BT6: Thu gọn về hàng đẳng...
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BT6: Thu gọn về hàng đẳng thức
\(3,\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(4,\left(3x-5\right)^2-2\left(3x-5\right)\left(3x+5\right)+\left(3x+5\right)^2\)
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H
6 tháng 7 2023
\(1,\left(2x+1\right)^2+2\left(2x+1\right)+1\\ =\left(2x+1\right)^2+2.\left(2x+1\right).1+1^2\\ =\left[\left(2x+1\right)+1\right]^2\\ b,\left(3x-2y\right)^2+4\left(3x-2y\right)+4\\ =\left(3x-2y\right)^2+2.\left(3x-2y\right).2+2^2\\ =\left[\left(3x-2y\right)+2\right]^2\)
6 tháng 7 2023
1) \(\left(2x+1\right)^2+2\left(2x+1\right)+1\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\cdot1+1^2\)
\(=\left[\left(2x+1\right)+1\right]^2\)
\(=\left(2x+2\right)^2\)
2) \(\left(3x+2y\right)^2+4\left(3x+2y\right)+4\)
\(=\left(3x+2y\right)^2+2\cdot\left(3x+2y\right)\cdot2+2^2\)
\(=\left[\left(3x+2y\right)+2\right]^2\)
\(=\left(3x+2y+2\right)^2\)
6 tháng 7 2023
5) \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)^2-2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left[\left(x-y\right)-\left(x+y\right)\right]^2\)
\(=\left(x-y-x-y\right)^2\)
\(=\left(-2y^2\right)\)
\(=4y^2\)
6) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)
\(=\left(x-5\right)^2-2\left(x-5\right)\left(x+5\right)+\left(x+5\right)^2\)
\(=\left[\left(x-5\right)-\left(x+5\right)\right]^2\)
\(=\left(x-5-x-5\right)^2\)
\(=\left(-10\right)^2=100\)
7) \(\left(x-2\right)^2+\left(x+1\right)^2+2\left(x-2\right)\left(-1-x\right)\)
\(=\left(x-2\right)^2-2\left(x-2\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left[\left(x-2\right)-\left(x+1\right)\right]^2\)
\(=\left(-3\right)^2=9\)
8) \(-\left(2x+3y\right)^2+\left(2x-3y\right)^2-2\left(4x^2-9y^2\right)\)
\(=\left(2x-3y\right)^2+2\left(2x+3y\right)\left(2x-3y\right)+\left(2x+3y\right)^2\)
\(=\left[\left(2x+3y\right)+\left(2x-3y\right)\right]^2\)
\(=\left(4x\right)^2=16x^2\)
H
12 tháng 6 2018
a)\(9x^2+30x+25+9x^2-30x+25-\left(9x^2-2^2\right)\)
=\(9x^2+54\)=\(9\left(x^2+6\right)\)
b)\(2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)
=\(8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)
=\(x^3-16x^2+25x\)
c)\(\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2\)
=\(\left(x+y-z-\left(x+y\right)\right)^2\)=\(\left(-z\right)^2\)
DN
12 tháng 6 2018
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
25 tháng 8 2020
a) ( x - 5 )( 2x + 3 ) + 2x( 1 - x )
= 2x2 - 7x - 15 + 2x - 2x2
= -5x - 15
= -5( x + 3 )
b) ( 3x - 5 )2 - ( x + 5 )( 5 - x ) - 5/2( -2x )2
= 9x2 - 30x + 25 + ( x + 5 )( x - 5 ) - 5/2.4x2
= 9x2 - 30x + 25 + x2 - 25 - 10x2
= -30x
c) ( 3x + 2 )( 4 - 6x + 9x2 ) - 3x( 3x - 2 )2 + 12( -2/3 - 3x2 )
= ( 3x )3 + 23 - 3x( 9x2 - 12x + 4 ) - 8 - 36x2
= 27x3 + 8 - 27x3 + 36x2 - 12x - 8 - 36x2
= -12x
25 tháng 2 2023
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
25 tháng 8 2018
rút gọn biểu thức
a)2x(2x−1)2−3x(x+3)(x−3)−4x(x+1)2
=2x(4x2-4x+1)-3x.(x2-9)-4x(x2+2x+1)
=8x3-8x2+2x-3x3-27x-4x3-8x2-4x
=8x3-16x2-7x3-29x
25 tháng 7 2017
Câu 1: \(3x+2\left(5-x\right)=0\)
\(\Rightarrow3x+10-2x=0\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\).
Câu 2: \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=3\)
\(\Rightarrow2x\left(5-3x\right)-2x\left(5-3x\right)-3\left(x-7\right)=0\)
\(\Rightarrow\left(2x-2x\right)\left(5-3x\right)-3\left(x-7\right)=3\)
\(\Rightarrow-3\left(x-7\right)=3\)
\(\Rightarrow x-7=-1\)
\(\Rightarrow x=6.\)
25 tháng 7 2017
Câu 3:
Áp dụng hằng đẳng thức mở rộng có:
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=a^3+b^3+c^3-3abc.\)
Câu 4: \(3x^2\left(3x^2-2y^2\right)-\left(3x^2-2y^2\right)\left(3x^2+2y^2\right)\)
\(=\left(3x^2-2y^2\right)\left[3x^2-\left(3x^2+2y^2\right)\right]\)
\(=\left(3x^2-2y^2\right)\left(-2y^2\right)\)
\(=-6x^2y^2+4y^3.\)
Câu 5:
Ta có: \(R=\left(2x-3\right)\left(4+6x\right)-\left(6-3x\right)\left(4x-2\right)\)
\(=\left(8x-12+12x^2-18x\right)-\left(24x-12x^2-12+6x\right)\)
\(=12x^2-10x-12-24x+12x^2+12-6x\)
\(=24x^2-40x.\)
3) \(\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)
\(=\left(x+3\right)^2-2\left(x+3\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left[\left(x+3\right)-\left(x-2\right)\right]^2\)
\(=\left(x+3-x+2\right)^2\)
\(=5^2=25\)
4) \(\left(3x-5\right)^2-2\left(3x-5\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x-5\right)-\left(3x+5\right)\right]^2\)
\(=\left(3x-5-3x-5\right)^2\)
\(=\left(-10\right)^2\)
\(=100\)