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Bài 2
a, bạn tự vẽ
b, Hoành độ giao điểm tm pt
\(2x^2-2x+3=0\)
\(\Delta'=1-3.2=-5< 0\)
Vậy pt vô nghiệm hay (d) ko cắt (P)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\sqrt{\dfrac{\sqrt{5}}{8\sqrt{5}+3\sqrt{35}}}.\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\sqrt{\dfrac{\sqrt{5}}{8\sqrt{5}+3\sqrt{5}.\sqrt{7}}}.\left(3\sqrt{2}+\sqrt{2}.\sqrt{7}\right)\)
\(=\sqrt{\dfrac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}.\left[\sqrt{2}\left(3+\sqrt{7}\right)\right]\)
\(=\sqrt{\dfrac{1}{8+3\sqrt{7}}}.\left[\sqrt{2}\left(3+\sqrt{7}\right)\right]\)
\(=\dfrac{\sqrt{2}\left(3+\sqrt{7}\right)}{\sqrt{8+3\sqrt{7}}}\)
\(=\dfrac{\sqrt{2}.\sqrt{2}\left(3+\sqrt{7}\right)}{\sqrt{2}.\sqrt{8+3\sqrt{7}}}\) (Nhân \(\sqrt{2}\) cả tử và mẫu)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{\sqrt{16+6\sqrt{7}}}\)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{\sqrt{\left(3+\sqrt{7}\right)^2}}\)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{\left|3+\sqrt{7}\right|}\)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{3+\sqrt{7}}\)
\(=2=VP\left(dpcm\right)\)
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`A=1/(x+sqrtx)+(2sqrtx)/(x-1)-1/(x-sqrtx)`
`=(sqrtx-1+2x-sqrtx-1)/(sqrtx(x-1))`
`=(2x-2)/(sqrtx(x-1))`
`=2/sqrtx`
`b)A=1`
`<=>2/sqrtx=1`
`<=>sqrtx=2`
`<=>x=4(tm)`
![](https://rs.olm.vn/images/avt/0.png?1311)
1.4:
a: CH=16^2/24=256/24=32/3
BC=24+32/3=104/3
AC=căn 32/3*104/3=16/3*căn 13
b: BC=12^2/6=24
AC=căn 24^2-12^2=12*căn 3
CH=24-6=18
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bất đẳng thức Cosi ta có :
\(x^4+1\ge2x^2;x^2+1\ge\left|x\right|\Rightarrow x^4+3\ge4\left|x\right|\)
Tương tự : \(y^4+3\ge4\left|y\right|\)
\(\Rightarrow x^4+y^4+6\ge4\left(\left|x\right|+\left|y\right|\right)\left(1\right)\)
Từ (1) suy ra \(x^4+y^4+6\ge4\left(x-y\right)\Rightarrow P\le\dfrac{1}{4}\)
Dấu = xảy ra \(x=1;y=-1\)
Từ (1) suy ra \(x^4+y^4+6\ge4\left(y-x\right)\Rightarrow P\ge-\dfrac{1}{4}\)
Dấu = xảy ra \(x=-1;y=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\((n+1)^n-1=n[(n+1)^{n-1}+(n+1)^{n-2}+....+(n+1)+1]\)
Giờ ta chỉ cần cmr \((n+1)^{n-1}+(n+1)^{n-2}+...+(n+1)+1\vdots n\)
Thật vậy:
\((n+1)^{n-1}+(n+2)^{n-2}+...+(n+1)+1\equiv 1^{n-1}+1^{n-2}+...+1^1+1=n\equiv 0\pmod n\)
Do đó ta có đpcm.
Bài 2 em xem lại. Số $2^{n(2^n-1)}$ chỉ toàn ước có dạng $2^k$ với $k=0,1,..., n(2^n-1)$ trong khi đó $(2^n-1)^2$ là số lẻ.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{AC}{BC}:\dfrac{AB}{BC}=\dfrac{AC}{AB}=\tan\alpha\)
\(\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{AB}{BC}:\dfrac{AC}{BC}=\dfrac{AB}{AC}=\cot\alpha\)
\(\tan\alpha\cot\alpha=\dfrac{AC}{AB}\cdot\dfrac{AB}{AC}=1\)
\(\sin^2\alpha+\cos^2\alpha=\dfrac{AC^2}{BC^2}+\dfrac{AB^2}{BC^2}=\dfrac{AB^2+AC^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\left(pytago\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Phương trình hoành độ giao điểm d1 và d2:
\(-3x-7=2x+3\)
\(\Rightarrow-5x=10\Rightarrow x=-2\)
Thế vào \(y=-3x-7=-3.\left(-2\right)-7=-1\)
Vậy \(M\left(-2;-1\right)\)
Bài 1 :
a, \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
b, \(x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)
c, \(x^2-2005=\left(x-\sqrt{2005}\right)\left(x+\sqrt{2005}\right)\)
d, \(x^2-2\sqrt{2}x+2=\left(x-\sqrt{2}\right)^2\)
e, \(x^2-\sqrt{5}=\left(x-\sqrt{\sqrt{5}}\right)\left(x+\sqrt{\sqrt{5}}\right)\)
f, \(x^2-2\sqrt{12}x+12=\left(x-\sqrt{12}\right)^2\)
a, \(2\sqrt{x}-9=0\)ĐK : x > = 0
\(\Leftrightarrow2\sqrt{x}=9\Leftrightarrow\sqrt{x}=\frac{9}{2}\Leftrightarrow x=\frac{81}{4}\)
b, \(\sqrt{9-12x+4x^2}=0\Leftrightarrow\sqrt{\left(3-2x\right)^2}=0\)
\(\Leftrightarrow3-2x=0\Leftrightarrow x=\frac{3}{2}\)
c, \(\sqrt{x^2+6x+9}=3x-1\Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\Leftrightarrow\left|x+3\right|=3x-1\)
TH1 : \(x+3=3x-1\Leftrightarrow2x=4\Leftrightarrow x=2\)
TH2 : \(x+3=1-3x\Leftrightarrow4x=-2\Leftrightarrow x=-\frac{1}{2}\)