Hằng Đẳng Thức
(x+4)(x-6)+(x-7)2
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Những câu hỏi liên quan
16 tháng 8 2020
a) \(\left(5x-2\right)^2-\left(7-6x\right)^2=0\)
\(\Leftrightarrow\left(5x-2-7+6x\right)\left(5x-2+7-6x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}11x-9=0\\-x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{11}\\x=5\end{cases}}}\)
b) \(\left(3x-1\right)^2+\left(5x+2\right)^2=x+5\)
\(\Leftrightarrow9x^2+6x+1+25x^2+20x+4=x+5\)
\(\Leftrightarrow34x^2+26x+5=x+5\)
\(\Leftrightarrow34x^2+25x=0\)
\(\Leftrightarrow x\left(34x+25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\34x+25=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-25}{34}\end{cases}}}\)
c) Tự làm nốt
16 tháng 8 2020
a) ( 5x - 2 )2 - ( 7 - 6x )2 = 0
<=> [ 5x - 2 - ( 7 - 6x ) ][ 5x - 2 + ( 7 - 6x ) ] = 0
<=> [ 5x - 2 - 7 + 6x ][ 5x - 2 + 7 - 6x ] = 0
<=> [ 11x - 9 ][ 5 - x ] = 0
<=> \(\orbr{\begin{cases}11x-9=0\\5-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{11}\\x=5\end{cases}}\)
b) ( 3x - 1 )2 + ( 5x + 2 )2 = x + 5
<=> 9x2 - 6x + 1 + 25x2 + 20x + 4 = x + 5
<=> 34x2 + 14x + 5 = x + 5
<=> 34x2 + 14x + 5 - x - 5 = 0
<=> 34x2 + 13x = 0
<=> 13x( 34/13x + 1 ) = 0
<=> \(\orbr{\begin{cases}13x=0\\\frac{34}{13}x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{13}{34}\end{cases}}\)
c) ( x - 2 )2 - ( 3 + 2x )2 = 20x - 4
<=> x2 - 4x + 4 - ( 4x2 + 12x + 9 ) = 20x - 4
<=> x2 - 4x + 4 - 4x2 - 12x - 9 - 20x + 4 = 0
<=> -3x2 - 36x - 1 = 0
=> Vô nghiệm ( bấm EQN ra nghiệm vô tỉ )
18 tháng 7 2021
1, \(x^3+3^3=\left(x+3\right)\left(x^2-3x+9\right)\)
2, đề sai
3, \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
4, \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
5, \(1000-y^3=\left(10-y\right)=\left(100+10y+y^2\right)\)
tương tự ...
8, \(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
18 tháng 7 2021
Câu 2 đề ko sai nha bạn.
2) x2 - (\(\sqrt{y^3}\))2 ( y>0)
= ( x -\(\sqrt{y^3}\)) ( x +\(\sqrt{y^3}\))
LY
1
5 tháng 8 2018
\(X=7-2\sqrt{6}\)
\(X=\left(\sqrt{6}\right)^2-2\sqrt{6}.1+1\)
\(X=\left(\sqrt{6}-1\right)^2\)
6 tháng 10 2020
Phân tích thôi hả :v
( 3x - 7 )2 - 25 = ( 3x - 7 )2 - 52 = ( 3x - 7 - 5 )( 3x - 7 + 5 ) = ( 3x - 12 )( 3x - 2 ) = 3( x - 4 )( 3x - 2 )
( x - 1/2 )2 - 9/4 = ( x - 1/2 )2 - ( 3/2 )2 = ( x - 1/2 - 3/2 )( x - 1/2 + 3/2 ) = ( x - 2 )( x + 1 )
49 - ( x + 7 )2 = 72 - ( x + 7 )2 = [ 7 - ( x + 7 ) ][ 7 + ( x + 7 ) ] = ( 7 - x - 7 )( 7 + x + 7 ) = -x( x + 14 )
25 - ( x - 3 )2 = 52 - ( x - 3 )2 = [ 5 - ( x - 3 ) ][ 5 + ( x - 3 ) ] = ( 5 - x + 3 )( 5 + x - 3 ) = ( 8 - x )( x + 2 )
17 tháng 7 2018
1. \(\left(x+1\right)^3-125\)
\(=\left(x+1\right)^3-5^3\)
\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)
2. \(\left(x+4\right)^3-64\)
\(=\left(x+4\right)^3-4^3\)
\(=\left(x+4-4\right).\left[\left(x+4\right)^2+\left(x+4\right).4+4^2\right]\)
3. \(x^3-\left(y-1\right)^3\)
\(=(x^3-y+1).\left[\left(x^2\right)+x.\left(y+1\right)+\left(y+1\right)^2\right]\)
\(\)4. \(\left(a+b\right)^3-c^3\)
\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)
5. \(125-\left(x+2\right)^3\)
\(=5^3-\left(x+2\right)^3\)
\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)
6. \(\left(x+1\right)^3+\left(x-2\right)^3\)
\(=\left[\left(x+1\right)+\left(x-2\right)\right].\left[\left(x+1\right)^2-\left(x+1\right).\left(x-2\right)+\left(x-2\right)^2\right]\)
MD
2
27 tháng 8 2021
Mất dấu nên xét 2 th.
TH1
`x^2-4x+4=0`
`<=>x^2-2.x.2+2^2=0`
`<=>(x-2)^2=0`
`<=>x-2=0`
`<=>x=2`
`=>S={2}`
TH2
`x^2+4x+4=0`
`<=>x^2+2.x.2+2^2=0`
`<=>(x+2)^2=0`
`<=>x+2=0`
`<=>x=-2`
`=>S={-2}`
MD
3
27 tháng 8 2021
`x^2+4x+4=0`
`⇔x^2+2.x.2+2^2=0`
`⇔(x+2)^2=0`
`⇔x+2=0⇔x=−2`
Vậy `x=-2`.
DD
Đoàn Đức Hà
Giáo viên
1 tháng 7 2021
a) \(\left(x-3\right)^2+2\left(x-3\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left(x-3+x+2\right)^2\)
\(=\left(2x-1\right)^2\)
Hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b^2\).
b) \(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left(x+5\right)^2-2\left(x+5\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)
\(=11^2=121\)
Hằng đẳng thức: \(\left(a-b\right)^2=a^2-2ab+b^2\).
NC
1 tháng 7 2021
a.\(\left(x-3\right)^2+2\left(x-3\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left[\left(x-3\right)+\left(x+2\right)\right]^2\)
\(=\left(x-3+x+2\right)^2\)
\(=\left(2x-1\right)^2\)
b.\(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left(x+5\right)^2-2\left(x+5\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)
\(=\left(x+5-x+6\right)^2\)
\(\left(x+4\right)\left(x-6\right)+\left(x-7\right)^2\)
\(=x^2-2x-24+x^2-12x+49\)
\(=2x^2-14x+25\)