Bài 3: 

1: \(35^2=1225\)

2: \(25^2=625\)

3: \(75^2=5625\)

4: \(95^2=9025\)

5: \(101\cdot99=9999\)

6: \(36\cdot44=1584\)

7: \(72\cdot68=4896\)

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

1: \(x^2-2x+1=\left(x-1\right)^2\)

2: \(4x^2-4x+1=\left(2x-1\right)^2\)

3: \(16x^2+8x+1=\left(4x+1\right)^2\)

4: \(9x^2+12x+4=\left(3x+2\right)^2\)

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

Tick cho mình nhé

a) \(99^3=\left(100-1\right)^3=100^3-3.100^2+3.100-1=1000000-30000+300-1=970299\)b) \(91^3+3.91^2.9+3.91.9^2+9^3=\left(91+9\right)^3=100^3=1000000\)

c) \(1001^3=\left(1000+1\right)^3=1000^3+3.1000^2+3.1000+1=1003003001\)d) \(102^3-6.102^2+24.102-8=\left(102-2\right)^3+12.102=100^3+1224=1001224\)

nhỏ quá bn ạ

a) (a+b)^3 +15a

b: T=(-1)+(-1)+...+(-1)

=-1011

làm giúp mình câu a nữa nhé 'v'

a)x=-2

b) y=10

c)x=-5

d) x=-12

\(a,\dfrac{x}{3}=\dfrac{6}{-9}\\ \Rightarrow x=-\dfrac{2}{3}.3\\ \Rightarrow x=-2\\ b,\dfrac{4}{y}=\dfrac{-2}{-5}\\ \Rightarrow y=4:\dfrac{2}{5}\\ \Rightarrow y=10\\ c,\dfrac{-2}{3}=\dfrac{x-1}{6}\\ \Rightarrow3x-3=-12\\ \Rightarrow3x=-9\\ \Rightarrow x=-3\\ d,\dfrac{3}{x}=\dfrac{6}{-24}\\ \Rightarrow x=3:-\dfrac{1}{4}\\ \Rightarrow x=-12\)

\(a,\dfrac{3}{4}=\dfrac{15}{20}\\ \dfrac{7}{10}=\dfrac{14}{20}\\ b,\dfrac{-5}{14}=\dfrac{-55}{154}\\ \dfrac{9}{22}=\dfrac{63}{154}\\ \dfrac{-4}{7}=\dfrac{-36}{63}\\ \dfrac{8}{9}=\dfrac{56}{63}\\ \dfrac{-10}{21}=\dfrac{-30}{63}\)

bạn ơi còn thiếu câu c

a: Để 5/n-1 là số nguyên thì \(n-1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{2;0;6;-4\right\}\)

b: Để n+8/n+1 là số nguyên thì \(n+1\in\left\{1;-1;7;-7\right\}\)

hay \(n\in\left\{0;-2;6;-8\right\}\)

Để 5/n-1 nhận giá trị là số nguyên thì: 

n+1 thuộc Ư(5)= {-1;1;-5;5}

Lập bảng:

=> n thuộc {0;-2;-6;4} thì n sẽ nhận giá trị là số nguyên

câu b làm tương tượng nhưng lấy n+1 thuộc Ư(7)