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\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
a) \(=5x^2+40x+80+4\left(x^2-10x+25\right)-9\left(x+4\right)\left(x-4\right)\)
\(=5x^2+40x+80+4x^2-40x+100-9x^2+144\)
\(=9x^2-9x^2+40x-40x+324\)
\(=324\)
b) \(=x^2+4xy+4y^2+4x^2-4xy+y^2-5x^2+5y^2-10y^2+90\)
\(=5x^2-5x^2+10y^2-10y^2+\left(4xy-4xy\right)+90\)
\(=90\)
c)
\(=a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\)
\(=\left(2a^2-2a^2\right)+\left(2b^2-2b^2\right)+2c^2+4ab-4ab+2\left(ac+bc-ac-bc\right)\)
\(=2c^2\)
a) 5( x + 4 )2 + 4( x - 5 )2 - 9( 4 + x )( x - 4 )
= 5( x2 + 8x + 16 ) + 4( x2 - 10x + 25 ) - 9( x2 - 16 )
= 5x2 + 40x + 80 + 4x2 - 40x + 100 - 9x2 + 144
= ( 5x2 + 4x2 - 9x2 ) + ( 40x - 40x ) + ( 80 + 100 + 144 )
= 324
b) ( x + 2y )2 + ( 2x - y )2 - 5( x + y )( x - y ) - 10( y + 3 )( y - 3 )
= x2 + 4xy + 4y2 + 4x2 - 4xy + y2 - 5( x2 - y2 ) - 10( y2 - 9 )
= x2 + 4xy + 4y2 + 4x2 - 4xy + y2 - 5x2 + 5y2 - 10y2 + 90
= ( x2 + 4x2 - 5x2 ) + ( 4xy - 4xy ) + ( 4x2 + y2 + 5y2 - 10y2 ) + 90
= 90
c) ( a + b + c )2 + ( a + b - c )2 - 2( a + b )2
= [ ( a + b ) + c ]2 + [ ( a + b ) - c ]2 - 2( a + b )2
= ( a + b )2 + 2( a + b )c + c2 + ( a + b )2 - 2( a + b )c + c2 - 2( a + b )2
= [ ( a + b )2 + ( a + b )2 - 2( a + b )2 ] + [ 2( a + b )c - 2( a + b )c ] + ( c2 + c2 )
= 2c2
a) A \(=\)\(\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)\(=\)\(\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\)\(\frac{2\left(x-2\right)}{x+2}\)\(=\)\(\frac{2x-4}{x+2}\)
Tại x = \(\frac{1}{2}\)thì:
A = \(\frac{2.\frac{1}{2}-4}{\frac{1}{2}+2}\)\(=\)\(\frac{-3}{\frac{5}{2}}\)\(=\)\(\frac{-6}{5}\)
mình biết câu b rồi nhưng câu a thì chưa!
b) x^3(x+y)-x^2(x^2+xy)-x(x-y)
=x^4+x^3y-x^4-x^3y-x^2+xy
=-x^2+xy tại x=10,y=-5 ta có;
=-10^2+10(-5)
= 50
a)\(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-b+c\right)}=\dfrac{a+b-c}{a-b+c}\)Giá trị của biểu thức trên tại \(a=4;b=-5;c=6\) là:
\(\dfrac{4-5-6}{4-\left(-5\right)+6}=-\dfrac{7}{15}\)
b: \(=\dfrac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\dfrac{2x-5y}{x-3y}\)
Đặt x/10=y/3=k
=>x=10k; y=3k
\(A=\dfrac{2\cdot10k-5\cdot3k}{10k-3\cdot3k}=\dfrac{5k}{k}=5\)
c: \(C=\left(\dfrac{x^3-y^3-x^3-y^3}{\left(x+y\right)\left(x-y\right)}\right):\dfrac{x^2-y^2-x^2}{x+y}\)
\(=\dfrac{-2y^3}{\left(x+y\right)\left(x-y\right)}\cdot\dfrac{x+y}{-y^2}=\dfrac{2y}{x-y}\)
\(=\dfrac{20}{9-10}=-20\)
Đk: x, y \(\ne\)0
Ta có: P = \(\frac{2}{x}-\left(\frac{x^2}{x^2+xy}+\frac{y^2-x^2}{xy}-\frac{y^2}{xy+y^2}\right)\cdot\frac{x+y}{x^2+xy+y^2}\)
P = \(\frac{2}{x}-\left(\frac{x^3+\left(y^2-x^2\right)\left(x+y\right)-y^3}{xy\left(x+y\right)}\right)\cdot\frac{x+y}{x^2+xy+y^2}\)
P = \(\frac{2}{x}-\frac{\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x-y\right)\left(x+y\right)^2}{xy\left(x+y\right)}\cdot\frac{x+y}{x^2+xy+y^2}\)
P = \(\frac{2}{x}-\frac{\left(x-y\right)\left(x^2+xy+y^2-x^2-2xy-y^2\right)}{xy\left(x^2+xy+y^2\right)}\)
P = \(\frac{2}{x}-\frac{-xy\left(x-y\right)}{xy\left(x^2+xy+y^2\right)}=\frac{2}{x}+\frac{x-y}{x^2+xy+y^2}=\frac{2x^2+2xy+2y^2+x^2-xy}{x\left(x^2+xy+y^2\right)}\)
P = \(\frac{3x^2+xy+2y^2}{x\left(x^2+xy+y^2\right)}\)
b) Ta có: x2 + y2 + 10 = 2x - 6y
<=> x2 - 2x + 1 + y2 + 6y + 9 = 0
<=> (x - 1)2 + (y + 3)2 = 0
<=> \(\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\) <=> \(\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
Do đó: P = \(\frac{3.1^2-3.1+2.\left(-3\right)^2}{1\left(1^2-3+\left(-3\right)^2\right)}=\frac{18}{7}\)
giải
Áp dụng hàng đẳng thức đáng nhớ :
a ) \(5.\left(x+2\right)\left(x-2\right)-\left(3-4x\right)^2\)
\(=5\left(x^2-2^2\right)-\left(9-24x+16x^2\right)\)
\(=5x^2-20-9+24x-16x^2\)
\(=-11x^2+24x-29\)
b ) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2\left(x^2-y^2\right)+\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)
\(=2x^2-2y^2+2x ^2+2y^2=4x^2\)
Chúc bạn học tốt !!!
Bài làm
a) 5 . ( x + 2 )( x - 2 ) - ( 3 - 4x )2
= 5 . ( x2 - 22 ) - [ 32 - 2.3.4x + ( 4x )2 ]
= 5x2 - 20 - ( 9 - 24x + 16x2 )
= 5x2 - 20 - 9 + 24x - 16x2
= ( 5x2 - 16x2 ) - ( 20 + 9 ) + 24x
= -11x2 - 29 + 24x
b) 2( x - y )( x + y ) + ( x + y )2 + ( x - y )2
= 2( x2 - y2 ) + x2 + 2xy + y2 + x2 - 2xy + y2
= 2x2 - 2y2 + x2 + 2xy + y2 + x2 - 2xy + y2
= ( 2x2 + x2 + x2 ) + ( -2y2 + y2 + y2 ) + ( 2xy - 2xy )
= 4x2
# Học tốt #
