\(cho\sqrt{4a-b^2}-\sqrt{b+2}=\sqrt{4a^2+b}Tínha^2+b^2\)
Những câu hỏi liên quan
Tìm GTNN:
A = \(\sqrt{m^2+2m+1}+\sqrt{m^2-2m+1}\)
B = \(\sqrt{4a^2-4a+1}+\sqrt{4a^2-12a+9}\)
A=|m+1|+|m-1|=|m+1|+|1-m|>=|m+1+1-m|=2
Dấu = xảy ra khi -1<=m<=1
B=|2a-1|+|2a-3|=|2a-1|+|3-2a|>=|2a-1+3-2a|=2
Dấu = xảy ra khi 1/2<=a<=3/2
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a)Adfrac{1}{2a-1}sqrt{5a^2left(1-4a+4a^2right)} với adfrac{1}{2} b)Adfrac{sqrt{x-2sqrt{x-1}}}{sqrt{x-1}-1}+dfrac{sqrt{x+2sqrt{x-1}}}{sqrt{x-1+1}} với x2 c)dfrac{a+b}{b^2}sqrt{dfrac{a^2b^4}{a^2+2ab+b^2}} với a+b0; b≠0d)Aleft(sqrt{dfrac{1-asqrt{a}}{1-sqrt{a}}}+sqrt{a}right)left(dfrac{1-sqrt{a}}{1-a}right)^2 với a≥0; a≠1e)Adfrac{x-1}{sqrt{y}-1}sqrt{dfrac{left(y-2sqrt{y}+1right)}{left(x-1right)^4}} với x≠1; y≠1; yof)Asqrt{dfrac{m}{1-2x+x^2}}sqrt{dfrac{4m-8mx+4mx^2}{81}} với m0; x≠4g)Aleft(dfrac{...
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a)A=\(\dfrac{1}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\) với a>\(\dfrac{1}{2}\)
b)A=\(\dfrac{\sqrt{x-2\sqrt{x-1}}}{\sqrt{x-1}-1}\)+\(\dfrac{\sqrt{x+2\sqrt{x-1}}}{\sqrt{x-1+1}}\) với x>2
c)\(\dfrac{a+b}{b^2}\)\(\sqrt{\dfrac{a^2b^4}{a^2+2ab+b^2}}\) với a+b>0; b≠0
d)A=\(\left(\sqrt{\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\) với a≥0; a≠1
e)A=\(\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)}{\left(x-1\right)^4}}\) với x≠1; y≠1; y>o
f)A=\(\sqrt{\dfrac{m}{1-2x+x^2}}\)\(\sqrt{\dfrac{4m-8mx+4mx^2}{81}}\) với m>0; x≠4
g)A=\(\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\)\(\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\) với x>0; x≠4
h)\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)\(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\) với a≥0; a≠1
a: \(A=\dfrac{1}{2a-1}\cdot\sqrt{5a^2}\cdot\left|2a-1\right|\)
\(=\dfrac{2a-1}{2a-1}\cdot a\sqrt{5}=a\sqrt{5}\)(do a>1/2)
b: \(A=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x-1}+1}\)
\(=\dfrac{\left|\sqrt{x-1}-1\right|}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}+1}\)
\(=\dfrac{\sqrt{x-1}-1}{\sqrt{x-1}-1}+1=1+1=2\)
c:
\(=\dfrac{a+b}{b^2}\cdot\dfrac{ab^2}{a+b}=a\)
d: Sửa đề: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\left(\dfrac{1}{1+\sqrt{a}}\right)^2\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)
e:
\(A=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{x-1}\)
f:
\(A=\sqrt{\dfrac{m}{\left(1-x\right)^2}\cdot\dfrac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\sqrt{\dfrac{m}{\left(x-1\right)^2}\cdot\dfrac{4m\left(x-1\right)^2}{81}}\)
\(=\sqrt{\dfrac{4m^2}{81}}=\dfrac{2m}{9}\)
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1) cho: 4a^3-3a+(b-1)\(\sqrt{2b+1}\)=0
biết \(\frac{-1}{2}\)=<b=<0 . Cmr: \(\sqrt{2b+1}\)+2a=0
2)cho (4a^2+1)a+(b-3)\(\sqrt{5-2b}\)=0
biết a>=0 Cmr: 2b+4a^2=5
cho a, b,c > 0 , \(a^2+b^2=2\) . tìm GTLN của
M = \(a\sqrt{9b\left(4a+5b\right)}+b\sqrt{9a\left(4b+5a\right)}\)
2M\(\le\)a(9b+4a+5b)+b(9a+4b+5a) (AM-GM)
=4(a2+b2)+28ab\(\le\)4(a2+b2)+14(a2+b2) (AM-GM)
=36 (do a2+b2=2)
=> M \(\le\)18
Dấu bằng có <=> a=b=1
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EM ĐAG CẦN GẤP GIÚP EM VỚI.
tìm GTNN của biểu thức
A=\(\sqrt{m^2+2m+1}\sqrt{m^2-2m+1}\)
B=\(\sqrt{4a^2-4a+1}+\sqrt{4a^2-12a+9}\)
Em thử nha!Sai thì thôi:((
\(A=\left|m+1\right|+\left|m-1\right|=\left|m+1\right|+\left|1-m\right|\ge\left|m+1+1-m\right|=2\)
Dấu"=" xảy ra khi \(\left(m+1\right)\left(1-m\right)\ge0\Leftrightarrow-m^2+1\Leftrightarrow-1\le m\le1\)
\(B=\sqrt{\left(2a\right)^2-2.2a.1+1}+\sqrt{4a^2-2.2a.3+9}\)
\(=\left|2a-1\right|+\left|2a-3\right|=\left|2a-1\right|+\left|3-2a\right|\ge2\)
Dấu "=" xảy ra khi...
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Cho a,b>0 tm: a+b=4ab
CMR: \(\frac{\sqrt{a^2+4b^2}}{ab}+\frac{\sqrt{b^2+4a^2}}{ab}\ge4\sqrt{5}\)
Câu 1: Cho A = (sqrt(x) + 1)/(sqrt(x) - 1) B = (sqrt(x) + 2)/(sqrt(x) - 2) - 3/(sqrt(x) + 2) + 12/(4 - x) với x >= 0 x ne1; x = 4
a) Tính giá trị biểu thức A khi x = 16 .
b) Chứng minh B = (sqrt(x) - 1)/(sqrt(x) - 2)
c) Biết P =A.B Tính giá trị nguyên của x để P lớn nhất.
a: Khi x=16 thì \(A=\dfrac{4+1}{4-1}=\dfrac{5}{3}\)
b: \(P=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{x-4}=\dfrac{x+\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=1+\dfrac{3}{\sqrt{x}-2}\)
Để P lớn nhất thì căn x-2=1
=>căn x=3
=>x=9
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cho a,b,c >0. chứng minh \(\frac{a}{\sqrt{4a^2+5bc}}+\frac{b}{\sqrt{4b^2+5ac}}+\frac{c}{\sqrt{4c^2+5ab}}\le1.\)
Cho a>b>0 .So sánh \(\sqrt{4a+1}-2\sqrt{a}\) và \(\sqrt{4b+1}-2\sqrt{b}\)
\(\sqrt{4a+1}-2\sqrt{a}=\frac{4a+1-4a}{\sqrt{4a+1}+2\sqrt{a}}=\frac{1}{\sqrt{4a+1}+2\sqrt{a}}\)
\(\sqrt{4b+1}-2\sqrt{b}=\frac{1}{\sqrt{4b+1}+2\sqrt{b}}\)
Mà \(a>b\Rightarrow\left\{{}\begin{matrix}\sqrt{4a+1}>\sqrt{4b+1}\\2\sqrt{a}>2\sqrt{b}\end{matrix}\right.\) \(\Rightarrow\sqrt{4a+1}+2\sqrt{a}>\sqrt{4b+1}+2\sqrt{b}\)
\(\Rightarrow\frac{1}{\sqrt{4a+1}+2\sqrt{a}}< \frac{1}{\sqrt{4b+1}+2\sqrt{b}}\)
\(\Rightarrow\sqrt{4a+1}-2\sqrt{a}< \sqrt{4b+1}-2\sqrt{b}\)