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Gọi $d$ là ước số (chung) của $a,b$
Đặt \(\left\{\begin{matrix} a=md\\ b=nd\end{matrix}\right.(m,n\in\mathbb{Z}^+)\)
Ta có:
\(\frac{a+1}{b}+\frac{b+1}{a}\in\mathbb{Z}\)
\(\Leftrightarrow \frac{a^2+b^2+a+b}{ab}\in\mathbb{Z}\)
\(\Leftrightarrow \frac{(a+b)^2+(a+b)-2ab}{ab}\in\mathbb{Z}\Leftrightarrow \frac{(a+b)^2+(a+b)}{ab}\in\mathbb{Z}\)
\(\Leftrightarrow (a+b)^2+a+b\vdots ab\)
\(\Leftrightarrow (md+nd)^2+md+nd\vdots mnd^2\)
\(\Leftrightarrow d(m+n)^2+m+n\vdots mnd\)
\(\Rightarrow d(m+n)^2+m+n\vdots d\Rightarrow m+n\vdots d\)
Mà \(m+n\neq 0\). Do đó suy ra \(m+n\geq d\)
\(\Rightarrow d(m+n)\geq d^2\) hay \(a+b\geq d^2\Rightarrow d\leq \sqrt{a+b}\)
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Ta có:
\(a+b+c=abc\Rightarrow a^2+ab+ac=a^2bc\)
\(\Rightarrow a^2+ab+ac+bc=a^2bc+bc\)
\(\Leftrightarrow (a+b)(a+c)=bc(a^2+1)\)
Tương tự: \(\left\{\begin{matrix} ac(b^2+1)=(b+c)(b+a)\\ ab(c^2+1)=(c+a)(c+b)\end{matrix}\right.\)
Do đó: \(S=\frac{a}{\sqrt{(a+b)(a+c)}}+\frac{b}{\sqrt{(b+c)(b+a)}}+\frac{c}{\sqrt{(c+a)(c+b)}}\)
Áp dụng BĐT AM-GM:
\(A\leq \frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)+\frac{1}{2}\left(\frac{b}{b+a}+\frac{b}{b+c}\right)+\frac{1}{2}\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\)
\(\Leftrightarrow S\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{a+c}{a+c}+\frac{b+c}{b+c}\right)=\frac{3}{2}\)
Vậy \(S_{\max}=\frac{3}{2}\)
Dấu bằng xảy ra khi \(a=b=c=\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 5:
a: Để C nguyên thì \(\sqrt{x}-2+5⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{1;-1;5;-5\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{3;1;7\right\}\)
hay \(x\in\left\{9;1;49\right\}\)
b: Để D nguyên thì \(2\sqrt{x}+6-7⋮\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)
\(\Leftrightarrow\sqrt{x}=4\)
hay x=16
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1
mình khỏi vẽ hình nha
dễ thấy
\(CH=BH\cdot\dfrac{\sqrt{5}}{\sqrt{7}}\)
mà \(CH+BH=BC=\sqrt{11}\)
\(\Rightarrow\left\{{}\begin{matrix}CH\approx1.519146459\\BH\approx1.797478331\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}AB\approx2.441630847\\AC\approx2.2446467\end{matrix}\right.\)
\(\Rightarrow P_{ABC}\approx8.002902337\)
câu 3
\(B=\dfrac{988153}{30127}=32+\dfrac{1}{\dfrac{30127}{24089}}=32+\dfrac{1}{1+\dfrac{1}{\dfrac{24089}{6038}}}=32+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{\dfrac{6038}{5975}}}}=32+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{\dfrac{5975}{63}}}}}\)
\(B=32+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{94+\dfrac{1}{\dfrac{63}{53}}}}}}=32+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{94+\dfrac{1}{1+\dfrac{1}{\dfrac{53}{10}}}}}}}\)
\(B=32+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{94+\dfrac{1}{1+\dfrac{1}{5+\dfrac{1}{\dfrac{10}{3}}}}}}}}\)
\(B=32+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{94+\dfrac{1}{1+\dfrac{1}{5+\dfrac{1}{3+\dfrac{1}{3}}}}}}}}\)
vậy \(\left\{b_1;b_2;...;b_n\right\}=\left\{32;1;3;1;94;1;5;3;3\right\}\)
sức mình đến đây thôi
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
a)
\(5\sqrt{x}-2=13\Rightarrow 5\sqrt{x}=15\Rightarrow \sqrt{x}=3\)
\(\Rightarrow x=3^2=9\)
b)
\(\sqrt{8x}+7\sqrt{18x}=9-\sqrt{50x}\)
\(\Leftrightarrow \sqrt{4}.\sqrt{2x}+7\sqrt{9}.\sqrt{2x}=9-\sqrt{25}.\sqrt{2x}\)
\(\Leftrightarrow 2\sqrt{2x}+21\sqrt{2x}=9-5\sqrt{2x}\)
\(\Leftrightarrow 28\sqrt{2x}=9\Rightarrow \sqrt{2x}=\frac{9}{28}\)
\(\Rightarrow 2x=(\frac{9}{28})^2\Rightarrow x=\frac{1}{2}.(\frac{9}{28})^2\)
Câu 2:
a) \(Q=\frac{2\sqrt{x}-9}{(\sqrt{x}-2)(\sqrt{x}-3)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\frac{2\sqrt{x}-9}{(\sqrt{x}-2)(\sqrt{x}-3)}-\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}+\frac{(2\sqrt{x}+1)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}\)
\(=\frac{2\sqrt{x}-9-(x-9)+(2x-3\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-3)}\)
\(=\frac{x-\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{(\sqrt{x}+1)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) Để \(Q=2\Leftrightarrow \frac{\sqrt{x}+1}{\sqrt{x}-3}=2\Rightarrow \sqrt{x}+1=2\sqrt{x}-6\)
\(\sqrt{x}=7\Rightarrow x=49\)
c)
\(Q\in \mathbb{Z}\Leftrightarrow \frac{\sqrt{x}+1}{\sqrt{x}-3}\in \mathbb{Z}\Rightarrow \sqrt{x}+1\vdots \sqrt{x}-3\)
\(\Leftrightarrow \sqrt{x}-3+4\vdots \sqrt{x}-3\)
\(\Leftrightarrow 4\vdots \sqrt{x}-3\Rightarrow \sqrt{x}-3\in \left\{\pm 1; \pm 2; \pm 4\right\}\)
\(\Rightarrow \sqrt{x}\in \left\{2;4;1; 5; 7\right\}\)
\(\Rightarrow \sqrt{x}\in \left\{4; 16; 1; 25; 49\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn nào làm được bài này thì giúp mình với ạ ! mình đang cần gấp
Bài 4:
\(AH=\sqrt{9\cdot16}=12\left(cm\right)\)
\(AB=\sqrt{9\cdot25}=15\left(cm\right)\)
AC=căn(25^2-15^2)=20(cm)
Xét ΔABC vuông tại A có sin ABC=AC/BC=4/5
nên góc ABC=53 độ
![](https://rs.olm.vn/images/avt/0.png?1311)
giải tạm 1 bài z -,-
2) Cauchy-Schwarz dạng Engel :
\(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}=\dfrac{6}{2}=3\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=2\)
Chúc bạn học tốt ~
4/ Ta có: \(6=a+b+c+ab+bc+ca\ge3\left(\sqrt[3]{\left(abc\right)^2}+\sqrt[3]{abc}\right)\)
Đặt \(\sqrt[3]{abc}=t\Rightarrow t^2+t\le2\Rightarrow t\le1\Rightarrow t^3=C=abc\le1\)
Vậy...
5/ \(D\le\left(\frac{a+b+c}{3}\right)^3.\left[\frac{2\left(a+b+c\right)}{3}\right]^3=\frac{512}{729}\)
Vậy ...
P/s: Em không chắc
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(a-\dfrac{a^2}{a+b^2}=\dfrac{ab^2}{a+b^2}\le\dfrac{ab^2}{2b\sqrt{a}}=\dfrac{b\sqrt{a}}{2}\)
Tương tự cho các BĐT còn lại cũng có:
\(b-\dfrac{b^2}{b+c^2}\le\dfrac{c\sqrt{b}}{2};c-\dfrac{c^2}{c+a^2}\le\dfrac{a\sqrt{c}}{2}\)
Sau đó cộng theo vế các BĐT trên
\(\dfrac{a^2}{a+b^2}+\dfrac{b^2}{b+c^2}+\dfrac{c^2}{c+a^2}\ge3-\dfrac{1}{2}\left(b\sqrt{a}+c\sqrt{b}+a\sqrt{c}\right)\)
\(\ge3-\dfrac{1}{2}\sqrt{\left(a+b+c\right)\left(ab+bc+ca\right)}\)
\(\ge3-\dfrac{1}{2}\sqrt{\left(a+b+c\right)\cdot\dfrac{\left(a+b+c\right)^2}{3}}=3-\dfrac{3}{2}=\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(a=b=c=1\)
Bài 2:
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a}{\sqrt{2b^2+2c^2-a^2}}=\dfrac{\sqrt{3}a^2}{\sqrt{3a^2\left(2b^2+2c^2-a^2\right)}}\)
\(\ge\dfrac{\sqrt{3}a^2}{\dfrac{3a^2+2b^2+2c^2-a^2}{2}}=\dfrac{\sqrt{3}a^2}{a^2+b^2+c^2}\)
Tương tự cho các BĐT còn lại ta có:
\(\dfrac{b}{\sqrt{2a^2+2c^2-b^2}}\ge\dfrac{\sqrt{3}b^2}{a^2+b^2+c^2};\dfrac{c}{\sqrt{2a^2+2b^2-c^2}}\ge\dfrac{\sqrt{3}c^2}{a^2+b^2+c^2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=\sqrt{3}=VP\)
Đẳng thức xảy ra khi \(a=b=c\)