1).(x+4)].[(x+2).(x+3)]+1
P=(x^2+5x+4)(x^2+5x+6)+1
P=[(x^2+5x+5)-1].[(x^2+5x+5)+1]+1
P=(x^2+5x+5)^2-1+1
P=\(\left(x^2+5x+5\right)^2\) \(\ge\)0 với mọi x
Câu b thì thay x vào rồi bấm máy ra ra kết quả
a. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}\)
b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)
=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)
check giùm mik
\(B=\dfrac{xy}{xy}+\dfrac{\left(x-y\right)x}{x\left(x-y\right)}-\dfrac{y\left(x-y\right)}{y\left(x-y\right)}=1\)
Bài 1 :
a )\(A=\frac{3-\sqrt{3}}{\sqrt{3}-1}+\frac{\sqrt{35}-\sqrt{15}}{\sqrt{5}}-\sqrt{28}\)
\(A=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{5}}-\sqrt{28}\)
\(A=\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{28}\)
\(A=\sqrt{7}-\sqrt{28}\)
\(A=\sqrt{7}-2\sqrt{7}=-\sqrt{7}\)
Vậy \(A=-\sqrt{7}\)
b)\(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\left(a,b>0;a\ne b\right)\)
\(B=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\)
\(B=\left(\sqrt{a}+\sqrt{b}\right).\frac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(B=a-b\)
Vậy \(B=a-b\left(a,b>0;a\ne b\right)\)
_Minh ngụy_
Bài 2 :
a )\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\left(x>0\right)\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Vậy \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)
b) \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)
Ta có : \(B>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)
Vì : \(\sqrt{x}\ge0\forall x\Rightarrow\)để \(B>O\)cần \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)( thỏa mãn \(x>0\))
Vậy \(x>1\)thì \(B>0\)
_Minh ngụy_
Lời giải:
ĐKXĐ: $x>0; x\neq 1$
\(P=\frac{1}{\sqrt{x}+1}+\frac{x}{\sqrt{x}(1-\sqrt{x})}=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\)
\(=\frac{1-\sqrt{x}+\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(1-\sqrt{x})}=\frac{x+1}{1-x}\)
b. Khi $x=\frac{1}{\sqrt{2}}$ thì:
\(P=\frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{2}\)
\(A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{2x-6\sqrt{x}+x+\sqrt{x+}3\sqrt{x}+3+3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{3x-13\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
Áp dụng BĐT Cauchy ta có:
P = \(x^2+y^2+\dfrac{33}{xy}\) \(\ge\) \(\dfrac{\left(x+y\right)^2}{2}+\dfrac{33}{x+y}\) = \(\dfrac{4^2}{2}+\dfrac{33}{4}=\dfrac{65}{4}\)
=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}x+y=4\\x=y\end{matrix}\right.\) <=> \(x=y=2\)
Vậy ...............................
\(A=\dfrac{1-\sqrt{x}+x}{\sqrt{x}}\\ =\dfrac{\left(1-\sqrt{x}+x\right)\sqrt{x}}{x}\\ =\dfrac{\sqrt{x}-x+x\sqrt{x}}{x}\)
lm tiếp...
bạn làm gùm mk luôn vs