để đây làm cho

3

Bài 3: 

Xét ΔIAB có 

\(\widehat{AIB}+\widehat{IAB}+\widehat{IBA}=180^0\)

\(\Leftrightarrow\widehat{IAB}+\widehat{IBA}=115^0\)

hay \(\widehat{DAB}+\widehat{ABC}=230^0\)

Xét tứ giác ABCD có 

\(\widehat{D}+\widehat{C}+\widehat{DAB}+\widehat{CBA}=360^0\)

\(\Leftrightarrow\widehat{D}+\widehat{C}=150^0\)

mà \(\widehat{C}-\widehat{D}=10^0\)

nên \(2\cdot\widehat{C}=160^0\)

\(\Leftrightarrow\widehat{C}=80^0\)

\(\Leftrightarrow\widehat{D}=70^0\)

2/

a/

\(\dfrac{4n+2}{n+1}=\dfrac{4n+4-2}{n+1}=\dfrac{4\left(n+1\right)-2}{n+1}=4-\dfrac{2}{n+1}\)

\(\Rightarrow4n+2⋮n+1\) Khi \(n+1=\left\{-2;-1;1;2\right\}\Rightarrow n=\left\{-3;-2;0;1\right\}\)

b/

\(\Rightarrow a+2=\dfrac{8}{b-1}\left(b\ne1\right)\) (1)

a nguyên => a+2 nguyên \(\Rightarrow8⋮\left(b-1\right)\)

\(\Rightarrow b-1=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

\(\Rightarrow b=\left\{-7;-3;-1;0;2;3;5;9\right\}\) Thay các giá trị của b vào (1) để tìm a

3/

Gọi số tiền mẹ cho Chi là A \(\Rightarrow100000\le A\le200000\)

Nếu bớt số tiền mẹ cho Chi đi 5000 đồng thì 

\(A-5000⋮15000;A-5000⋮8000\)

\(\Rightarrow A-5000=UC\left(8000;15000\right)\) Và \(95000\le A-5000\le195000\)

\(\Rightarrow A-5000=120000\Rightarrow A=125000\)

1: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

=>\(x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)

=>\(-9x^2+27x+9x^2+18x+9=15\)

=>45x=6

=>\(x=\frac{6}{45}=\frac{2}{15}\)

2: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

=>\(x\left(x^2-25\right)-\left(x^3+8\right)=3\)

=>\(x^3-25x-x^3-8=3\)

=>-25x=11

=>\(x=-\frac{11}{25}\)

3: \(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x-5\right)\left(x+5\right)=264\)

=>\(x^3+64-x\left(x^2-25\right)=264\)

=>\(x^3+64-x^3+25x=264\)

=>25x=200

=>x=8

4: \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)+6\left(x-2\right)\left(x+2\right)=60\)

=>\(x^3-6x^2+12x-8-\left(x^3-8\right)+6\left(x^2-4\right)=60\)

=>\(-6x^2+12x+6x^2-24=60\)

=>12x-24=60

=>12x=84

=>x=7

5: \(\left(x+3\right)^4-\left(x-3\right)^4-24x^3=108\)

=>\(\left\lbrack\left(x+3\right)^2-\left(x-3\right)^2\right\rbrack\left\lbrack\left(x+3\right)^2+\left(x-3\right)^2\right\rbrack-24x^3=108\)

=>\(\left(x^2+6x+9-x^2+6x-9\right)\left(x^2+6x+9+x^2-6x+9\right)-24x^3=108\)

=>\(12x\left(2x^3+18\right)-24x^3=108\)

=>\(24x^3+216x-24x^3=108\)

=>216x=108

=>\(x=\frac{108}{216}=\frac12\)

7: \(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)

=>\(25x^2-10x+1-\left(25x^2-16\right)=7\)

=>\(25x^2-10x+1-25x^2+16=7\)

=>-10x=7-17=-10

=>x=1

8: \(\left(4x+1\right)^2-\left(2x+3\right)^2+5\left(x+2\right)^2+3\left(x-2\right)\left(x+2\right)=500\)

=>\(16x^2+8x+1-\left(4x^2+12x+9\right)+5\left(x^2+4x+4\right)+3\left(x^2-4\right)\) =500

=>\(16x^2+8x+1-4x^2-12x-9+5x^2+20x+20+3x^2-12=500\)

=>\(20x^2+16x-500=0\)

=>\(x^2+\frac45x-25=0\)

=>\(x^2+2\cdot x\cdot\frac25+\frac{4}{25}-25-\frac{4}{25}=0\)

=>\(\left(x+\frac25\right)^2=25+\frac{4}{25}=\frac{629}{25}\)

=>\(\left[\begin{array}{l}x+\frac25=\frac{\sqrt{629}}{5}\\ x+\frac25=-\frac{\sqrt{629}}{5}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{629}-2}{5}\left(nhận\right)\\ x=\frac{-\sqrt{629}-2}{5}\left(nhận\right)\end{array}\right.\)

9: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

=>\(x^3-27+x\left(4-x^2\right)=1\)

=>4x-27=1

=>4x=28

=>x=7

10: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

=>\(x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\)

=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)

=>12x-6=-10

=>12x=-4

=>x=-4/12=-1/3

Bài 3: 

Xét ΔIAB có 

\(\widehat{AIB}+\widehat{IAB}+\widehat{IBA}=180^0\)

\(\Leftrightarrow\widehat{IAB}+\widehat{IBA}=115^0\)

hay \(\widehat{DAB}+\widehat{ABC}=230^0\)

Xét tứ giác ABCD có 

\(\widehat{D}+\widehat{C}+\widehat{DAB}+\widehat{CBA}=360^0\)

\(\Leftrightarrow\widehat{D}+\widehat{C}=150^0\)

mà \(\widehat{C}-\widehat{D}=10^0\)

nên \(2\cdot\widehat{C}=160^0\)

\(\Leftrightarrow\widehat{C}=80^0\)

\(\Leftrightarrow\widehat{D}=70^0\)

Bài 10: AE//BD

=>\(\hat{EAB}=\hat{ABD}\) (hai góc so le trong) và \(\hat{CBD}=\hat{BEA}\) (hai góc đồng vị)

mà \(\hat{ABD}=\hat{CBD}\) (BD là phân giác của góc ABC)

nên \(\hat{EAB}=\hat{BEA}\)

\(a,\Leftrightarrow3x^2+24x-x^2-2x^2-2x=2\Leftrightarrow22x=2\Leftrightarrow x=\dfrac{1}{11}\\ b,\Leftrightarrow\left[{}\begin{matrix}5-x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)