a) Cho a ≥ 0, b ≥ 0. Chứng minh bất đẳng thức Cauchy:
b) Cho a, b, c > 0. Chứng minh rằng:
c) Cho a, b > 0 và 3a + 5b = 12. Tìm giá trị lớn nhất của tích P = ab.
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\(ĐKXĐ:\hept{\begin{cases}x-4\ne0\\3-\sqrt{x}\ne0\\x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne4\\\sqrt{x}\ne3\\x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne4\\x\ne9\\x\ge0\end{cases}}\)
Rút gọn
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-1\right):\left(\frac{4-x}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-\frac{x-4}{x-4}\right):\left(\frac{4-x}{x+2\sqrt{x}-3\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{x-2\sqrt{x}-x+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-\left(x+4\sqrt{x}+4\right)-\left(x-6\sqrt{x}+9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-x^2-4\sqrt{x}-4-x^2+6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{-2x^2-x-2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\frac{\left(-2\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(-2x^2-x-2\sqrt{x}-9\right)}\)
\(D=\frac{\left(-2\right)\left(\sqrt{x}-3\right)\left(x^2-4\right)}{\left(x-4\right)\left(-2x^2-x-2\sqrt{x}-9\right)}\)
Sai thui nhé !!!!
\(\text{Đat: A=biêu thuc cần tính}\Rightarrow\sqrt{2}A=\sqrt{28+10\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(\Rightarrow2\sqrt{A}=\sqrt{5^2+2.5\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1^2}\)
\(\Rightarrow2\sqrt{A}=\sqrt{\left(5+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=4+2\sqrt{3}\Rightarrow A=\sqrt{8}+\sqrt{6}\)
\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}\)\(-\sqrt{3-2.\sqrt{3}.\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)\(-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=|\sqrt{3}+\sqrt{2}|-|\sqrt{3}-\sqrt{2}|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
#)Giải :
Ta có : \(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=4a^2b^2-\left(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2\right)\)
\(=4a^2b^2-a^4-b^4-c^4-2a^2b^2+2b^2c^2+2c^2a^2\)
\(=2a^2b^2-a^4-b^4-c^4+2b^2c^2+2c^2a^2\)
\(=-a^4+2a^2b^2-b^4-c^2+2b^2c^2+2c^2a^2\)
\(=-\left(a^2-b^2\right)^2-c^4+2b^2c^2+2c^2c^2\)
\(=-\left(a^2-b^2\right)^2-c\left(c^2-2b^2+2a^2\right)>0\)
\(\Rightarrow A>0\left(đpcm\right)\)
\(A=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
=>\(A=\left(a+b-c\right)\left(a+b+c\right)\left(c-a+b\right)\left(a-b+c\right)\)
do a,b,c la do dai 3 canh tam giac => A>0=>dpcm
#)Giải :
\(x^2+y^2+z^2=4x-2y+6z-14\)
\(\Leftrightarrow x^2+y^2+z^2-4x-2y+6z-14=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-2=0\\y+1=0\\z-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\\z=3\end{cases}}}\)
Vậy x = 2; y = -1; z = 3
a) Kẻ đường cao : BH , AI , CK
Ta có: sinA = BH / c ; sinB = AI / c
=> sinA/sinB = BH / AI (1)
Mà BH = a.sinC ; AI = b.sinC
=> BH/AI = a/b (2)
Từ (1) và (2)
=> sinA/sinB = a/b => a/sinA = b/sinB
CMTT ta có:
b/sinB = c/sinC ; c/sinC = a/sinA
Từ đó suy ra a /sinA = b / sinB = c /sinC
Có \(a+1+1\ge3\sqrt[3]{a}\)
\(b+1+1\ge3\sqrt[3]{b}\)
\(\Rightarrow a+b+1+1+1+1\ge3\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\)
\(\Rightarrow3\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\le6\)
\(\Rightarrow\sqrt[3]{a}+\sqrt[3]{b}\le2\)
"=" tại a=b=1