TK(/hoi-dap/detail/40878446416.html) 

Có a,b = b,a x  3 + 1,3

=> ab = ba x 3 + 1,3

=> 10a + b = 30b + 3a + 13

=> 7a - 29b = 13

=> 7a = 29b + 13

Mà \(b>0=>29b+13>13\)

=> 7a > 13 

=> a \(\ge2\)

TH1: a = 2 => b = \(\dfrac{1}{29}\left(l\right)\)

TH2: a = 3 => b = \(\dfrac{8}{29}\left(l\right)\)

TH3: a = 4 => b = \(\dfrac{15}{29}\left(l\right)\)

TH4: a = 5 => \(b=\dfrac{22}{29}\) (l)

TH5: a = 6 => \(b=1\) (thỏa mãn)

TH6: a = 7 => \(b=\dfrac{36}{29}\left(l\right)\)

TH7: a = 8 => b = \(\dfrac{43}{29}\left(l\right)\)

TH8: a = 9 => \(b=\dfrac{50}{29}\left(l\right)\)

Vậy a = 6, b = 1

a)\(\left[{}\begin{matrix}\dfrac{5}{2}-x=\dfrac{1}{3}\\\dfrac{5}{2}-x=-\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=\dfrac{17}{6}\end{matrix}\right.\)

b) 8/6-x-1/5=0

9/6-x=1/5

x=13/10

a = 1,2

b = 1,8

a) \(\left|2,5-x\right|=1,3\)

\(\Leftrightarrow\left[{}\begin{matrix}2,5-x=1,3\\2,5-x=-1,3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)

b) \(1,6-\left|x-0,2\right|=0\)

\(\Leftrightarrow\left|x-0,2\right|=1,6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)

Tìm AB :

AB = BA x 3 + 6

=> 10A + B = (10B + A) x 3 + 6

=> 10A + B = 30B + 3A + 6

=> 10A + B - 6 = 30B + 3A

Cùng bớt 10A + b ta được :

6 = 20B + 2A

=> AB ko tồn tại vì a,b khác 0.

 Bài 1: Thực hiện phép tính:

\(\text{a, 1,3 + 2,5 – 4,7 + 5,6 – 4,3=}0,4\)

\(\text{b, - 5,7 + 4,2 – 8,2 + 11,7}=2\\ \)

\(\text{c, 25.(- 0,8).4.(-0,5).0,224}=8,96\)

Bài 3: Tính nhanh:

\(\text{a, -13,45 – 7,98 – 8,55}=-29,98\)

\(\text{b, 9,72 + 8,38 + 3,62}=21,72\)

\(c,\)\(\)\(\left| {45,37} \right| - \left| { - 29,73} \right| + \left( { - 12,27} \right)\)\(=45,37-29,73-12,27=3,38\)

\(\text{d, 31,71 – 7,41.15 – 2,59.15}=-118,29\)

Bài 1:

a) \(1,3+2,5-4,7+5,6-4,3\) 

\(=\left(1,3+2,5+5,6\right)+\left(-4,7-4,3\right)\) 

\(=9,4+\left(-9\right)\) 

\(=0,4\) 

b) \(-5,7+4,2-8,2+11,7\) 

\(=\left(-5,7+11,7\right)+\left(4,2-8.2\right)\) 

\(=6+\left(-4\right)\) 

\(=2\) 

c) \(25.\left(-0,8\right).4.\left(-0,5\right).0,224\) 

\(=\left(25.4\right).\left(-0,8.-0,5\right).0,224\) 

\(=100.0,4.0,224\) 

\(=40.0,224\) 

\(=8,96\)

2,35+0,23+0,05=2,63

Ta có 

a²+b²+c² = ab+bc+ca

<=> 2(a²+b²+c²)= 2(ab+bc+ca)

<=> (a - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ac + a²) = 0

<=> (a - b)² + (b - c)² + (c - a)² = 0

<=> a = b = c

Thế vào pt thứ (2) ta được

a⁸ + b⁸ + c⁸ = 3

<=> 3a⁸ = 3

<=> a⁸ = 1

<=> a = b = c = 1(3) hoặc a = b = c = - 1(4)

Từ (3) => P = 1 + 1 - 1 = 1

Từ (4) => P = - 1 + 1 + 1 = 1

ta có   :\(a^2+b^2+c^2=ab+bc+ca\)

\(\Rightarrow2.\left(a^2+b^2+c^2\right)=2.\left(ab+bc+ca\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

mà ta có:  \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\)   \(\forall a,b,c\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)  \(\forall a,b,c\)

dấu  \("="\) xảy ra \(\Leftrightarrow a=b=c\)

lại có:\(a^8+b^8+c^8=3\)  mà \(a=b=c\)

\(\Rightarrow a^8+a^8+a^8=3\)

\(\Leftrightarrow a^8=1\)

\(\Leftrightarrow a=1\)

vậy \(a=b=c=1\)

bạn ghost mantits cóp ở đâu thế thấy sai sai

Ta co a,b = a + (1/10)b va b,a = b + (1/10) a 
Tu gia thiet => a + (1/10) b = 3[b +(1/10) a] + 1,3 
=> 10a + b = 30b + 3a + 13 => 7a = 29b + 13 
Vi a < 10 => 29b + 13 < 70 => b < 2 => b = 1 => a = 6 
Vay a = 6; b = 1

tích nha