Bài 1:Viết mỗi biểu thức sau dưới dạng bình phương một đa thức:
a,-(2x^2+2x+1)(2x^2-2x+1)+(2x^2+1^2)
b,(x^2+x+2)^2+(x-1)^2-2(x^2+x+2)(x-1)
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TT
2 tháng 9 2020
Bài 1 :
a) \(\left(x-4\right)\left(x+4\right)=x^2-16\)
b) \(\left(x-5\right)\left(x+5\right)=x^2-25\)
Bài 2 :
a) \(x^2-2x+1=\left(x-1\right)^2\)
b) \(x^2+2x+1=\left(x+1\right)^2\)
c) \(x^2-6x+9=\left(x-3\right)^2\)
XO
2 tháng 9 2020
1) a. (x - 4)(x + 4) = x2 - 4x + 4x - 16 = x2 - 16
b. (x - 5)(x + 5) = x2 - 5x + 5x - 25 = x2 - 25
2. x2 - 2x + 1 = x2 - x - x + 1 = x(x - 1) - (x - 1) = (x - 1)2
(x2 + 2x + 1) = x2 + x + x + 1 = x(x + 1) + (x + 1) = (x + 1)2
x2 - 6x + 9 = x2 - 3x - 3x + 9 = x(x - 3) -3(x - 3) = (x - 3)2
24 tháng 8 2021
a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)
\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)
\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)
\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)
\(=\left(x^2+9x+19\right)^2\)
NT
24 tháng 8 2021
b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)
\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)
\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)
c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)
\(=x^2-2x\left(y+2\right)+\left(y+2\right)^2\)
\(=\left(x-y-2\right)^2\)
d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)
\(=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+y+1\right)^2\)
17 tháng 7 2021
a) Ta có: \(\left(x^2+9x+18\right)^2+2\left(x^2+9x\right)+37\)
\(=\left(x^2+9x+18\right)^2+2\cdot\left(x^2+9x+18\right)-36+37\)
\(=\left(x^2+9x+19\right)^2\)
b) Ta có: \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)
\(=\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+2\left(x+1\right)\left(y+1\right)\)
\(=\left(x^2+2x+2+y^2+2y\right)^2\)
18 tháng 9 2021
a) \(2x^2-2x-x^2+6=0\)
\(\Leftrightarrow x^2-2x+1+5=0\)
\(\Leftrightarrow\left(x-1\right)^2=-5\) ( vô lý)
Vậy không có x thoả mãn \(2x.\left(x-1\right)-x^2+6=0\)
b) \(x^4-2x^2.\left(3+2x^2\right)+3x^2.\left(x^2+1\right)=-3\)
\(\Leftrightarrow x^4-6x^2-4x^4+3x^4+3x^2+3=0\)
\(\Leftrightarrow3-3x^2=0\)
\(\Leftrightarrow3x^2=3\Leftrightarrow x^2=1\) \(\Leftrightarrow x\in\left\{-1;1\right\}\)
Vậy \(x\in\left\{-1;1\right\}\)
c) \(\left(x+1\right).\left(x^2-x+1\right)-2x=x.\left(x-2\right).\left(x+2\right)\)
\(\Leftrightarrow x^3+1-2x-x.\left(x^2-4\right)=0\)
\(\Leftrightarrow x^3+1-2x-x^3+4x=0\)
\(\Leftrightarrow1+2x=0\Leftrightarrow x=\dfrac{-1}{2}\)
Vậy x=\(\dfrac{-1}{2}\)
d) \(\left(x+3\right).\left(x^2-3x+9\right)-x.\left(x-2\right).\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x.\left(x^2-4\right)-15=0\)
\(\Leftrightarrow x^3-27-x^3+4x-15=0\)
\(\Leftrightarrow4x-42=0\)
\(\Leftrightarrow x=10,5\)
Vậy x=10,5
17 tháng 8 2021
\(a)\)
\(\frac{1}{x+1}-\frac{x-1}{x}=\frac{3x+1}{x\left(x+1\right)}\)
\(\Leftrightarrow x-x^2+1=3x+1\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b)\)
\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)
\(\Leftrightarrow x^2+2x+1=x^2+10\)
\(\Leftrightarrow2x-9=0\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{2}{9}\)
20 tháng 10 2020
Bài 1.
a) -2x( -3x + 2 ) - ( x + 2 )2
= 6x2 - 4x - ( x2 + 4x + 4 )
= 6x2 - 4x - x2 - 4x - 4
= 5x2 - 8x - 4
b) ( x + 2 )( x2 - 2x + 4 ) - 2( x + 1 )( 1 - x )
= x3 + 8 + 2( x + 1 )( x - 1 )
= x3 + 8 + 2( x2 - 1 )
= x3 + 8 + 2x2 - 2
= x3 + 2x2 + 6
c) ( 2x - 1 )2 - 2( 4x2 - 1 ) + ( 2x + 1 )2
= 4x2 - 4x + 1 - 8x2 + 2 + 4x2 + 4x + 1
= 4
d) x2 - 3x + xy - 3y
= x( x - 3 ) + y( x - 3 )
= ( x - 3 )( x + y )
Bài 2.
a) 4x2 - 4xy + y2 = ( 2x - y )2
b) 9x3 - 9x2y - 4x + 4y
= 9x2( x - y ) - 4( x - y )
= ( x - y )( 9x2 - 4 )
= ( x - y )( 3x - 2 )( 3x + 2 )
c) x3 + 2 + 3( x3 - 2 )
= x3 + 2 + 3x3 - 6
= 4x3 - 4
= 4( x3 - 1 )
= 4( x - 1 )( x2 + x + 1 )
Bài 3.
2( x - 2 ) = x2 - 4x + 4
⇔ ( x - 2 )2 - 2( x - 2 ) = 0
⇔ ( x - 2 )( x - 2 - 2 ) = 0
⇔ ( x - 2 )( x - 4 ) = 0
⇔ x = 2 hoặc x = 4
BA
30 tháng 7 2019
\(\left|2x-\frac{1}{2}\right|+1=3x\)
\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}=3x-1\\2x-\frac{1}{2}=1-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1+\frac{1}{2}\\2x+3x=1+\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)
b) \(\left(x^2+x+2\right)^2+\left(x-1\right)^2-2\left(x^2+x+2\right)\left(x-1\right)\)
\(=\left(x^2+x+2\right)^2-2\left(x^2+x+2\right)\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(x^2+x+2-x+1\right)^2\)
\(=\left(x^2+3\right)^2\)