Làm tính nhân
a) (1/2x+y).(1/2x+y)
b) (x-1/2y).(x-1/2y)
c) (a+b)^2
d) (a-b)^2
e)(a+b).(a-b)
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a: \(\left(3x+y\right)^2=\left(3x\right)^2+2\cdot3x\cdot y+y^2=9x^2+6xy+y^2\)
b: \(\left(x+2y\right)^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)
\(=x^2+4xy+4y^2\)
c: \(\left(\dfrac{2}{3}x-1\right)^2=\left(\dfrac{2}{3}x\right)^2-2\cdot\dfrac{2}{3}x\cdot1+1^2\)
\(=\dfrac{4}{9}x^2-\dfrac{4}{3}x+1\)
d: \(\left(\dfrac{1}{3}x+2\right)^2=\left(\dfrac{1}{3}x\right)^2+2\cdot\dfrac{1}{3}x\cdot2+2^2\)
\(=\dfrac{1}{9}x^2+\dfrac{4}{3}x+4\)
e: \(\left(4x-2y\right)^2=\left(4x\right)^2-2\cdot4x\cdot2y+\left(2y\right)^2\)
\(=16x^2-16xy+4y^2\)
a) \(\left(3x+y\right)^2\)
\(=\left(3x\right)^2+2\cdot3x\cdot y+y^2\)
\(=9x^2+6xy+y^2\)
b) \(\left(x+2y\right)^2\)
\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)
\(=x^2+4xy+4y^2\)
c) \(\left(\dfrac{2}{3}x-1\right)^2\)
\(=\left(\dfrac{2}{3}x\right)^2-2\cdot\dfrac{2}{3}x\cdot1+1^2\)
\(=\dfrac{4}{9}x^2-\dfrac{4}{3}x+1\)
d) \(\left(\dfrac{1}{3}x+2\right)^2\)
\(=\left(\dfrac{1}{3}x\right)^2+2\cdot\dfrac{1}{3}x\cdot2+2^2\)
\(=\dfrac{x^2}{9}+\dfrac{4}{3}x+4\)
e) \(\left(4x-2y\right)^2\)
\(=\left(4x\right)^2-2\cdot4x\cdot2y+\left(2y\right)^2\)
\(=16x^2-16xy+4y^2\)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
\(a,A=5x^2a-10xya+5y^2a\)
\(=5a\left(x^2-2xy+y^2\right)\)
\(=5a\left(x-y\right)^2\)
Thay x = 124; y=24;a=2 ta có
\(5.2\left(124-24\right)^2=10.100^2=100000\)
\(b,B=2x^2+2y^2-x^2z+z-y^2z-2\)
\(=2\left(x^2+y^2-1\right)-z\left(x^2+y^2-1\right)\)
\(=\left(x^2+y^2-1\right)\left(2-z\right)\)
Thay x = 1 ; y = 1; z= -1 ta có
\(\left(1^2+1^2-1\right)\left(2-\left(-1\right)\right)=\left(1+1-1\right)\left(2+1\right)=1.3=3\)
\(c,C=x^2-y^2+2y-1\)
\(=x^2-\left(y^2-2y+1\right)=x^2-\left(y-1\right)^2=\left(x-y+1\right)\left(x+y-1\right)\)
Thay x = 75; y = 26 ta có
\(\left(75-26+1\right)\left(75+26-1\right)=50.100=5000\)
1) \(\frac{x-y}{x+y}=\frac{z-x}{z+x}\)
\(\Leftrightarrow\left(x-y\right)\left(z+x\right)=\left(z-x\right)\left(x+y\right)\)
\(\Leftrightarrow z\left(x-y\right)+x\left(x-y\right)=x\left(z-x\right)+y\left(z-x\right)\)
\(\Leftrightarrow xz-zy+x^2-xy=xz-x^2+yz-xy\)
\(\Leftrightarrow-zy+x^2=-x^2+yz\)
\(\Leftrightarrow-2x^2=-2zy\)
\(\Leftrightarrow x^2=yz\)(đpcm)
1a : x = -1
2a : x = 10
còn mấy bài khác mình không biết giải nha
Trả lời:
Bài 4:
b, B = ( x + 1 ) ( x7 - x6 + x5 - x4 + x3 - x2 + x - 1 )
= x8 - x7 + x6 - x5 + x4 - x3 + x2 - x + x7 - x6 + x5 - x4 + x3 - x2 + x - 1
= x8 - 1
Thay x = 2 vào biểu thức B, ta có:
28 - 1 = 255
c, C = ( x + 1 ) ( x6 - x5 + x4 - x3 + x2 - x + 1 )
= x7 - x6 + x5 - x4 + x3 - x2 + x + x6 - x5 + x4 - x3 + x2 - x + 1
= x7 + 1
Thay x = 2 vào biểu thức C, ta có:
27 + 1 = 129
d, D = 2x ( 10x2 - 5x - 2 ) - 5x ( 4x2 - 2x - 1 )
= 20x3 - 10x2 - 4x - 20x3 + 10x2 + 5x
= x
Thay x = - 5 vào biểu thức D, ta có:
D = - 5
Bài 5:
a, A = ( x3 - x2y + xy2 - y3 ) ( x + y )
= x4 + x3y - x3y - x2y2 + x2y2 + xy3 - xy3 - y4
= x4 - y4
Thay x = 2; y = - 1/2 vào biểu thức A, ta có:
A = 24 - ( - 1/2 )4 = 16 - 1/16 = 255/16
b, B = ( a - b ) ( a4 + a3b + a2b2 + ab3 + b4 )
= a5 + a4b + a3b2 + a2b3 + ab4 - ab4 - a3b2 - a2b3 - ab4 - b5
= a5 + a4b - ab4 - b5
Thay a = 3; b = - 2 vào biểu thức B, ta có:
B = 35 + 34.( - 2 ) - 3.( - 2 )4 - ( - 2 )5 = 243 - 162 - 48 + 32 = 65
c, ( x2 - 2xy + 2y2 ) ( x2 + y2 ) + 2x3y - 3x2y2 + 2xy3
= x4 + x2y2 - 2x3y - 2xy3 + 2x2y2 + 2y4 + 2x3y - 3x2y2 + 2xy3
= x4 + 2y4
Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:
( - 1/2 )4 + 2.( - 1/2 )4 = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16
a)\(\left(x+y\right)^2:\left(x+y\right)=x+y\)
b)\(\left(x-y\right)^5:\left(y-x\right)^4=\left(x-y\right)^5:\left(x-y\right)^4=x-y\)
c)\(\left(5x^4-3x^3+x^2\right):3x^2=\frac{5}{3}x^2-x+\frac{1}{3}^{ }\)
d)\(\left(x^3y^3-\frac{1}{2}x^2y^3+x^3y^2\right):\frac{1}{2}x^2y^2=2xy-y+x\)
Làm theo hằng đẳng thức nhé :) Tính thì bạn tự tính :))
\(a)\) \(\left(\frac{1}{2}x+y\right)\left(\frac{1}{2}x+y\right)=\left(\frac{1}{2}x+y\right)^2=\frac{1}{4}x^2+xy+y^2\)
\(b)\) \(\left(x-\frac{1}{2}y\right)\left(x-\frac{1}{2}y\right)=\left(x-\frac{1}{2}y\right)^2=x^2-xy+y^2\)
\(c)\) \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(d)\) \(\left(a-b\right)^2=a^2-2ab+b^2\)
\(e)\) \(\left(a+b\right)\left(a-b\right)=a^2-b^2\)
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