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a) \(29.31=\left(30-1\right)\left(30+1\right)=30^2-1=900-1=899\)
b) \(33,3^2-2.33,3.3,3+3,3^2=\left(33,3-3,3\right)^2=30^2=900\)
c) \(20,1.19,9=\left(20+0,1\right)\left(20-0,1\right)=20^2-0,1^2=400-0.01=399,99\)
d) \(37^2+2.37.13+13^2=\left(37+13\right)^2=50^2=2500\)
a. y4 - 14y2 + 49
Gọi y2 là t, ta có:
t2 - 14t + 49
<=> t2 - 14t + 72
<=> (t - 7)2
Thay x2 = t
<=> (x2 - 7)2
b. \(\dfrac{1}{4}-x^2\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^2-x^2\)
\(\Leftrightarrow\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)
c. x4 - 16
<=> (x2)2 - 42
<=> (x2 - 4)(x2 + 4)
d. x2 - 9
<=> x2 - 32
<=> (x - 3)(x + 3)
`a, 4x^2 - 25y^2 = (2x-5y)(2x+5y)`.
`b, 8x^3 +27 = (2x+3)(4x^2 - 6x + 9)`.
`c, 125x^3 - 64y^3 = (5x)^3 - (4y)^3 = (5x-4y)(25x^2 + 20xy + 16y^2)`.
\(a,\\ 4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\\ b,\\ 8x^3+27=\left(2x\right)^3+3^3=\left(2x+3\right)\left(4x^2+6x+9\right)\\ c,\\ 125x^3-64y^3=\left(5x\right)^3-\left(4y\right)^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)
a: \(=\dfrac{x-2x-1}{x+1}=\dfrac{-\left(x+1\right)}{x+1}=-1\)
b: \(=\dfrac{2+2x}{x\left(x+1\right)}=\dfrac{2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x}\)
c: \(=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
2:
a: \(=\left(2x^2-xy\right)+\left(2xz-yz\right)\)
\(=x\left(2x-y\right)+z\left(x-2y\right)=\left(x-2y\right)\left(x+z\right)\)
b: \(=\left(x^2-4y^2\right)-\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y-1\right)\)
c: \(=\left(y^2+10y+25\right)-9z^2\)
\(=\left(y+5\right)^2-\left(3z\right)^2\)
\(=\left(y+5+3z\right)\left(y+5-3z\right)\)
d: \(=\left(x+2y\right)^3-\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x+2y\right)\left[\left(x+2y\right)^2-\left(x-2y\right)\right]\)
\(=\left(x+2y\right)\left(x^2+4xy+4y^2-x+2y\right)\)
1:
a: \(x\left(3-4x\right)+5\left(3-4x\right)=\left(3-4x\right)\left(x+5\right)\)
b: \(2y\left(5y-6\right)-4\left(6-5y\right)\)
\(=2y\left(5y-6\right)+4\left(5y-6\right)\)
\(=2\left(5y-6\right)\left(y+2\right)\)
c: \(=27\left(x-2\right)^3-3x\left(x-2\right)^2\)
\(=3\left(x-2\right)^2\cdot\left[9\left(x-2\right)-x\right]\)
\(=3\left(x-2\right)^2\left(8x-18\right)=6\left(x-2\right)^2\cdot\left(4x-9\right)\)
d: \(=6y\left(x-y\right)\left(x+y\right)-8y\left(x+y\right)^2\)
\(=2y\left(x+y\right)\left[3\left(x-y\right)-4\left(x+y\right)\right]\)
\(=2y\left(x+y\right)\left(3x-3y-4x-4y\right)\)
\(=2y\left(x+y\right)\left(-x-7y\right)\)
Bài 1
a) x(3 - 4x) + 5(3 - 4x)
= (3 - 4x)(x + 5)
b) 2y(5y - 6) - 4(6- 5y)
= 2y(5y - 6) + 4(5y - 6)
= (5y - 6)(2y + 4)
= 2(5y - 6)(y + 2)
c) 27(x - 2)³ - 3x(2 - x)²
= 27(x - 2)³ - 3x(x - 2)²
= 3(x - 2)²[9(x - 2) - x]
= 3(x - 2)²(9x - 18 - x)
= 3(x - 2)²(8x - 18)
= 6(x - 2)²(4x - 9)
d) 6y(x² - y²) - 8y(x + y)²
= 6y(x - y)(x + y) - 8y(x + y)²
= 2y(x + y)[3(x - y) - 4(x + y)]
= 2y(x + y)(3x - 3y - 4x - 4y)
= 2y(x + y)(-x - 7y)
= -2y(x + y)(x + 7y)
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
a, \(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
b, \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
c, \(x^2-10x+25=x^2-2.5x+5^2=\left(x-5\right)^2\)
1. \(25+10x+x^2\\ \Leftrightarrow5^2+2\cdot5\cdot x+x^2\\ \Leftrightarrow\left(5+x\right)^2\\ \Leftrightarrow\left(5+x\right)\left(5+x\right)\)
2. \(8x^3-\dfrac{1}{8}\\ \Leftrightarrow\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[4x^2+x+\dfrac{1}{4}\right]\)
3. \(x^2-10x+25\\ \Leftrightarrow x^2-2\cdot5\cdot x+5^2\\ \Leftrightarrow\left(x-5\right)^2\\ \Leftrightarrow\left(x-5\right)\left(x-5\right)\)
Bài 9:
a) Ta có: \(A=\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)
\(=4x^2+4xy+y^2-4x^2+y^2-xy-y^2\)
\(=3xy-y^2\)
\(=3\cdot\left(-2\right)\cdot3-3^2=-18-9=-27\)
b) Ta có: \(B=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)
\(=\dfrac{31}{2}\)
Bài 7:
a) \(498^2=\left(500-2\right)^2=250000-2000+4=248004\)
b) \(93\cdot107=100^2-7^2=10000-49=9951\)
c) \(163^2+74\cdot163+37^2=\left(163+37\right)^2=200^2=40000\)
d) \(1995^2-1994\cdot1996=1995^2-1995^2+1=1\)
e) \(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=18^8-18^8+1=1\)
f) \(125^2-2\cdot125\cdot25+25^2=\left(125-25\right)^2=100^2=10000\)
a)\(=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
b)\(=\left(x+9\right)^2-\left(6x\right)^2=\left(x+9-6x\right)\left(x+9+6x\right)=\left(-5x+9\right)\left(7x+9\right)\)
c)\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t\right)^2\\ =\left(x-y+z-t\right)\left(x-y-z+t\right)\)
Bài 1:
a: \(\left(3x+2\right)^2-4=3x\left(3x+4\right)\)
Bài 2:
a: \(4x^2+4x+1\)
b: \(9x^2+9x+\dfrac{9}{4}\)