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9 tháng 9 2017

Ta có : \(3^2+6^2+9^2+...+30^2=3\left(1^2+2^2+3^2+...+10^2\right)=3.385=1155\)

Ta có :

\(3^2+6^2+9^2+.......+30^2=9\left(1^2+2^2+3^2+.........+10^2\right)\)

\(=9.385=3465\)

15 tháng 9 2016

bạn bấm máy tính là ra kq cho nhanh

15 tháng 9 2016

Ta có    P=32+62+92+...30

                   = (3x1)2+ (3+2)+ (3x3)2+.....+ (3x10)2

               = 32x12+ 32x 22+ 32x3+......+ 3x 102

                    =  3(12+  22+ 3+......+ 102)

                     =  3x 385

                   = 9 x 385

                  =3465

 P=32+62+92+...30= 3465

3 tháng 7 2017

\(P=3^2+6^2+...+30^2\)

\(=1.3^2+2^2.3^2+...+3^2.10^2\)

\(=3^2\left(1+2^2+...+10^2\right)\)

\(=9.385=3465\)

Vậy P = 3465

24 tháng 10 2017

3465

1 tháng 2 2017

đặt A= 22+42+62+82+.......+202

\(\Leftrightarrow A\)= 12.22+22.22+22.32+...+22.102

\(\Leftrightarrow\)A= 22(12+22+32+...+102)

\(\Leftrightarrow\)A=4.385=1540

1 tháng 2 2017

Đặt T = \(1^2+2^2+....+10^2=385\)

\(\Rightarrow T\times2^2=1^2.2^2+2^2.2^2....+10^2.2^2=385.2^2\)

\(\Rightarrow T\times2^2=\left(1.2\right)^2+\left(2.2\right)^2....+\left(10.2\right)^2=385.2^2\)

\(\Rightarrow T\times2^2=\left(2\right)^2+\left(4\right)^2+....\left(20\right)^2=385.2^2\)

\(\Rightarrow T\times2^2=S=385.2^2\)

\(\Rightarrow S=385\times4\)

6 tháng 10 2015

nhiều v~~~, dễ mà lp 8 ? 

1,(2x + 3 ) \(^{^{ }2}\)=\(\left(2x\right)^2+2.2x.3+3^2\)

=\(4x^2+12x+9\)

2, ( 3x + 2y )\(^2=\left(3x\right)^2+2.3x.2y+\left(2y\right)^2\)

=\(9x^2+12xy+4y^2\)

3,(3a -1 )\(^2=\left(3a\right)^2-2.3a.1+1^2\)

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

4, (a - 2 )\(^2=a^2-2.a.2+2^2\)

=\(a^2-4a+4\)

5, ( 1 - 5a )\(^2=1^2-2.1.5a+\left(5a\right)^2\)

=\(1-10a+25a\)

6, ( x - 4 )\(^3=x^3-3x^24+3x4^2-4^3\)

=\(x^3-12x^2+48x-64\)

22 tháng 7 2017

1. \(125x^3+y^6=\left(5x\right)^3+\left(y^2\right)^3\)

\(=\left(5x+y^2\right)\left[\left(5x\right)^2-5x.y^2+\left(y^2\right)^2\right]\)

\(=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)

2. \(4x\left(x-2y\right)+8y\left(2y-x\right)\)

\(=4x\left(x-2y\right)-8y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(4x-8y\right)\)

3. \(25\left(x-y\right)^2-16\left(x+y\right)^2\)

\(=\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2\)

\(=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)

\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)\)

\(=\left(x-9y\right)\left(9x-y\right)\)

4. \(x^4-x^3-x^2+1\)

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

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

\(=\left(x-1\right)\left(x^3-x-1\right)\)

5. \(a^3x-ab+b-x\)

\(=a^3x-x-ab+b\)

\(=x\left(a^3-1\right)-b\left(a-1\right)\)

\(=x\left(a-1\right)\left(a^2+a+1\right)-b\left(a-1\right)\)

\(=\left(a-1\right)\left[x\left(a^2+a+1\right)-b\right]\)

6. \(x^3-64=x^3-4^3\)

\(=\left(x-4\right)\left(x^2+4x+16\right)\)

7. \(0,125\left(a+1\right)^3-1\)

\(=\left[0,5\left(a+1\right)\right]^3-1^3\)

\(=\left[0,5\left(a+1\right)-1\right]\left\{\left[0,5\left(a+1\right)\right]^2+\left[0,5\left(a+1\right).1\right]+1^2\right\}\)

\(=\left[0,5\left(a+1-2\right)\right]\left[0,25a^2+0,5a+0,25+0,5a+0,5+1\right]\)

\(=\left[0,5\left(a-1\right)\right]\left(0,25a^2+a+1,75\right)\)

8. \(9\left(x+5\right)^2-\left(x-7\right)^2\)

\(=\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2\)

\(=\left(3x+15-x+7\right)\left(3x+15+x-7\right)\)

\(=\left(2x+22\right)\left(4x+8\right)\)

9. \(49\left(y-4\right)^2-9\left(y+2\right)^2\)

\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)

\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)

\(=\left(4y-34\right)\left(10y-22\right)\)

10. \(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(xy-1\right)\)

11. \(x^3+3x^2+3x+1-27z^3\)

\(=\left(x+1\right)^3-\left(3z\right)^3\)

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

12. \(x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-1\right)\)

31 tháng 5 2018

3) \(x^2-7x+6=0\)

\(\Leftrightarrow x^2-6x-x+6=0\)

\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)

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

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

S=\(\left\{6;1\right\}\)

\(\)