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\(a,\left(x+2\right)^2=x^2+4x+4\\ b,\left(x-1\right)^2=x^2-2x+1\\ c,\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)
a) (x - 1/2x²y)²
= x² - 2x . 1/2 x²y + (1/2x²y)²
= x² - x³y + 1/4 x⁴y²
b) (2xy² - 1)(1 + 2xy²)
= (2xy²)² - 1²
= 4x²y⁴ - 1
c) (x - y + 2)²
= (x - y)² + 2(x - y).2 + 2²
= x² - 2xy + y² + 4x - 4y + 4
= x² + y² - 2xy + 4x - 4y + 4
d) (x + 1/2)(1/2 - x)
= (1/2)² - x²
= 1/4 - x²
e) (x² - 1/3)²
= (x²)² - 2x².1/3 + (1/3)²
= x⁴ - 2/3 x² + 1/9
m) \(\dfrac{1}{4}x^2-4x^2=\left(\dfrac{1}{2}x-2x\right)\left(\dfrac{1}{2}x+2x\right)\)
n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)
o) \(\left(x-3\right)\left(x+3\right)=x^2-9\)
a) (x² + 2)²
= (x²)² + 2.x².2 + 2²
= x⁴ + 4x² + 4
b) (x + y + z)²
= [(x + y) + z]²
= (x + y)² + 2(x + y).z + z²
= x² + 2xy + y² + 2xz + 2yz + z²
= x² + y² + z² + 2xy + 2xz + 2yz
1) \(\left[\left(a+b\right)-c\right]^2=\left(a+b\right)^2-2c\left(a+b\right)+c^2\)
\(=\left(a^2+2ab+b^2\right)-2ac-2bc+c^2\)
\(=a^2+b^2+c^2+2ab-2ac-2bc\)
2)Phần này tg tự
3)\(\left(x+y+z\right)\left(x+y-z\right)=\left(x+y\right)^2-z^2=x^2+2xy+y^2-z^2\)
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)
\(\left(2x+1\right)\left(x+3\right)+\left(x+1\right)^2\left(x+2\right)+\left(x+5\right)\left(x+1\right)\)
\(=2x^2+6x+x+3+x^3+2x^2+x+2x^2+4x+2+x^2+x+5x+5\)
\(=x^3+7x^2+18x+10\)
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\(\left(x-2\right)^2=x^2-2.x.2+2^2=x^2-4x+4.\)
\(\left(x-2\right)^2=x^2-4x+4\)