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3(x-3)(x+7)+(x-4)^2+48=3x^2+12x-63+x^2-8x+63
=4x^2+4x=4x(x+1)
![](https://rs.olm.vn/images/avt/0.png?1311)
thay x=0,5 vào biểu thức ta có:
3.(0,5-3).(0,5+7)+(0,5-4)^2+48
=3.(-2,5).7,5+(-3,5)^2+48
=3.(-2,5).7,5+12,25+48
=-7,5.(7,5)+12,25+48
=-56,25+12,25+48
=-44+48
=4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(=\left(3x-9\right)\left(x+7\right)+\left(x^2-8x+16\right)+48\)
\(=3x^2+21x-9x-63+x^2-8x+16+48\)
\(=4x^2+4x+1\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)
\(=\left(2x+1\right)^2\)
Thay x = 0,5 vào biểu thức ta có :
\(\left(2\cdot0,5+1\right)^2\)
\(=\left(1+1\right)^2\)
\(=2^2\)
\(=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-2xy-4z^2+y^2=\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)=\left(6+4-2.45\right)\left(6+4+2.45\right)=-8000\)b) \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48=3\left(x^2+4x-21\right)+\left(x^2-8x+16\right)+48=4x^2+4x+1=\left(2x+1\right)^2=\left(2.0,5+1\right)^2=4\)
a: Ta có: \(x^2-2xy+y^2-4z^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2\)
\(=\left(x-y-2z\right)\left(x-y+2z\right)\)
\(=\left(6+4-2\cdot45\right)\left(6+4+2\cdot45\right)\)
\(=-8000\)
b: Ta có: \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(=3\left(x^2+4x-21\right)+\left(x-4\right)^2+48\)
\(=3x^2+12x-63+x^2-8x+16+48\)
\(=2x^2+4x+1\)
\(=2\cdot\dfrac{1}{4}+4\cdot\dfrac{1}{2}+1\)
\(=\dfrac{7}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,5x^2 - 10xy + 5y^2 - 20z^2
=5(x^2 -2xy +y^2-4z^2 )
=5[(x-y)^2-(2z)^2 ]
=5 .(x-y-2z)(x-y+2z)
b,.= (5x^2+5xy)-(x+y)
=5x(x+y)-(x+y)
=(x+y)(5x-1)
d,x2 - 4x + 3 = x2 - x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)
e,x2 - x - 6 = x2 +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
f,x2 - x - 6 = x2 +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
g,2x^2(3x - 5)
= 2x^2 x 3x - 2x^2 x 5
= 6x^3 - 10x^2
\(\text{1) }\)
\(\text{a) }5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)
\(\text{b) }5x^2+5xy-x-y\)
\(=\left(5x^2-x\right)+\left(5xy-y\right)\)
\(=x\left(5x-1\right)+y\left(5x-1\right)\)
\(=\left(5x-1\right)\left(x+y\right)\)
\(\text{c) }2\left(x+4\right)-x^2+16\)
\(=2\left(x+4\right)-\left(x^2-16\right)\)
\(=2\left(x+4\right)-\left(x+4\right)\left(x-4\right)\)
\(=\left(x+4\right)\left(2-x+4\right)\)
\(=\left(x+4\right)\left(6-x\right)\)
\(\text{d) }x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=\left(x^2+3x\right)+\left(x+3\right)\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+3\right)\left(x+1\right)\)
\(\text{e) }x^2+5x-6\)
\(=x^2+6x-x-6\)
\(=\left(x^2+6x\right)-\left(x+6\right)\)
\(=x\left(x+6\right)-\left(x+6\right)\)
\(=\left(x+6\right)\left(x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3( x - 3 )( x + 7) + ( x - 4)2 + 48=(3x-9)(x+7)+x2-2.x.4+42+48
=3x2+21x-9x-63+x2-8x+16+48
=4x2+4x+1
=4(x2+x)+1
Thay x=0,5 ta có:
3( x - 3 )( x + 7) + ( x - 4)2 + 48=4(0,52+0,5)+1=4(0,25+0,5)+1=4.0,75+1=3+1=4
Mình ko chắc lắm đâu nha, bạn bấm thử máy tính xem, sai thì cho mình xin lỗi
3( x - 3 )( x + 7) + ( x - 4)2 + 48=(3x-9)(x+7)+x2-2.x.4+42+48
=3x2+21x-9x-63+x2-8x+16+48
=4x2+4x+1
=4(x2+x)+1
Thay x=0,5 ta có:
3( x - 3 )( x + 7) + ( x - 4)2 + 48=4(0,52+0,5)+1=4(0,25+0,5)+1=4.0,75+1=3+1=4
Mình ko chắc lắm đâu nha, bạn bấm thử máy tính xem, sai thì cho mình xin lỗi
![](https://rs.olm.vn/images/avt/0.png?1311)
Có:\(3\left(x+3\right)\left(x-7\right)+\left(x+4\right)^2+48\)
\(=3\left(x^2-7x+3x-21\right)+\left(x^2+8x+16\right)+48\)
\(=3x^2-21x+9x-63+x^2+8x+16+48\)
\(=4x^2-4x+1\)
\(=\left(2x+1\right)^2\)
Với x=0,5 ta co:\(\left(2x+1\right)^2=\left(2\cdot0,5+1\right)^2=\left(1+1\right)^2=4\)
3(x-3)(x+7)+(x-4)2+48
=3x2+12x-63+x2-8x+16+48
=4x2+4x+1
=(2x+1)2
Thay x=0,5 ta có:
(2.0,5+1)2=(1+1)2=22=4
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(\Rightarrow\left(3x-9\right).\left(x+7\right)+x^2-4.2.x+4^2+48\)
\(\Rightarrow3x\left(x+7\right)-9\left(x+7\right)+x^2-8x+16+48\)
\(\Rightarrow9x^2+21x-9x+63+x^2-8x+64\)
\(\Rightarrow\left(9x^2+x^2\right)+\left(21x-9x-8x\right)+63+64=10x^2+4x+127\)
Tại x = 0,5
\(\Rightarrow10.0,5^2+4.0,5+127=131,5\)
1) \(2x^2+5x-3=0\)
\(\Leftrightarrow2x^2+6x-x-3=0\)
\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}}\)
\(2x^2+5x-3=0\)
\(\Leftrightarrow2x^2+2x+3x-3=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)