\(2x^2+xy-y^2-5x+y+2=0\)

...">

K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

Bài 2:a. \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\) \(\Leftrightarrow\left|y+3\right|=6x-2x^2-2xy-y^2-9\) \(\Leftrightarrow\left|y+3\right|=-x^2-2xy-y^2-x^2+6x-9\) \(\Leftrightarrow\left|y+3\right|=-\left(x+y\right)^2-\left(x-3\right)^2\) \(\Leftrightarrow\left|y+3\right|=-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]\) Có: \(\left|y+3\right|\ge0\) \(-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]\le0\) Do...
Đọc tiếp

Bài 2:

a. \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\) 

\(\Leftrightarrow\left|y+3\right|=6x-2x^2-2xy-y^2-9\) 

\(\Leftrightarrow\left|y+3\right|=-x^2-2xy-y^2-x^2+6x-9\) 

\(\Leftrightarrow\left|y+3\right|=-\left(x+y\right)^2-\left(x-3\right)^2\) 

\(\Leftrightarrow\left|y+3\right|=-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]\) 

Có: \(\left|y+3\right|\ge0\) 

\(-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]\le0\) 

Do đó: \(\left|y+3\right|=-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]=0\) 

\(\Leftrightarrow\hept{\begin{cases}y+3=0\\x+y=0\\x-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\) 

b. \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\) 

\(\Leftrightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+\left[2\left(x^2-5x-2012\right)\right]^2=0\) 

\(\Leftrightarrow\left(2x^2+x-2013-2x^2+10x+4024\right)^2=0\) 

\(\Leftrightarrow\left(11x+2011\right)^2=0\) 

\(\Leftrightarrow11x+2011=0\) 

\(\Leftrightarrow x=-\frac{2011}{11}\) 

0
27 tháng 7 2018

Chỗ dấu bằng thứ hai sai nên bạn làm cũng chưa đúng

x^6 -y^6 = (x^2-y^2)(x^4 +x^2 .y^2 + y^4)

Bạn hiểu ra chỗ sai của mình chưa.Chúc bạn học tốt.

27 tháng 7 2018

cho minh xin de

22 tháng 10 2019

toi ko bt

22 tháng 10 2019

\(x^2+y^2=0\)

Mà \(x^2\ge0;y^2\ge0\)nên \(x^2+y^2\ge0\)

(Dấu "="\(\Leftrightarrow x=y=0\))

đăng lên làm j z

21 tháng 12 2016

mơn em iu nhìu nhắm nak.

21 tháng 12 2016

shit ~ pate tăng động -_-

3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\) \(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\) \(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\) \(\Leftrightarrow x^2+5x-2x-10-5=1x+5\) \(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0 \(\Leftrightarrow\) \(x^2+2x-20=0\) \(\Leftrightarrow x^2+2x-10x-20=0\) \(\Leftrightarrow\) (x\(^2\) + 2x) - (10x +...
Đọc tiếp

3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\)

\(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)

\(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\)

\(\Leftrightarrow x^2+5x-2x-10-5=1x+5\)

\(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0

\(\Leftrightarrow\) \(x^2+2x-20=0\)

\(\Leftrightarrow x^2+2x-10x-20=0\)

\(\Leftrightarrow\) (x\(^2\) + 2x) - (10x + 20) = 0

\(\Leftrightarrow\) x.(x + 2) - 10.(x + 2) = 0

\(\Leftrightarrow\)

4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)

\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x\left(x+7\right)}\)

\(\Leftrightarrow\frac{\left(x-4\right).\left(x+7\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)

\(\Leftrightarrow\) \(x^2+7x-4x-28-x-7=-7\)

\(\Leftrightarrow x^2+7x-4x-x-28-7+7=0\)

\(\Leftrightarrow\) x\(^2\) + 2x - 28 = 0

\(\Leftrightarrow\) x\(^2\) + 2x - 14x - 28 = 0

\(\Leftrightarrow\) (x\(^2\) + 2x) - (14x + 28) = 0

\(\Leftrightarrow\) x.(x + 2) - 14.(x + 2) = 0

\(\Leftrightarrow\) (x - 14) = 0 hoặc (x + 2) = 0

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = -2 (Loại)

5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)

\(\Leftrightarrow\) \(\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)

\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)

\(\Leftrightarrow\) \(x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)

\(\Leftrightarrow2x^2-8x+8=0\)

\(\Leftrightarrow\) 2x\(^2\) - 2x - 8x + 8 = 0

\(\Leftrightarrow\) 2x(x - 1) - 8(x - 1) = 0

\(\Leftrightarrow\) 2x - 8 = 0 hoặc x - 1 = 0

\(\Leftrightarrow\) 2x = 8 hoặc x = 1

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = 1 (Nhận)

Vậy S = {4; 1}

6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)

\(\Leftrightarrow\) x\(^2\) + x + x + 1 - x\(^2\) + x + x - 1 = 4

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

\(\Leftrightarrow\) 4 (x - 1) =0

\(\Leftrightarrow\) x - 1 = 0 / 4 = 0

\(\Leftrightarrow\) x = 1 (Nhận)

Vậy S = {1}

7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x+1\right)}\)

\(\Leftrightarrow x^2+x+x+1-4x=x^2-x-x+1\)

\(\Leftrightarrow\) 0

Vậy S ={\(\varnothing\)}

0
AH
Akai Haruma
Giáo viên
24 tháng 11 2018

a)

\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)

c)

\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)

d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)

AH
Akai Haruma
Giáo viên
24 tháng 11 2018

e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)

f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)

g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)

\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)

2 tháng 7 2019

\(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)

\(=20x^3-10x^2+5x-20x^3+10x^2+4x\)

\(=9x\)

Thay x=15 \(\Rightarrow A=9.15=135\)

4 tháng 7 2019

\(B=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)

\(=6x^2y^2-6xy^3-8x^3+8x^2y^2+5x^2y^2-5xy^3\)

\(=19x^2y^2-11xy^3-8x^3\)

Thay x=1/2 ; y=2 vào B \(\Rightarrow19.\left(\frac{1}{2}\right)^2.2^2-11\cdot\frac{1}{2}\cdot2^3-8\cdot\left(\frac{1}{2}\right)^3\)

\(=19-44-1\)

\(=-26\)

7 tháng 11 2017

1)

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)

d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)

e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)

f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)

\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)

2)

a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)

b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)

\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)

c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )

d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)

\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)

e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)

\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)