1.

Đặt \(x+y=a\Rightarrow y=a-x\)

\(\Rightarrow x^2+2x\left(a-x\right)-14\left(a-x\right)-10x+3\left(a-x\right)^2+27=0\)

\(\Leftrightarrow2x^2-4\left(a+1\right)x+3a^2-10a+27=0\)

\(\Delta'=4\left(a+1\right)^2-2\left(3a^2-10a+27\right)\ge0\)

\(\Leftrightarrow-a^2+14a-25\ge0\)

\(\Rightarrow7-2\sqrt{6}\le a\le7+2\sqrt{6}\)

\(\Rightarrow-10-2\sqrt{6}\le P\le-10+2\sqrt{6}\)

2. Chắc đề là \(a;b>0\) (đảm bảo mẫu dương) chứ ko phải \(a.b>4\)

\(M\ge\dfrac{\left(a+b\right)^2}{a+b-8}=\dfrac{\left(a+b-8+8\right)^2}{a+b-8}=\dfrac{\left(a+b-8\right)^2+16\left(a+b-8\right)+64}{a+b-8}\)

\(M\ge a+b-8+\dfrac{64}{a+b-8}+16\ge2\sqrt{\dfrac{64\left(a+b-8\right)}{a+b-8}}+16=32\)

Dấu "=" xảy ra khi \(a=b=8\)

\(\Rightarrow\left|3x+1\right|=4+\dfrac{1}{2}=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}3x+1=\dfrac{9}{2}\\3x+1=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=\dfrac{7}{2}\\3x=-\dfrac{11}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{11}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left|3x+1\right|=\dfrac{9}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=\dfrac{9}{2}\\3x+1=-\dfrac{9}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{11}{6}\end{matrix}\right.\)

a/

\(\frac{3a-b}{a+b}=\frac{3\left(a+b\right)-4b}{a+b}=3-\frac{4b}{a+b}=\frac{3}{4}.\)

\(\Rightarrow\frac{4b}{a+b}=\frac{9}{4}\Rightarrow9a+9b=16b\Rightarrow9a=7b\Rightarrow\frac{a}{b}=\frac{7}{9}\)

b/

\(\frac{a}{b}=\frac{3}{7}\Rightarrow\frac{a}{3}=\frac{b}{7}=\frac{3a}{9}=\frac{4b}{28}=\frac{3a-4b}{9-28}=\frac{3a-4b}{-19}\)

\(\frac{a}{3}=\frac{b}{7}\Rightarrow\frac{2a}{6}=\frac{3b}{21}\Rightarrow\frac{2a+3b}{6+21}=\frac{2a+3b}{27}\)

\(\Rightarrow\frac{3a-4b}{-19}=\frac{2a+3b}{27}\Rightarrow\frac{3a-4b}{2a+3b}=-\frac{19}{27}\)

1

ab=83,ba=38

2

ab=62,ba=26

câu 3 sai đề

\(ab\:là\:a.b\:hay\:là\:\overline{ab}\)

a, chứng tỏ A chia hết cho 40

a: A=3(1+3+3^2+3^3)+...+3^129(1+3+3^2+3^3)

=40(3+...+3^129) chia hết cho 40

b: A=(3+3^2+3^3)+....+3^129(3+3^2+3^3)

=39(1+...+3^129) chia hết cho 39

c: A chia hết cho 40

A chia hết cho 3

=>A chia hết cho BCNN(40;3)=120

D

D

B

a) ab+ba=121
=) 10a+b+10b+a=121
=) 11a+11b=121
=) 11.(a+b)=121
=) a+b=121:11=11
Mà ab-ba=*5
=) 10a+b-10b-a=*5
=) 9a+(-9b)=*5 =) 9a-9b=*5
=) 9.(a-b)=*5
Mà \(9.\left(a-b\right)⋮9\)=) *5\(⋮9\)=) (*+5)\(⋮9\)
=) * = 4
=) 9.(a-b)=45
=) a-b=45:9=5
Có a+b=11,a-b=5 =) a=(11+5):2=8
=) b=8-5=3
=) ab=83 với *=4
b) Vì ab:5(dư 1) =) b:5(dư 1 )
=) b=1,6
Mà ab-ba=*8
=) 10a+b-10b-a=*8
=) 9a+(-9b)=*8
=) 9a-9b=*8
=) 9.(a-b)=*8
Vì \(9.\left(a-b\right)⋮9\)=) *8\(⋮9\)=) *+8\(⋮9\)
=) * = 1
=) 9.(a-b)=18
=) a-b=18:9=2
Với b=1 =) a-1=2 =) a=2+1=3
Với b=6 =) a-6=2 =) a=2+6=8
Vậy ab=31,86 với *=1
c) Vì ab:5(dư 3) =) b:5(dư 3) =) b=3,8
Mà ab=ba+5*
=) ab-ba=5*
=) 10a+b-10b-a=5*
=) 9a+(-9b)=5*
=) 9a-9b=5*
=) 9.(a-b)=5*
Vì \(9.\left(a-b\right)⋮9\)=)5*\(⋮9\)=) 5 + * \(⋮9\)
=) * = 4
=) 9.(a-b)=54 =) a-b = 54:9 = 6
Với b = 3 =) a-3=6 =) a= 6+3 = 9
Với b = 8 =) a-8=6 =) a= 6+8 = 14 ( loại vì a có 1 chữ số )
Vậy ab=93 với * = 4