con A bn bấm nhầm đúng ko mik sửa lại nhé

A= 2014² - 4018. 1014 + 1014²

= (2014 - 1014)²

= 1000²

= 1000000

B= 9x² - 6x + 2013

= 9x² - 6x + 1 + 2012

= (3x - 1)² + 2012

thay x = \(\dfrac{200001}{3}\)vào biểu thức B ta có:

B = (3.\(\dfrac{200001}{3}\)- 1)² + 2012

= (200001 - 1)² + 2012

= 200000² + 2012

= 40000002012

mik chỉ làm đc đến đây thôi nhưng mong bn ủng hộ!

a \(=9x^2-6x+1+2012\)

\(=\left(3x-1\right)^2+2012\)

\(=200000^2+2012\)

b: \(=2014^2-2\cdot2014\cdot1014+1014^2\)

\(=\left(2014-1014\right)^2=1000^2=10^6\)

c: \(x^2+3y^2=4xy\)

=>x^2-4xy+3y^2=0

=>(x-y)*(x-3y)=0

=>x=y hoặc x=3y

KHi x=y thì \(C=\dfrac{2x+2013x}{x-2x}=-2015\)

Khi x=3y thì \(C=\dfrac{6y+2013y}{3y-2y}=2019\)

\(x\left(x+5\right)=9x\)

\(\Leftrightarrow xx+5x=9x\)

\(\Leftrightarrow xx+5x-9x=0\)

\(\Leftrightarrow xx-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=4\)

Bài 1:
a) (2x - y) + (2x - y) + (2x - y) + 3y
= 3(2x - y) + 3y
= 3(2x - y + 3y)
= 3(2x + 2y)
= 3.2(x + y)
= 6(x + y)

b) (x + 2y) + (x - 2y) + (8x - 3y)
= x + 2y + x - 2y + 8x - 3y
= 9x - 3y
= 3(3x - y)

c) (x + 2y) - 2(x - 2y) - (2x - 3y)
= x + 2y - 2x + 4y - 2x + 3y
= 9y - 3x
= 3(3y - x)

Bài 2:
M + 2(x² - 4y²) + Q = 6x² - 4xy + 5y² + P
M + 2x² - 8y²   -3x² + 7xy - 2y² = 6x² - 4xy + 5y² + 9x² - 6xy + 3y²
M + 2x² - 3x² - 6x² - 9x² - 8y² - 2y² - 5y² - 3y² + 7xy + 4xy + 6xy = 0
M - 16x² - 18y² + 17xy = 0
M = 16x² + 18y² - 17xy

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

x/y+3.y/x=4

đặt b=y/x<1

1/b+3b=4

3b^2-4b+1=0

b=1loia

b=1/3

(2+5b)/(1-2.b)

\(P=\frac{2+5.\frac{1}{3}}{1-2.\frac{1}{3}}=\frac{\frac{11}{3}}{\frac{1}{3}}=11\)

11 nha bạn

Chúc các bạn học giỏi

Tết vui vẻ nha

a) Thay \(x=2,y=\frac{1}{2}\), ta được \(B=2^2-4.2.\frac{1}{2}+4.\left(\frac{1}{2}\right)^2=4-4+1=1\)

b) Thay \(x=1,\left|y\right|=2.5\Leftrightarrow x=1,y=2,5\), ta được \(B=1^2-4.1.2,5+4.\left(2,5\right)^2=1-10+25=16\)

c) Thay \(2x=3y,x+2y=-7\Leftrightarrow\left\{{}\begin{matrix}2x-3y=0\\x+2y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\), ta được \(B=\left(-3\right)^2-4\left(-3\right)\left(-2\right)+4\left(-2\right)^2=9-24+16=1\)

d) Thay $x=2y$, ta được \(B=\left(2y\right)^2-4\left(2y\right)y+4y^2=4y^2-8y^2+4y^2=0\)

B=x²-4xy+4.y².

B=(x²-2xy)+4y²-2xy.

B=x(x-2y)+2y(2y-x)

B=x(x-2y)-2y(2y-x)=(x-2y)².

a)Thay x=2;y=1/2, ta được:

B=(2-1)²=1

b)TH1:y=2,5

B=(x-2y)²=(1-2.2,5)²=(-4)²=16.

TH2:y=-2,5

B=(x-2y)²=(1+2,5.2)²=6²=36

Vậy B=16 hoặc 36.

c)x=\(\frac{3}{2}\)y ⇒y(\(\frac{3}{2}\)+2)=-7

y.\(\frac{7}{2}\)=-7⇒y=-2

x=(-2).\(\frac{3}{2}\)=-3

B=[-3-2.(-2)]²=1²=1

d)B=(x-2y)²=0²=0.