\(\dfrac{1+x}{3-x}-\dfrac{1-2x}{3+x}-\dfrac{x\left(1-x\right)}{9-x^2}\)

\(=\dfrac{x^2+4x+3-\left(3-x\right)\left(1-2x\right)}{\left(3-x\right)\left(3+x\right)}-\dfrac{x\left(1-x\right)}{9-x^2}\)

\(=\dfrac{x^2+4x+3-3+7x-2x^2-x+x^2}{9-x^2}\)

\(=\dfrac{10x}{9-x^2}\)

\(\dfrac{5a+7b}{6a+5b}=\dfrac{29}{28}\)

\(\Rightarrow140a+196b=174a+145b\)

\(\Rightarrow51b=34a\)

\(\Rightarrow3b=2a\)

Do \(\left(a;b\right)=1\)

\(\Rightarrow a=3,b=2\)

Vậy...

Bài 2:

a, \(x^2-6x+10=x^2-6x+9+1\)

\(=\left(x-3\right)^2+1\ge1>0\)

\(\Rightarrowđpcm\)

b, \(x^2-4xy+4y^2+1=\left(x-2y\right)^2+1>0\)

\(\Rightarrowđpcm\)

c, \(x^2-4x+7=x^2-4x+4+3\)

\(=\left(x-2\right)^2+3\ge3\)

\(\Rightarrowđpcm\)

d, \(x^2+y^2-2x+4y+5\)

\(=x^2-2x+1+y^2+4y+4\)

\(=\left(x-1\right)^2+\left(y+2\right)^2\ge0\)

\(\Rightarrowđpcm\)

Bài 3:

Áp dụng bất đẳng thức AM - GM có:
\(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge2\sqrt{x.\dfrac{1}{x}}+2\sqrt{y.\dfrac{1}{y}}+2\sqrt{z.\dfrac{1}{z}}\)

\(=2+2+2=6\)

Dấu " = " khi x = y = z = 1

Vậy...

\(\sqrt{\left(x-1\right)}=x-3\)

\(\Rightarrow\left|x-1\right|=\left(x-3\right)^2\)

\(\Rightarrow\left|x-1\right|=x^2-6x+9\)

+) Xét \(x\ge1\) có:
\(x-1=x^2-6x+9\)

\(\Leftrightarrow x^2-7x+10=0\)

\(\Leftrightarrow x^2-5x-2x+10=0\)

\(\Leftrightarrow x\left(x-5\right)-2\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Thử lại: x = 5 ( t/m ), x = 2 ( không t/m )

+) Xét x < 1 có:
\(1-x=x^2-6x+9\)

\(\Leftrightarrow x^2+5x+8=0\)

\(\Leftrightarrow x^2+5x+\dfrac{25}{4}+\dfrac{7}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0\)

Do \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}>0\)

\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0\) ( vô lí )

Vậy x = 5

\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\Rightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\\left[{}\begin{matrix}x-7=-1\\x-7=1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=6\\x=8\end{matrix}\right.\)

Vậy x = 7 hoặc x = 6 hoặc x = 8

a)15 chia hết cho x+1

=>x+1\(\in\)Ư(15)={1;-1;3;-3;5;-5;15;-15}

=>x\(\in\){0;-2;2;-4;4;-6;14;-16}

Mà n là số tự nhiên

=>n\(\in\){0;2;4;14}

b)x+6 là bội x+3

=>x+6 chia hết cho x+3

Mà x+3 chia hết cho x+3

=>x+6-x-3 chia hết cho x+3

=>3 chia hết cho x+3

=>x+3\(\in\)Ư(3)={1;-1;3;-3}

=>x\(\in\){-2;-3;0;-6}

Mà x là số tự nhên nên x=0

c)x+6 là ước của 5x+79

=>5x+79 chia hết cho x+6

Mà x+6 chia hết cho x+6 =>5x+30 chia hết cho x+6

=>5x+79-5x-30 chia hết cho x+6

=>49 chia hết cho x+6

=>x+6 \(\in\)Ư(49)={1;-1;49;-49}

=>x\(\in\){-5;-7;43;-55}

Mà x là số tự nhiên nên x=43

Ta có: \(A=a^3+3ab+b^3\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\)

\(=a^2-ab+b^2+3ab\) ( do a + b = 1 )

\(=a^2+2ab+b^2=\left(a+b\right)^2=1\) ( do a + b = 1)

Vậy A = 1

Giải:
Ta có: \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}\)

\(\Rightarrow24+48y=18+72y\)

\(\Rightarrow6=24y\Rightarrow y=\dfrac{1}{4}\)

\(\Rightarrow\dfrac{1+\dfrac{1}{2}}{18}=\dfrac{1+\dfrac{3}{2}}{6x}\)

\(\Rightarrow9x=45\Rightarrow x=5\)

Vậy x = 5

\(\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+...+3^2+1^2\right)\)

\(=\left(20-19\right)\left(20+19\right)+\left(18+17\right)\left(18-17\right)+...+\left(4+3\right)\left(4-3\right)+\left(2+1\right)\left(2-1\right)\)

\(=39+35+...+7+3\)

\(=210\)