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\(a,A=-1+3-5+7-9+...-2013+2015-2017=\left(-1+3\right)+\left(-5+7\right)+...+\left(-2013+2015\right)-2017\)\(=2+2+..+2-2017\)
\(=2.504-2017=-1009\)
\(b,B=2-4+6-8+...+2014-2016+2018\)\(=2+\left(-4+6\right)+\left(-8+10\right)+...+\left(-2016+2018\right)==2+2+...+2\)\(=2+503.2=1008\)
a: \(=7x\left(xy-3\right)\)
d: \(=\left(x+1\right)\left(10x-8y\right)\)
\(=2\left(5x-4y\right)\left(x+1\right)\)
e: \(=\left(x-100\right)\cdot7x\)
f: \(=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)
Bài 2:
a: \(\left(2x+1\right)\left(1-2x\right)+\left(2x-1\right)^2=22\)
\(\Leftrightarrow\left(2x-1\right)\left(-2x-1\right)+\left(2x-1\right)^2=22\)
\(\Leftrightarrow\left(2x-1\right)\left(-2x-1+2x-1\right)=22\)
\(\Leftrightarrow2x-1=-11\)
=>2x=-10
hay x=-5
b: \(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow2x^2-10x+34-2x^2-4x-2=0\)
=>-14x+32=0
=>-14x=-32
hay x=16/7
c: \(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-4x+1-7\left(x^2-9\right)=36\)
\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\)
=>8x+76=36
=>8x=-40
hay x=-5
d: \(\Leftrightarrow\left(x^2-9\right)\left(x^2+9\right)-\left(x^4-4\right)-3x=15x-41\)
\(\Leftrightarrow x^4-81-x^4+4-3x-15x+41=0\)
=>-18x-36=0
hay x=-2
e: \(\Leftrightarrow x^2-14x+49-x^2-6x-9+x^2-10x+25=x^2-9\)
\(\Leftrightarrow x^2-30x+55=x^2-9\)
=>-30x+55=-9
=>-30x=-64
hay x=32/15
B1:
a) \(1001^2=\left(1000+1\right)^2\)
\(=1000^2+2.1000+1=1000000+2000+1\)
= \(1002001\)
b) \(29,9.30,1\)
= \(\left(30-0,1\right)\left(30+0,1\right)\)
= \(30^2-0,1^2=900-0,01=899,99\)
c) \(31,8^2-2.31,8.21,8+21,8^2\)
= \(\left(31,8-21,8\right)^2=10^2=100\)
B2:
a) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
b) \(a^6-b^3=\left(a^2\right)^3-b^3\)
= \(\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
c) \(8y^3-125=\left(2y\right)^3-5^3\)
= \(\left(2y-5\right)\left(4y^2+10y+25\right)\)
d) \(8z^3+27=\left(2z\right)^3+3^3\)
= \(\left(2z+3\right)\left(4z^2-6z+9\right)\)
B3:
a) A = \(x^2-20x+101\)
= \(x^2-20x+100+1\)
= \(\left(x-10\right)^2+1\ge1\) với mọi x
MinA = 1 khi và chỉ khi x = 10
b) B = \(4a^2+4a+2\)
= \(4a^2+4a+1+1\)
= \(\left(2a+1\right)^2+1\ge1\) với mọi x
MinB = 1 khi và chỉ khi a = \(-\dfrac{1}{2}\)
Bài 1. Tính:
a) \(x^2\left(x-2x^3\right)\)
\(=x^3-2x^5\)
b) \(\left(x^2+1\right)\left(5-x\right)\)
\(=5x^2-x^3+5-x\)
c. \(\left(x-2\right)\left(x^2+3x-4\right)\)
\(=x^3+3x^2-4x-2x^2-6x+8\)
\(=x^3+x^2-10x+8\)
d) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
e) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
f) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2+3x-6-6x^3-x^2+2x\)
\(=17x^2+5x-6-6x^3\)
g) \(\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
h) \(\left(xy-2\right)\left(x^3-2x-6\right)\)
\(=x^4y-2x^2y-6xy-2x^3+4x+12\)
i) \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)
\(=20x^3-5x^4+10x^3-4x^4+x^3-2x^2+8x^3-2x^2+4x-12x^2+3x-6\)
\(=39x^3-9x^4-16x^2+7x-6\)
Bài 5: Tìm x, biết
1) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2-9\right)-6=0\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\dfrac{-7}{-4}=\dfrac{7}{4}\)
Vậy \(x=\dfrac{7}{4}\)
2) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)-10=0\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\)
\(\Leftrightarrow-24x=-27\)
\(\Leftrightarrow x=\dfrac{-27}{-24}=\dfrac{9}{8}\)
Vậy \(x=\dfrac{9}{8}\)
4) \(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(x^2-4\right)-6=0\)
\(\Leftrightarrow x^2-8x+16-x^2+4-6=0\)
\(\Leftrightarrow-8x+14=0\)
\(\Leftrightarrow-8x=-14\)
\(\Leftrightarrow x=\dfrac{-14}{-8}=\dfrac{7}{4}\)
Vậy \(x=\dfrac{7}{4}\)
5) \(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(\Leftrightarrow9\left(x^2+2x+1\right)-\left(9x^2-4\right)-10=0\)
\(\Leftrightarrow9x^2+18x+9-9x^2+4-10=0\)
\(\Leftrightarrow18x+3=0\)
\(\Leftrightarrow18x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{18}=\dfrac{-1}{6}\)
Vậy \(x=\dfrac{-1}{6}\)
Bài 1.
a) ( x3 - 8) : ( x2 + 2x + 4 )
= ( x - 2)( x2 + 2x + 4 ) : ( x2 + 2x + 4 )
= x - 2
b) ( 3x2 - 6x ) : ( 2 - x)
= 3x( x - 2) : ( 2 - x)
= -3x( 2 - x ) : ( 2 - x)
= - 3x
Bài 2 .
\(\dfrac{2x-1}{x^2-x}\)
a) Để A có nghĩa tức là A xác định :
ĐKXĐ : x( x - 1) # 0
=> x # 0 ; x # 1
Vậy,...
b) Vì : x = 0 không thỏa mãn ĐKXĐ nên tại x = 0 giá trị của A không xác định
Vì : x = 3 thỏa mãn ĐKXĐ nên ta thay x = 3 vào A , ta có :
\(A=\dfrac{2.3-1}{3^2-3}=\dfrac{5}{6}\)
Vậy , tại : x = 3 thì A = \(\dfrac{5}{6}\)
Bài 3 .
a) ( 6x + 1)2 + ( 6x - 1)2 - 2( 1 + 6x )( 6x - 1)
= ( 6x + 1)2 - 2( 1 + 6x )( 6x - 1) + ( 6x - 1)2
= ( 6x + 1 - 6x + 1)2
= 1
b) 3( 22 + 1)( 24 + 1)( 28 + 1)( 216 + 1)
= ( 22 - 1)( 22 + 1)( 24 + 1)( 28 + 1)( 216 + 1)
= ( 24 - 1)( 24 + 1)( 28 + 1)( 216 + 1)
= ( 28 - 1)( 28 + 1)( 216 + 1)
= ( 216 - 1)( 216 + 1)
= 232 - 1
c) x( 2x2 - 3) - x2( 5x + 3 ) + 3x2
= 2x3 - 3x - 5x3 - 3x2 + 3x2
= - 3x3 - 3x
d) 3x( x - 2) - 5x( 1 - x) - 8( x2 - 3)
= 3x2 - 6x - 5x + 5x2 - 8x2 + 24
= -11x + 24
Ta có:
P = x 3 - 3 x 2 + 3 x - 1 + 1 = x - 1 3 + 1 T h a y x = 101 v à o P t a đ ư ợ c P = 101 - 1 3 + 1 = 100 3 + 1
Đáp án cần chọn là :A