cho B =\(\dfrac{4\sqrt{a}}{2a+1}\) so sánh B với 2
Bài 5: Bảng căn bậc hai
ĐKXĐ: a>=0
\(B-2=\dfrac{4\sqrt{a}-4a-2}{2a+1}=\dfrac{-\left(4a-4\sqrt{a}+2\right)}{2a+1}\)
\(=\dfrac{-\left(2\sqrt{a}-1\right)^2-1}{2a+1}< 0\)
=>B<2
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\(\sqrt{7x}+1=8x-3+\sqrt{2-x}\)
chứng minh rằng: điều kiện:a> hoặc =0 , b>0
a,\(\dfrac{\sqrt{ab}-b}{\sqrt{b}}-\sqrt{\dfrac{a}{b}}< 0\)
Cho a,b,c là số dương thỏa mãn a+b+c=3. Chứng minh rằng: \(a^2b+b^2c+c^2a\ge\dfrac{9a^2b^2c^2}{1+2a^2b^2c^2}\)
Lời giải:
Ta có: \(a^2b+b^2c+c^2a\geq \frac{9a^2b^2c^2}{1+2a^2b^2c^2}\)
\(\Leftrightarrow (a^2b+b^2c+c^2a)(1+2a^2b^2c^2)\geq 9a^2b^2c^2\)
\(\Leftrightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\geq 3a^2b^2c^2(a+b+c)(*)\)
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Áp dụng BĐT AM-GM ta có:
\(a^2b+a^4b^3c^2+a^3b^2c^4\geq 3\sqrt[3]{a^9b^6c^6}=3a^3b^2c^2\)
\(b^2c+a^2b^4c^3+a^4b^3c^2\geq 3a^2b^3c^2\)
\(c^2a+a^3b^2c^4+a^2b^4c^3\geq 3a^2b^2c^3\)
Cộng theo vế:
\(\Rightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\geq 3a^2b^2c^2(a+b+c)\)
Vậy $(*)$ đúng
Do đó ta có đpcm
Dấu bằng xảy ra khi $a=b=c=1$
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không dung bảng số và máy tính hãy tính
a, tg83độ -cotg7 độ
b, sin\(_a\).cos\(a\) biết tg\(a\)+cotg\(a\)=3
a) ta có : \(tan83-cot7=\dfrac{sin83}{cos83}-\dfrac{cos7}{sin7}=\dfrac{sin83}{cos83}-\dfrac{sin83}{cos83}=0\)
b) ta có : \(tana+cota=3\Leftrightarrow\dfrac{sina}{cosa}+\dfrac{cosa}{sina}=3\)
\(\Leftrightarrow\dfrac{sin^2a+cos^2a}{sina.cosa}=3\Leftrightarrow\dfrac{1}{sina.cosa}=3\Leftrightarrow sina.cosa=\dfrac{1}{3}\)
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Cho M=\(\dfrac{(\sqrt{1+\sqrt{1-x^2})}(\sqrt{\left(1+x^2\right)}-\sqrt{\left(1-x^2\right)})}{2+\sqrt{1-x^2}}\)
Rút gọn M
P= \((\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1})\cdot(\dfrac{1-x}{\sqrt{2}})^2\)
(Với x≥0;x≠1)
a)Rút Gọn P
b)Chứng Minh rằng nếu 0<x<1 thì p>0
a) ta có : \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)
\(\Leftrightarrow P=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)
\(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)
\(\Leftrightarrow P=\left(\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\) \(\Leftrightarrow P=\left(\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\) \(\Leftrightarrow P=\sqrt{x}-x\)b) ta có : \(x< 1\Leftrightarrow x-1< 0\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)< 0\)
\(\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow x-\sqrt{x}< 0\Leftrightarrow\sqrt{x}-x>0\)
\(\Leftrightarrow P>0\left(đpcm\right)\)
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Tính
a) (sinα + cos α)2 + (sin α + cos α)2
\(=2\left(sina+cosa\right)^2\)
\(=2\left(sin^2a+cos^2a+2sina\cdot cosa\right)\)
\(=2\left(1+sin2a\right)=2+2sin2a\)
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chứnng minh rằng :
a, điều kiện : a> hoặc = 0, a#1\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\)
Ta có:
\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{a+\sqrt{a}+\sqrt{a}+1}{\sqrt{a}+1}\right)\left(\dfrac{\sqrt{a}-1-a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{a+2\sqrt{a}+1}{\sqrt{a}+1}.\dfrac{-\left(a-2\sqrt{a}+1\right)}{\sqrt{a}-1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.\dfrac{-\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}\)
\(=-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)\)
\(=-\left(a-1\right)\)
\(=1-a\)
\(\rightarrowđpcm\)
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\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\) \(=\left[1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right].\left[1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right]\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
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chứng minh răng :
a,\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{a-\sqrt{a}}{a-1}\right)=1-a\left(a>hoaăặc=0,a\right)\left(a#1\right)\)b, \(\dfrac{\sqrt{ab}-b}{\sqrt{b}}-\sqrt{\dfrac{a}{b}}< 0\left(a>hoac=0,b>0\right)\)
\(=\dfrac{\sqrt{ab}-b-\sqrt{a}}{\sqrt{b}}\)
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