chứng minh √x/x+√x+1<1/3
chứng minh √x/x+√x+1<1/3
Cho \(A=\dfrac{\sqrt{x^3}-x}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x-1}-\sqrt{x}}+\dfrac{1}{\sqrt{x-1}+\sqrt{x}}\).
a) Rút gọn biểu thức A.
b) Tính giá trị của biểu thức A khi \(\left|2020-x\right|=2015\).
a, Rút gọn:
\(A=\dfrac{\sqrt{x^3}-x}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x-1}-\sqrt{x}}+\dfrac{1}{\sqrt{x-1}+\sqrt{x}}\\ =\dfrac{x\sqrt{x}-x}{\sqrt{x}-1}-\left(\sqrt{x-1}+\sqrt{x}\right)-\left(\sqrt{x-1}-\sqrt{x}\right)\\ =\dfrac{x\sqrt{x}-x}{\sqrt{x}-1}-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}+\sqrt{x}\\ =\dfrac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-2\sqrt{x-1}\\ =x-2\sqrt{x-1}\)
\(\dfrac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\)
\(\dfrac{\left(2+\sqrt{3}\right).\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\)
= \(\dfrac{\left(2+\sqrt{3}\right).\sqrt{2-\sqrt{3}}.\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}.}\sqrt{2+\sqrt{3}}}\)
= \(\dfrac{\left(2+\sqrt{3}\right).\sqrt{\left(2-\sqrt{3}\right).\left(2-\sqrt{3}\right)}}{\sqrt{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}}\)
= \(\dfrac{\left(2+\sqrt{3}\right).\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)
= \(\dfrac{\left(2+\sqrt{3}\right).\left|2-\sqrt{3}\right|}{\sqrt{4-3}}\)
= \(\dfrac{\left(2+\sqrt{3}\right).\left(2-\sqrt{3}\right)}{\sqrt{1}}\)
= \(\dfrac{2^2-\left(\sqrt{3}\right)^2}{1}\)
= \(4-3\)
= 1
a)Rút gọn biểu thứcP=\((\dfrac{\sqrt{a-2}+2}{3})(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}):(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\)
b)Cho các số dương a,b thỏa mãn a+b=\(\sqrt{2017-a^2}+\sqrt{2017-b^2}.Chứng\) Minh \(a^2+b^2=2017\)
a) điều kiện xác định : \(a>2;a\ne11\)
ta có : \(P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)
\(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{3\sqrt{a-2}+1}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}-\dfrac{1}{\sqrt{a-2}}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}\left(3-\sqrt{a-2}\right)+a+7}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{3\sqrt{a-2}+1-\sqrt{a-2}+3}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{3\left(\sqrt{a-2}+3\right)}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{2\sqrt{a-2}+4}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{3}{\left(3-\sqrt{a-2}\right)}\right)\left(\dfrac{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}{2\left(\sqrt{a-2}+2\right)}\right)\) \(\Leftrightarrow P=\dfrac{-\sqrt{a-2}}{2}\)
ta có : \(a+b=\sqrt{2017-a^2}+\sqrt{2017-b^2}\)
\(\Leftrightarrow\left(a+b\right)\left(\sqrt{2017-a^2}-\sqrt{2017-b^2}\right)=b^2-a^2\)
\(\Leftrightarrow b-a=\sqrt{2017-a^2}-\sqrt{2017-b^2}\)
\(\Leftrightarrow2b=2\sqrt{2017-a^2}\Leftrightarrow b^2=2017-a^2\Rightarrow\left(đpcm\right)\)
b, ta giả sử \(a+b=\sqrt{2017-a^2}+\sqrt{2017-b^2}\rightarrow a^2+b^2=2017\)
thật vậy khi ta thay \(a^2+b^2=2017\) vào biểu thức thì thấy thoả mãn
vậy dpcm
a)Tính giá trị biểu thức:p= \(\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{3}+\sqrt{2}}\)
b)Chứng minh rằng nếu a,b,c là các số dương thỏa mãn a+c =2b thì ta luôn có
\(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}=\dfrac{2}{\sqrt{a}+\sqrt{c}}\)
\(P=\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{3}+\sqrt{2}}\)
=\(\dfrac{\left(3+2\sqrt{2.3}+2\right)\sqrt{3-2\sqrt{3.2}+2}}{\sqrt{3}+\sqrt{2}}\)
=\(\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}+\sqrt{2}}\)
=\(\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
=\(3-2=1\)
ta có : \(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\ge2\sqrt{\dfrac{1}{\sqrt{ab}+\sqrt{ac}+\sqrt{bc}+b}}\)
\(\ge\dfrac{2}{\sqrt{a+b+c+b}}=\dfrac{2}{\sqrt{4b}}=\dfrac{2}{2\sqrt{b}}=\dfrac{1}{\sqrt{b}}=\dfrac{2}{\sqrt{a+c}}\ge\dfrac{2}{\sqrt{a}+\sqrt{b}}\)
dấu "=" xảy ra khi \(a=b=c\Leftrightarrow a+c=2b\Rightarrow\left(đpcm\right)\)
rut gon P
P=\(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(P=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1=\dfrac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-2\sqrt{a}-1+1=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)
tim GTNN cua
\(a-\sqrt{a}\)
\(a-\sqrt{a}=a-\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}\\ =\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\forall x\ge0\)
Dấu "=" xảy ra khi:
\(\sqrt{a}-\dfrac{1}{2}=0\\ \Leftrightarrow a=\dfrac{1}{4}\)
Vậy.............
Cho biểu thức: \(Q=\dfrac{1}{2\sqrt{x}-2}+\dfrac{1}{2\sqrt{x}+2}+\dfrac{x}{1-x}\)
a) Tìm điều kiện để Q có nghĩa
b) Rút gọn Q
c) Tìm giá trị của Q khi x = \(\dfrac{4}{9}\)
d) Tìm x để A = \(\dfrac{-1}{2}\)
e) Tìm những giá trị nguyên của x đẻ giá trị của Q nguyên.
Giúp mình câu e với nhé!
a) điều kiện xác định : \(x\ge0;x\ne1\)
b) ta có : \(Q=\dfrac{1}{2\sqrt{x}-2}+\dfrac{1}{2\sqrt{x}+2}+\dfrac{x}{1-x}\)
\(\Leftrightarrow Q=\dfrac{1}{2\left(\sqrt{x}-1\right)}+\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow Q=\dfrac{\sqrt{x}+1+\sqrt{x}-1-2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow Q=\dfrac{-2\sqrt{x}}{\sqrt{x}+1}\)
c) thay \(x=\dfrac{4}{9}\) vào \(Q\) ta có \(Q=\dfrac{-2\sqrt{\dfrac{4}{9}}}{\sqrt{\dfrac{4}{9}}+1}=\dfrac{-4}{5}\)
d) để \(A=\dfrac{-1}{2}\Leftrightarrow\dfrac{-2\sqrt{x}}{\sqrt{x}+1}=\dfrac{-1}{2}\Leftrightarrow-\sqrt{x}-1=-4\sqrt{x}\)
\(\Leftrightarrow3\sqrt{x}=1\Leftrightarrow\sqrt{x}=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{9}\)
e) ta có \(Q=\dfrac{-2\sqrt{x}}{\sqrt{x}+1}=\dfrac{-2\sqrt{x}-2+2}{\sqrt{x}+1}=-2+\dfrac{2}{\sqrt{x}+1}\)
\(\Rightarrow\) \(Q\) nguyên \(\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\) nguyên \(\Leftrightarrow\sqrt{x}+1\) thuộc ước của \(2\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\in\left\{\pm1;\pm2\right\}\) \(\Leftrightarrow x\in\left\{0;1\right\}\)
biết \(cos^2\alpha-2sin^2\alpha=\dfrac{1}{4}\). tính \(\alpha\)
+) ta có nếu : \(cos\alpha=0\) thì \(pt\Leftrightarrow sin^2x=-\dfrac{1}{8}\left(vôlí\right)\) (vì \(sin^2\alpha+cos^2\alpha=1\)) \(\Rightarrow cosx\ne0\)
+) ta có : \(cos^2\alpha-2sin^2\alpha=\dfrac{1}{4}\) \(\Leftrightarrow1-2tan^2\alpha=\dfrac{1}{4}\left(\dfrac{1}{cos^2\alpha}\right)\)
\(\Leftrightarrow1-2tan^2\alpha=\dfrac{1}{4}\left(1+tan^2\alpha\right)\) \(\Leftrightarrow\dfrac{9}{4}tan^2\alpha-\dfrac{3}{4}\Leftrightarrow tan^2\alpha=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}tan\alpha=\dfrac{1}{\sqrt{3}}\\tan\alpha=\dfrac{-1}{\sqrt{3}}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}tan\alpha=tan30\\tan\alpha=tan\left(-30\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\alpha=30+k360\\\alpha=-30+k360\end{matrix}\right.\)
Giải phương trình:
a) \(x^2+x=36-12\sqrt{x+1}\)
b) \(4\sqrt{x+1}=x^2-5x+14\)
c) \(\dfrac{\left(\sqrt{x}-1\right)^2}{2}=2\sqrt{x}-4-\sqrt{x-9}\)
Câu b : \(x^2-5x+14=4\sqrt{x+1}\) ( ĐK : \(x\ge-1\) )
\(\Leftrightarrow x^2-5x+14-4\sqrt{x+1}=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left[\left(x+1\right)-4\sqrt{x+1}+4\right]=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
Do : \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(\sqrt{x+1}-2\right)^2\ge0\end{matrix}\right.\Rightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(\sqrt{x+1}-2\right)^2=0\end{matrix}\right.\Leftrightarrow x=3\)
Vậy \(x=3\)
a. Ta có : x2 + x = 36 - 12\(\sqrt{x+1}\)
⇌ x2 + 2x + 1 = 36 - 12\(\sqrt{x+1}\) + x + 1
⇌ (x+1)2 = ( \(\sqrt{x+1}\) -6)2
⇌ (x+1)2 - ( \(\sqrt{x+1}\) -6)2 = 0
còn lại tự làm nha