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Cho b+c=10, chứng minh hằng đẳng thức:
(10a+b)(10a+c)=100a(a+1)+bc
Áp dụng để tính nhẩm:62. 68; 43. 47
![](https://rs.olm.vn/images/avt/0.png?1311)
b+c=10 => b=10-c
Ta có:
(10a+b)(10a+c)
\(=\left(10a+10-c\right)\left(10a+c\right)\)
\(=100a^2+10ac+100a+10c-10ac-c^2\)
\(=100a^2+100a+10c-c^2\) (1)
Ta lại có:
\(100a\left(a+1\right)+bc=100a\left(a+1\right)+\left(10-c\right)c\)
\(=100a^2+100a+10c-c^2\) (2)
Từ (1)(2) suy ra (10a+b)(10a+c)=100a(a+1)+bc
Ta có:
\(62.68=\left(10.6+2\right)\left(10.6+8\right)=100.6.\left(6+1\right)+2.8=4216\)
\(43.47=\left(10.4+3\right)\left(10.4+7\right)=100.4.\left(4+1\right)+3.7=2021\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(x+1\right)^2\\ b,=\left(y-2\right)^2\\ c,=\left(x-3\right)^2\\ d,=\left(a-7\right)^2\\ e,=\left(m-2\right)^2\\ f,=\left(2x-1\right)^2\\ g,=\left(a+5\right)^2\\ h,=\left(z-10^2\right)\\ i,=\left(x+3y\right)^2\\ j,=\left(2x-5b\right)^2\\ k,=\left(a+5\right)^2\\ l,=\left(x^2+1\right)^2\\ m,=\left(y^3-1\right)^2=\left(y-1\right)^2\left(y^2+y+1\right)^2\\ n,=\left(c^5-5\right)^2\\ o,=\left(3x^2+2y\right)^2\\ p,=5m^2n^3\left(5m^2n^3-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{\left|x-2y\right|}\) + |\(x\) + 2y| = 4
Hay \(\dfrac{1}{\left|x-2y\right|+\left|x+2y\right|}\) = 4 vậy em nhỉ
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)\)
\(=\left(2x+y\right)^2:\left(2x+y\right)\)
\(=2x+y\)
b ) \(\left(27x^3+1\right):\left(3x+1\right)\)
\(=\left(3x+1\right)\left(9x^2-3x+1\right):\left(3x+1\right)\)
\(=9x^2-3x+1\)
c ) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)\)
\(=\left(x-3y\right)^2:\left(3y-x\right)\)
\(=\left(3y-x\right)^2:\left(3y-x\right)\)
\(=3y-x\)
d ) \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)
\(=\left(2x-1\right)\left(4x^2+2x+1\right):\left(4x^2+2x+1\right)\)
\(=2x-1\)
:D
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-y\right)^2+4xy\)
\(=x^2-2xy+y^2+4xy\)
\(=x^2+2xy+y^2\)
\(=\left(x+y\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x + 1)(x + 2)(x + 5) − x2(x + 8) = 27
x2 + 2x + x + 2(x + 5) − x3 − 8x2 = 27
x2(x + 5) + 2x(x + 5) + x(x + 5) + 2(x + 5) − x3 − 8x2 = 27
x3 + 5x2 + 2x2 + 10x + x2 + 5x + 2x + 10 − x3 − 8x2 = 27
17x + 10 = 27
17x = 17
x = 17 : 17
x = 1
Vậy x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(10a-25-a^2=-\left(a^2-10a+25\right)=-\left(a^2-2.a.5+5^2\right)=-\left(a-5\right)^2\)