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a: Ta có: \(AB^2+CH^2\)
\(=AH^2+BH^2+AC^2-AH^2\)
\(=AC^2+BH^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+3}{\sqrt{x}-1}=1+\dfrac{3}{\sqrt{x}-1}\)
Để \(A\in Z\Rightarrow\dfrac{3}{\sqrt{x}-1}\in Z\Rightarrow3⋮\sqrt{x}-1\Rightarrow\sqrt{x}-1\in\left\{3;1;-1;-3\right\}\)
\(\Rightarrow x\in\left\{16;4;0\right\}\)
\(B=\dfrac{2\sqrt{x}-3}{\sqrt{x}+2}=\dfrac{2\left(\sqrt{x}+2\right)-7}{\sqrt{x}+2}=2-\dfrac{7}{\sqrt{x}+2}\)
Để \(B\in Z\Rightarrow\dfrac{7}{\sqrt{x}+2}\in Z\Rightarrow7⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2\in\left\{1;7;-1;-7\right\}\)
\(\Rightarrow x=25\)
Lời giải:
a.
\(A=\frac{\sqrt{x}+2}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)
Với $x$ nguyên, để $A$ nguyên thì $\sqrt{x}-1$ phải là ước của $3$
$\Rightarrow \sqrt{x}-1\in\left\{\pm 1;\pm 3\right\}$
$\Rightarrow \sqrt{x}\in\left\{0; 2; -2; 4\right\}$
Vì $\sqrt{x}\geq 0$ nên $\sqrt{x}\in\left\{0;2;4\right\}$
$\Rightarrow x\in\left\{0;4;16\right\}$
b.
$B=\frac{2(\sqrt{x}+2)-7}{\sqrt{x}+2}=2-\frac{7}{\sqrt{x}+2}$
Để $B$ nguyên thì $\sqrt{x}+2$ là ước của $7$. Mà $\sqrt{x}+2\geq 2$ nên $\sqrt{x}+2\in\left\{7\right\}$
$\Rightarrow x=25$
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 18
a, Với \(a>0;a\ne1;4\)
\(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b, Thay a = 9 => căn a = 3
\(A=\dfrac{3-2}{3.3}=\dfrac{1}{9}\)
c, Ta có : \(A.B=\dfrac{\sqrt{a}-2}{3\sqrt{a}}.\dfrac{3\sqrt{a}}{\sqrt{a}+1}=\dfrac{\sqrt{a}-2}{\sqrt{a}+1}< 0\)
Vì \(\sqrt{a}+1>\sqrt{a}-2\)
\(\left\{{}\begin{matrix}\sqrt{a}+1>0\\\sqrt{a}-2< 0\end{matrix}\right.\Leftrightarrow a< 4\)
Kết hợp với đk vậy \(0< a< 4;a\ne1\)
Bài 18:
1) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
2) Thay a=9 vào B, ta được:
\(B=\dfrac{3\cdot3}{3+1}=\dfrac{9}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)ĐK : \(x>0;x\ne1\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b, \(A=\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\Rightarrow3\sqrt{x}-3=\sqrt{x}\Leftrightarrow2\sqrt{x}=3\)
\(\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\)
c, \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=\dfrac{\sqrt{x}-1-9x}{\sqrt{x}}\)
\(=1-\dfrac{1}{\sqrt{x}}-9\sqrt{x}\)Đặt \(\sqrt{x}=t^2\left(t>0\right)\)
\(1-t-9t^2=-\left(9t^2-t-1\right)=-\left(9t^2-2.3.\dfrac{1}{6}.t+\dfrac{1}{36}-\dfrac{37}{36}\right)\)
\(=-\left(3t-\dfrac{1}{6}\right)+\dfrac{37}{36}\le\dfrac{37}{36}\)
Dấu ''='' xảy ra khi t = 1/18 => t^2 = 1/324 => \(\sqrt{x}=\dfrac{1}{324}\Rightarrow x=\dfrac{1}{104876}\)
Vậy GTLN P là 37/36 khi x = 1/104876
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\(\dfrac{\sqrt{3}-3}{\sqrt{3}+1}=\dfrac{\left(\sqrt{3}-3\right)\left(\sqrt{3}-1\right)}{2}=\dfrac{3-\sqrt{3}-3\sqrt{3}+3}{2}=\dfrac{6-4\sqrt{3}}{2}=3-2\sqrt{3}\)
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\(\left\{{}\begin{matrix}\dfrac{9}{\sqrt{2x-1}}+\dfrac{3}{y+1}=2\\\dfrac{4}{\sqrt{2x-1}}-\dfrac{1}{y+1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{\sqrt{2x-1}}+\dfrac{3}{y+1}=2\left(1\right)\\\dfrac{12}{\sqrt{2x-1}}-\dfrac{3}{y+1}=3\left(2\right)\end{matrix}\right.\)
Lấy \(\left(2\right)+\left(1\right)\) ta được:
\(\dfrac{21}{\sqrt{2x-1}}=5\\ \Leftrightarrow5\sqrt{2x-1}=21\\ \Leftrightarrow25\left(2x-1\right)=441\\ \Leftrightarrow50x-25=441\\ \Leftrightarrow50x=466\Leftrightarrow x=\dfrac{233}{25}\)
Thay x vào (1)
\(\dfrac{9}{\sqrt{2\cdot\dfrac{233}{25}-1}}+\dfrac{3}{y+1}=2\\ \Leftrightarrow\dfrac{9}{\sqrt{\dfrac{441}{25}}}+\dfrac{3}{y+1}=2\\ \Leftrightarrow\dfrac{9}{\dfrac{21}{5}}+\dfrac{3}{y+1}=2\\ \Leftrightarrow\dfrac{15}{7}+\dfrac{3}{y+1}=2\\ \Leftrightarrow15\left(y+1\right)+21=14\left(y+1\right)\\ \Leftrightarrow15y+15+21=14y+14\\ \Leftrightarrow y=-22\)
Vậy pt có tập nghiệm \(\left(x;y\right)=\left(\dfrac{233}{25};-22\right)\)
\(\left\{{}\begin{matrix}\dfrac{9}{\sqrt{2x-1}}+\dfrac{3}{y+1}=2\\\dfrac{4}{\sqrt{2x-1}}-\dfrac{1}{y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{\sqrt{2x-1}}+\dfrac{12}{y+1}=8\\\dfrac{36}{\sqrt{2x-1}}-\dfrac{9}{y+1}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{y+1}=-1\\\dfrac{4}{\sqrt{2x-1}}-\dfrac{1}{y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=-21\\\dfrac{4}{\sqrt{2x-1}}=\dfrac{20}{21}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-22\\2x-1=\dfrac{441}{25}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{233}{25}\\y=-22\end{matrix}\right.\)
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2x^2-x-2020=0
=>x=(1+căn 16161)/4 hoặc x=(1-căn 16161)/4
Gọi A(1+căn 16161/4;0); B(1-căn 16161/4;0); N(0;b)
\(AB=\dfrac{\sqrt{2\cdot\sqrt{16161}}}{2};AN=\sqrt{\left(0-\dfrac{1+\sqrt{16161}}{4}\right)^2+\left(b-0\right)^2}\)
\(BN=\sqrt{\left(0-\dfrac{1-\sqrt{16161}}{4}\right)^2+\left(b-0\right)^2}\)
ΔABN vuông tại N
=>NA^2+NB^2=AB^2
=>\(\left(\dfrac{1+\sqrt{16161}}{4}\right)^2+b^2+\left(\dfrac{1-\sqrt{16161}}{4}\right)^2+b^2=\left(\dfrac{1+\sqrt{16161}}{4}-\dfrac{1-\sqrt{16161}}{4}\right)^2\)
=>b^2=-2(1-16161)/16*2=1010
=>b=căn 1010
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\(\widehat{BME}=\widehat{BMK}\) (do K đối xứng E qua MB)
Mà \(\widehat{BMK}=\widehat{BCM}\) (cùng phụ \(\widehat{MBC}\))
\(\Rightarrow\widehat{BME}=\widehat{BCM}\)
\(\Rightarrow ME\) là tiếp tuyến của (O) tại M
Tương tự, ta có MF là tiếp tuyến của (O) tại M
\(\Rightarrow M;E;F\) thẳng hàng
\(\Rightarrow S_{BEFC}=S_{BEMK}+S_{CFMK}=2S_{BMK}+2S_{CMK}=2S_{MBC}\)
Mà \(S_{MBC}=\dfrac{1}{2}MK.BC\Rightarrow S_{MBC-max}\) khi \(MK_{max}\)
\(\Rightarrow M\) nằm chính giữa cung BC \(\Rightarrow MK_{max}=R=4\left(cm\right)\)
\(\Rightarrow S_{BEFC-max}=2.\dfrac{1}{2}.4.8=32\left(cm^2\right)\)
a: Khi m=-1/2 thì (1) sẽ là;
\(x^2-3x-\dfrac{15}{4}=0\)
=>\(x=\dfrac{3\pm2\sqrt{6}}{2}\)
b: \(\text{Δ}=\left(2m+4\right)^2-4\left(m^2-4\right)\)
\(=4m^2+16m+16-4m^2+16=16m+32\)
Để (1) có hai nghiệm pb thì 16m+32>0
=>m>-2
\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)
\(=\left(2m+4\right)^2-4\left(m^2-4\right)\)
\(=4m^2+16m+16-4m^2+16m=16m+32\)
=>x1-x2=căn 16m+32 hoặc x1-x2=-căn 16m+32
\(\dfrac{x_1}{x_2}-\dfrac{x_2}{x_1}=2\)
=>\(\dfrac{x_1^2-x_2^2}{x_1x_2}=2\)
=>\(\left(x_1-x_2\right)\cdot\dfrac{2m+4}{m^2-4}=2\)
TH1: (x1-x2)=căn 16m+32
=>\(\sqrt{16m+32}=2:\dfrac{2m+4}{m^2-4}=\dfrac{2\left(m^2-4\right)}{2\left(m+2\right)}=m-2\)
=>16m+32=m^2-4m+4 và m>=2
=>m^2-20m-28=0 và m>=2
=>\(m=10+8\sqrt{2}\)
TH2: x1-x2=-căn 16m+32
=>\(\sqrt{16m+32}=2-m\)
=>m^2-4m+4=16m+32 và m<=2
=>\(m=10-8\sqrt{2}\)