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Trả lời:
a) P = ( 2x + 1 )2 - 4x6 - 2x5
= 4x2 + 4x + 1 - 4x6 - 2x5
= - 4x6 - 2x5 + 4x2 + 4x + 1
b) Q = ( x + 2 )5 - x5 - x4
= ( x + 2 )2( x + 2 )2( x + 2 ) - x5 - x4
= ( x2 + 4x + 4 )( x2 + 4x + 4 )( x + 2 ) - x5 - x4
= ( x4 + 4x3 + 4x2 + 4x3 + 16x2 + 16x + 4x2 + 16x + 16 )( x + 2 ) - x5 - x4
= ( x4 + 8x3 + 24x2 + 32x + 16 )( x + 2 ) - x5 - x4
= x5 + 2x4 + 8x4 + 16x3 + 24x3 + 48x2 + 32x2 + 64x + 16x + 32 - x5 - x4
= 9x4 + 40x3 + 80x2 + 80x + 32
Bài 1.
\(a, (3x-4)^2\)
\(=\left(3x\right)^2-2\cdot3x\cdot4+4^2\)
\(=9x^2-24x+16\)
\(b,\left(1+4x\right)^2\)
\(=1^2+2\cdot1\cdot4x+\left(4x\right)^2\)
\(=16x^2+8x+1\)
\(c,\left(2x+3\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
\(d,\left(5-2x\right)^3\)
\(=5^3-3\cdot5^2\cdot2x+3\cdot5\cdot\left(2x\right)^2-\left(2x\right)^3\)
\(=125-150x+60x^2-8x^3\)
\(e,49x^2-25\)
\(=\left(7x\right)^2-5^2\)
\(=\left(7x-5\right)\left(7x+5\right)\)
\(f,\dfrac{1}{25}-81y^2\)
\(=\left(\dfrac{1}{5}\right)^2-\left(9y\right)^2\)
\(=\left(\dfrac{1}{5}-9y\right)\left(\dfrac{1}{5}+9y\right)\)
Bài 2.
\(a,\left(x-5\right)^2-\left(x+7\right)\left(x-7\right)=8\)
\(\Rightarrow x^2-2\cdot x\cdot5+5^2-\left(x^2-7^2\right)=8\)
\(\Rightarrow x^2-10x+25-\left(x^2-49\right)=8\)
\(\Rightarrow x^2-10x+25-x^2+49=8\)
\(\Rightarrow\left(x^2-x^2\right)-10x=8-25-49\)
\(\Rightarrow-10x=-66\)
\(\Rightarrow x=\dfrac{33}{5}\)
\(b,\left(2x+5\right)^2-4\left(x+1\right)\left(x-1\right)=10\)
\(\Rightarrow\left(2x\right)^2+2\cdot2x\cdot5+5^2-4\left(x^2-1^2\right)=10\)
\(\Rightarrow4x^2+20x+25-4x^2+4=10\)
\(\Rightarrow\left(4x^2-4x^2\right)+20x=10-25-4\)
\(\Rightarrow20x=-19\)
\(\Rightarrow x=\dfrac{-19}{20}\)
#\(Toru\)
Bài 1
a) (3x - 4)²
= (3x)² - 2.3x.4 + 4²
= 9x² - 24x + 16
b) (1 + 4x)²
= 1² + 2.1.4x + (4x)²
= 1 + 8x + 16x²
c) (2x + 3)³
= (2x)³ + 3.(2x)².3 + 3.2x.3² + 3³
= 8x³ + 36x² + 54x + 27
d) (5 - 2x)³
= 5³ - 3.5².2x + 3.5.(2x)² - (2x)³
= 125 - 150x + 60x² - 8x³
e) 49x² - 25
= (7x)² - 5²
= (7x - 5)(7x + 5)
f) 1/25 - 81y²
= (1/5)² - (9y)²
= (1/5 - 9y)(1/5 + 9y)
Bài 1:
\(\left(3x+4\right)^2=9x^2+24x+16\)
\(\left(x-1\right)^2=x^2-2x+1\)
\(\left(x-3\right)^2=x^2-6x+9\)
\(\left(\dfrac{1}{2}x-5\right)^2=\dfrac{1}{4}x^2-5x+25\)
\(x^2-1=\left(x+1\right)\left(x-1\right)\)
\(x^2-y^2=\left(x+y\right)\left(x-y\right)\)
\(x^2-2=\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)\)
\(4x-\dfrac{1}{9}=\left(2\sqrt{x}+\dfrac{1}{3}\right)\left(2\sqrt{x}-\dfrac{1}{3}\right)\)
Bài 3:
\(x^2-2x+1=\left(x-1\right)^2\)
\(x^2-10x+25=\left(x-5\right)^2\)
\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)
\(\left(2x-3\right)\left(2x+3\right)=4x^2-9\)
\(\left(\dfrac{2}{3}x+5\right)\left(\dfrac{2}{3}x-5\right)=\dfrac{4}{9}x^2-25\)
a: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
b: \(=\dfrac{x^5-3x^4+5x^3-x^2+3x-5}{x^2-3x+5}=x^2-1\)
c: \(=\dfrac{2x^4-5x^3+2x^2+2x-1}{x^2-x-1}\)
\(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
Câu 1 :
\(\left(x-2\right)^2=x^2-4x+4\)
Câu 2:
\(2x^2\left(4x-5x^3\right)+10x^5-5x^3\)
\(=8x^3-10x^5+10x^5-5x^3\)
\(=3x^3\)
\(\left(x-2\right)\left(x^2-2x+4\right)+\left(x-4\right)\left(x-2\right)\)
\(=x^3-4x^2+8x-8+x^2-6x+8\)
\(=x^3-3x^2+2x\)
Còn lại tự làm nha dài lắm
Ta có : x2 + 3x
= x2 + \(2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)
\(=\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)