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a) Ta có: \(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{3a+3\sqrt{a}-\left(a-\sqrt{a}+2\sqrt{a}-2\right)}{\sqrt{a}}\)
\(=2+\dfrac{3a+3\sqrt{a}-a+\sqrt{a}-2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\left(a+2\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
b) Ta có: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2\left(a-\sqrt{a}+1\right)}{\sqrt{a}}>0\forall a\) thỏa mãn ĐKXĐ
hay P>6
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Lời giải:
$A=4n^3-36n^2+56n=4n(n^2-9n+14)=4n(n-2)(n-7)$
Vì $n-2,n-7$ khác tính chẵn lẻ nên 1 trong 2 số sẽ là số chẵn.
$\Rightarrow n(n-2)(n-7)\vdots 2$
$\Rightarrow =4n(n-2)(n-7)\vdots 8(*)$
Lại có:
Nếu $n$ chia hết cho $3$ thì $A=4n(n-2)(n-7)\vdots 3$
Nếu $n$ chia $3$ dư $1$ thì $n-7\vdots 3\Rightarrow A\vdots 3$
Nếu $n$ chia $3$ dư $2$ thì $n-2\vdots 3\Rightarrow A\vdots 3$
Tóm lại $A\vdots 3(**)$
Từ $(*); (**)$ mà $(3,8)=1$ nên $A\vdots 24$.
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: ΔABC vuông tại A
nên \(\widehat{B}+\widehat{C}=90^0\)
hay \(\widehat{C}=30^0\)
Xét ΔABC vuông tại A có
\(AB=BC\cdot\sin30^0\)
\(\Leftrightarrow BC=4:\dfrac{1}{2}=8\left(cm\right)\)
Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:
\(AB^2+AC^2=BC^2\)
\(\Leftrightarrow AC^2=8^2-4^2=48\)
hay \(AC=4\sqrt{3}\left(cm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}\)=\(\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
7. Ta có: \(\left(x+\sqrt{x^2+3}\right)\left(\sqrt{x^2+3}-x\right)=x^2+3-x=3\)
\(\Rightarrow\sqrt{x^2+3}-x=y+\sqrt{y^2+3}\Rightarrow x+y=\sqrt{x^2+3}-\sqrt{y^2+3}\left(1\right)\)
Lại có \(\left(y+\sqrt{y^2+3}\right)\left(\sqrt{y^2+3}-y\right)=y^2+3-y=3\)
\(\Rightarrow\sqrt{x^2+3}+x=\sqrt{y^2+3}-y\Rightarrow x+y=\sqrt{y^2+3}-\sqrt{x^2+3}\left(2\right)\)
Lấy \(\left(1\right)+\left(2\right)\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)
9. Ta có: \(\sqrt{55+\sqrt{109}}-\sqrt{55-\sqrt{109}}\)
\(=\sqrt{\dfrac{110+2\sqrt{109}}{2}}-\sqrt{\dfrac{110-2\sqrt{109}}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{109}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{109}-1\right)^2}{2}}=\dfrac{\sqrt{109}+1}{\sqrt{2}}-\dfrac{\sqrt{109}-1}{\sqrt{2}}\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
Lại có: \(\dfrac{\sqrt{2-\sqrt{4y-y^2}}}{y-2}.\sqrt{4+2\sqrt{4y-y^2}}\)
\(=\dfrac{\sqrt{4-2\sqrt{y\left(4-y\right)}}}{\sqrt{2}\left(y-2\right)}.\sqrt{\left(\sqrt{y}\right)^2+2\sqrt{y\left(4-y\right)}+\left(\sqrt{4-y}\right)^2}\)
\(\dfrac{\sqrt{\left(\sqrt{y}\right)^2-2\sqrt{y\left(4-y\right)}+\left(\sqrt{4-y}\right)^2}}{\sqrt{2}\left(y-2\right)}.\sqrt{\left(\sqrt{y}+\sqrt{4-y}\right)^2}\)
\(=\dfrac{\sqrt{\left(\sqrt{y}-\sqrt{4-y}\right)^2}}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|=\dfrac{\left|\sqrt{y}-\sqrt{4-y}\right|}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|\)
Vì \(y>2\Rightarrow\left\{{}\begin{matrix}\sqrt{y}>\sqrt{2}\\\sqrt{4-y}< \sqrt{2}\end{matrix}\right.\Rightarrow\sqrt{y}-\sqrt{4-y}>0\)
\(\Rightarrow\dfrac{\left|\sqrt{y}-\sqrt{4-y}\right|}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|=\dfrac{\left(\sqrt{y}-\sqrt{4-y}\right)\left(\sqrt{y}+\sqrt{4+y}\right)}{\sqrt{2}\left(y-2\right)}\)
\(=\dfrac{y-\left(4-y\right)}{\sqrt{2}\left(y-2\right)}=\dfrac{2y-4}{\sqrt{2}\left(y-2\right)}=\dfrac{2\left(y-2\right)}{\sqrt{2}\left(y-2\right)}=\sqrt{2}\)
\(\Rightarrow\dfrac{\sqrt{2-\sqrt{4y-y^2}}}{y-2}.\sqrt{4+2\sqrt{4y-y^2}}=\sqrt{55+\sqrt{109}}-\sqrt{55-\sqrt{109}}\)
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a: Δ=(m-2)^2-4(m-4)
=m^2-4m+4-4m+16
=m^2-8m+20
=m^2-8m+16+4
=(m-2)^2+4>=4>0
=>Phương trình luôn có 2 nghiệm pb
b: x1^2+x2^2
=(x1+x2)^2-2x1x2
=(m-2)^2-2(m-4)
=m^2-4m+4-2m+8
=m^2-6m+12
=(m-3)^2+3>=3
Dấu = xảy ra khi m=3
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b: Xét ΔAHB vuông tại H có HE là đường cao ứng với cạnh huyền AB
nên \(AE\cdot AB=AH^2\left(1\right)\)
Xét ΔAHC vuông tại H có HF là đường cao ứng với cạnh huyền AC
nên \(AF\cdot AC=AH^2\left(2\right)\)
Từ (1) và (2) suy ra \(AE\cdot AB=AF\cdot AC\)
\(9,=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2}{4}=\dfrac{1}{2}\\ 10,=\dfrac{\sqrt{5}+2+\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=2\sqrt{5}\\ 11,=\dfrac{8+6\sqrt{2}-8+6\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{12\sqrt{2}}{-2}=-6\sqrt{2}\\ 12,=\dfrac{2+\sqrt{6}+2-\sqrt{6}}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}=\dfrac{4}{-2}=-2\\ 13,=\sqrt{2}-1+\sqrt{2}+3=2\sqrt{2}+2\\ 14,=2-\sqrt{3}+\sqrt{3}-1=1\\ 15,=3-\sqrt{5}+\sqrt{5}-2=1\)