![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{a}{5}=\dfrac{b}{4}\Rightarrow\dfrac{a^2}{25}=\dfrac{b^2}{16}\)
Áp dụng tính chất DTSBN :
\(\dfrac{a^2}{25}=\dfrac{b^2}{16}=\dfrac{a^2-b^2}{25-16}=\dfrac{1}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{1}{9}\cdot25=\dfrac{25}{9}\\b^2=\dfrac{1}{9}\cdot16=\dfrac{16}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{5}{3};b=\dfrac{4}{3}\\a=\dfrac{-5}{3};b=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(\left(a;b\right)\in\left\{\left(\dfrac{5}{3};\dfrac{4}{3}\right);\left(-\dfrac{5}{3};-\dfrac{4}{3}\right)\right\}\)
b) \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)
Áp dụng tính chất DTSBN :
\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=4.4=16\\b^2=4.9=36\\c^2=4,16=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=4;=6;c=8\\a=-4;b=-6;c=-8\end{matrix}\right.\)
Vậy (a;b;c) \(\in\left\{\left(4;6;8\right);\left(-4;-6;-8\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Để \(\frac{2a+2b}{ab+1}\) là bình phương của 1 số nguyên thì 2a + 2b chia hết cho ab + 1; mà ab + 1 chia hết cho 2a + 2b => ab + 1 = 2b + 2a
=> \(\frac{2a+2b}{ab+1}\)=1 = 12
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
Theo tính chất dãy tỉ số bằng nhau ta có:
Ta có:
Mà nên a, b và c cùng dấu.
Vậy ta tìm được các số a1 = 4; b1 = 6; c1 = 8 hoặc a2 = -4; b2 = -6 và c2 = -8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)
\(\Rightarrow\left(a^{100}+b^{100}\right)\left(a^{102}+b^{102}\right)=\left(a^{101}+b^{101}\right)^2\)
\(\Rightarrow a^{202}+b^{202}+a^{100}b^{102}+a^{102}b^{100}=a^{202}+b^{202}+2a^{101}b^{101}\)
\(\Rightarrow a^{100}b^{100}\left(a^2+b^2\right)=a^{100}b^{100}\left(2ab\right)\)
\(\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow\left(a-b\right)^2=0\)
\(\Rightarrow a=b\)
Thế vào \(a^{100}+b^{100}=a^{101}+b^{101}\)
\(\Rightarrow a^{100}+a^{100}=a^{101}+a^{101}\)
\(\Rightarrow2a^{100}\left(a-1\right)=0\)
\(\Rightarrow a=1\Rightarrow b=1\)
\(\Rightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: a//b nên \(\widehat{A_2}+\widehat{B_2}=180^0\)(2 góc kề bù)
Mà \(\widehat{A_2}-\widehat{B_2}=30^0\left(gt\right)\)
\(\Rightarrow\widehat{A_2}=\left(180^0+30^0\right):2=105^0\)
Ta có: a//b(gt)
\(\Rightarrow\widehat{A_2}=\widehat{B_1}=105^0\)( 2 góc so le trong)
Ta có: \(\widehat{A_2}+\widehat{A_1}=180^0\)(2 góc kề bù)
\(\Rightarrow\widehat{A_1}=180^0-\widehat{A_2}=180^0-105^0=75^0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:a2+b2=13(1)
ab=6=>2ab=12(2)
Lấy (1)-(2), vế theo vế ta đc:
a2+b2-2ab=13-12
=>a2-2ab+b2=1
=>(a-b)2=1=>a-b=1=>a= b+1
Vậy |a+b|=|2b+1|
mk ko chắc nhé bn
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(=\left(a^2+b^2\right)+\left(2.6\right)\)
\(=13+12\)
\(=25\)
\(\Rightarrow\left(a+b\right)^2=25\)
\(\Rightarrow a+b=\pm5\)
\(\Rightarrow\left|a+b\right|=5\)
Vậy \(\left|a+b\right|=5\)
5