Tìm x,y biết :
\(\left[\left(x-2y\right)\left(x-7y\right)-x^2+4y^2\right]:\left(x-2y\right)=18\)
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
HS
Những câu hỏi liên quan
Thực hiện phép tính:a) dfrac{2}{5}xyleft(x^2y-5x+10yright)b) left(x^2-1right)left(x^2+2x+yright)c) left(x+3yright)^2d) left(4x-yright)^3e) left(x^2-2yright)left(x^2+2yright)g) 18x^4y^2z:10x^4yh) left(x^3y^3+dfrac{1}{2}x^2y^3-x^3y^2right):dfrac{1}{3}x^2y^2i) left(6x^3-7x^2-x+2right):left(2x+1right)k) dfrac{5x-1}{3x^2y}+dfrac{x+1}{3x^2y}l) dfrac{3x+1}{x^2-3x+1}+dfrac{x^2-6x}{x^2-3x+1}m) dfrac{2x+3}{10x-4}+dfrac{5-3x}{4-10x}n) dfrac{x}{x^2+2x+1}+dfrac{3}{5x^2-5}o) dfrac{x^2+2}{x^3-1}+dfrac{2}{x^2+x...
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Thực hiện phép tính:
a) \(\dfrac{2}{5}xy\left(x^2y-5x+10y\right)\)
b) \(\left(x^2-1\right)\left(x^2+2x+y\right)\)
c) \(\left(x+3y\right)^2\)
d) \(\left(4x-y\right)^3\)
e) \(\left(x^2-2y\right)\left(x^2+2y\right)\)
g) \(18x^4y^2z:10x^4y\)
h) \(\left(x^3y^3+\dfrac{1}{2}x^2y^3-x^3y^2\right):\dfrac{1}{3}x^2y^2\)
i) \(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\)
k) \(\dfrac{5x-1}{3x^2y}+\dfrac{x+1}{3x^2y}\)
l) \(\dfrac{3x+1}{x^2-3x+1}+\dfrac{x^2-6x}{x^2-3x+1}\)
m) \(\dfrac{2x+3}{10x-4}+\dfrac{5-3x}{4-10x}\)
n) \(\dfrac{x}{x^2+2x+1}+\dfrac{3}{5x^2-5}\)
o) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
p) \(\dfrac{4x+2}{15x^3y}\dfrac{5y-3}{9x^2y}+\dfrac{x+1}{5xy^3}\)
q) \(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)
r) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
x) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)\)
y) \(\dfrac{x^2-4}{3x+12}.\dfrac{x+4}{2x-4}\)
z) \(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)
t) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
w) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
c: \(=x^2+6xy+9y^2\)
e: \(=x^4-4y^2\)
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cho x+2y và 2x+y là 2 số thực dương khác 2.tìm Min của biểu thức:
\(P=\frac{\left(2x^2+y\right)\left(4x+y^2\right)}{\left(2x+y-2\right)^2}+\frac{\left(2y^2+x\right)\left(4y+x^2\right)}{\left(2y+x-2\right)^2}-3\left(x+y\right)\)
giải hệ phương trình:
1, \(\left\{{}\begin{matrix}2+6y=\frac{x}{y}-\sqrt{x-2y}\\\sqrt{x+\sqrt{x-2y}}=x+3y-2\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^2+y^2+xy+1=4y\\y\left(x+y\right)^2=2x^2-7y+2\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x^2\left(y+1\right)=6y-2\\x^4y^2+2x^2y^2+y\left(x^2+1\right)=12y^2-1\end{matrix}\right.\)
Giải hệ phương trình:
\(a,\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2\right)=13\\\left(x+y\right)\left(x^2-y^2\right)=25\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}+y\sqrt{x-1}=2\left(x-y\right)\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}x^2+y^2+xy+1=4y\\y\left(x+y\right)^2=2x^2+7y+2\end{matrix}\right.\)
a/ \(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2\right)=13\\\left(x-y\right)\left(x+y\right)^2=25\end{matrix}\right.\)
Do \(x=y;x=-y\) đều ko phải nghiệm
\(\Rightarrow\frac{x^2+y^2}{\left(x+y\right)^2}=\frac{13}{25}\Leftrightarrow25\left(x^2+y^2\right)=13\left(x+y\right)^2\)
\(\Leftrightarrow12x^2-26xy+12y^2=0\)
\(\Leftrightarrow\left(2x-3y\right)\left(3x-2y\right)=0\Rightarrow\left[{}\begin{matrix}y=\frac{2}{3}x\\y=\frac{3}{2}x\end{matrix}\right.\)
Thay vào 1 trong 2 pt ban đầu là xong
b/ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\y\ge0\end{matrix}\right.\) \(\Rightarrow x+y>0\)
\(xy+x+y+y^2=x^2-y^2\)
\(\Leftrightarrow x\left(y+1\right)+y\left(y+1\right)=\left(x-y\right)\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)\left(y+1\right)=\left(x+y\right)\left(x-y\right)\)
\(\Leftrightarrow y+1=x-y\Rightarrow x=2y+1\)
Thay vào pt dưới:
\(\left(2y+1\right)\sqrt{2y}+y\sqrt{2y}=2\left(y+1\right)\)
\(\Leftrightarrow\sqrt{2y}\left(3y+1\right)=2\left(y+1\right)\)
\(\Leftrightarrow y\left(9y^2+6y+1\right)=2\left(y^2+2y+1\right)\)
\(\Leftrightarrow9y^3+2y^2-3y-2=0\)
Nghiệm quá xấu, bạn coi lại đề
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c/ \(y=0\) không phải nghiệm
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1+y\left(x+y\right)=4y\\y\left(x+y\right)^2-2\left(x^2+1\right)=7y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x^2+1}{y}+x+y=4\\\left(x+y\right)^2-2\left(\frac{x^2+1}{y}\right)=7\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\\frac{x^2+1}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=4\\a^2-2b=7\end{matrix}\right.\) \(\Rightarrow a^2-2\left(4-a\right)=7\)
\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\Rightarrow b=1\\a=-5\Rightarrow b=9\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=3\\\frac{x^2+1}{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=3-x\\x^2+1-y=0\end{matrix}\right.\)
\(\Rightarrow x^2+1-\left(3-x\right)=0\Rightarrow...\)
TH2: làm tương tự
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giải hệ phương trình:
1, \(\left\{{}\begin{matrix}\left(x+y-3\right)^3=4y^3\left(x^2y^2+xy+\frac{45}{4}\right)\\x+4y-3=2xy^2\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3+7y=\left(x+y\right)^2+x^2y+7x+4\\3x^2+y^2+8y+4=8x\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}2x+5y=xy+2\\x^2+4y+21=y^2+10x\end{matrix}\right.\)
Cho \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\) và \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\). Tìm x và y
https://olm.vn/hoi-dap/detail/108858274535.html
Bài tương tự gưi link ib
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\(\hept{\begin{cases}\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\\\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\end{cases}}\)
<=> \(\hept{\begin{cases}x^3+8y^3=0\left(1\right)\\x^3-8y^3=16\left(2\right)\end{cases}}\)
Lấy (1) + (2) theo vế
=> 2x3 = 16
=> x3 = 8 = 23
=> x = 2
Thế x = 2 vào (1)
=> 23 + 8y3 = 0
=> 8 + 8y3 = 0
=> 8y3 = -8
=> y3 = -1 = (-1)3
=> y = -1
Vậy \(\hept{\begin{cases}x=2\\y=-1\end{cases}}\)
Từ \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\)
\(\Leftrightarrow x^3+\left(2y\right)^3=0\)\(\Leftrightarrow x^3+8y^3=0\)(1)
Từ \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\)
\(\Leftrightarrow x^3-\left(2y\right)^3=16\)\(\Leftrightarrow x^3-8y^3=16\)(2)
Cộng (1) với (2) ta được: \(\left(x^3+8y^3\right)+\left(x^3-8y^3\right)=16\)
\(\Leftrightarrow2x^3=16\)\(\Leftrightarrow x^3=8\)\(\Leftrightarrow x=2\)
Thay \(x=2\)vào (1) ta được:
\(2^3+8y^3=0\)\(\Leftrightarrow8y^3+8=0\)
\(\Leftrightarrow8y^3=-8\)\(\Leftrightarrow y^3=-1\)\(\Leftrightarrow y=-1\)
Vậy \(x=2\); \(y=-1\)
NT
21 tháng 12 2023 lúc 18:04
Bài 1: Phân tích đa thức thành nhân tử:1) 3x^3y^2-6xy2) left(x-2yright).left(x+3yright)-2.left(x-2yright)3) left(3x-1right).left(x-2yright)-5x.left(2y-xright)4) x^2-y^2-6y-95) left(3x-yright)^2-4y^26) 4x^2-9y^2-4x+1 8) x^2y-xy^2-2x+2y9) x^2-y^2-2x+2yBài 2: Tìm x:1) left(2x-1right)^2-4.left(2x-1right)02) 9x^3-x03) left(3-2xright)^2-2.left(2x-3right)04) left(2x-5right)left(x+5right)-10x+250
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Bài 1: Phân tích đa thức thành nhân tử:
1) \(3x^3y^2-6xy\)
2) \(\left(x-2y\right).\left(x+3y\right)-2.\left(x-2y\right)\)
3) \(\left(3x-1\right).\left(x-2y\right)-5x.\left(2y-x\right)\)
4) \(x^2-y^2-6y-9\)
5) \(\left(3x-y\right)^2-4y^2\)
6) \(4x^2-9y^2-4x+1\)
8) \(x^2y-xy^2-2x+2y\)
9) \(x^2-y^2-2x+2y\)
Bài 2: Tìm x:
1) \(\left(2x-1\right)^2-4.\left(2x-1\right)=0\)
2) \(9x^3-x=0\)
3) \(\left(3-2x\right)^2-2.\left(2x-3\right)=0\)
4) \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
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ND
21 tháng 1 2020 lúc 16:36
Giải các hệ phương trình :
a) \(\hept{\begin{cases}2x\left(x+1\right)\left(y+1\right)+xy=-6\\2y\left(y+1\right)\left(x+1\right)+yx=6\end{cases}x,y\inℝ}\)
b) \(\hept{\begin{cases}x^3+3x^2y-4y^3+x-y=0\\\left(x^2+3x+2\right)\left(y^2+7y+12\right)=24\end{cases}}\)
b, \(x^3+3x^2y-4y^3+x-y=0\)
\(\Leftrightarrow x^3-x^2y+4x^2y-4xy^2+4xy^2-4y^3+x-y=0\)
\(\Leftrightarrow x^2\left(x-y\right)+4xy\left(x-y\right)+4y^2\left(x-y\right)+\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+4xy+4y^2+1\right)=0\)
\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)
Khi đó pt (2) của hệ trở thành:
\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
\(\Leftrightarrow\left(x^2+5x+5\right)^2-1=24\)
\(\Leftrightarrow\left(x^2+5x+5\right)^2-5^2=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy hệ có nghiệm \(\left(x;y\right)\in\left\{\left(0;0\right),\left(-5;-5\right)\right\}\)
Cxleft[x^2-yright]xleft[x^3-2y^2right]xleft[x^4-3y^3right]xleft[x^5-4y^4right]tại x2,y-2Dx^2left[x+yright]-y^2left[x+yright]+left[x^2-y^2right]+2left[x+yright]+3 biết rằng x+y+10Mleft[x+yright]xleft[y+zright]xleft[x+zright]biết ranhwfx+y+zxyz2
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C=\(x\)\(\left[x^2-y\right]\)x\(\left[x^3-2y^2\right]\)x\(\left[x^4-3y^3\right]\)x\(\left[x^5-4y^4\right]\)tại \(x=2,y=-2\)
D=\(x^2\left[x+y\right]\)-\(y^2\)\(\left[x+y\right]\)+\(\left[x^2-y^2\right]\)+2\(\left[x+y\right]\)+3 biết rằng x+y+1=0
M=\(\left[x+y\right]\)x\(\left[y+z\right]\)x\(\left[x+z\right]\)biết ranhwfx+y+z=xyz=2
a: \(x^3-2y^2=2^3-2\cdot\left(-2\right)^2=8-2\cdot4=0\)
=>\(C=x\left(x^2-y\right)\left(x^3-2y^2\right)\left(x^4-3y^3\right)\left(x^5-4y^4\right)=0\)
b: x+y+1=0
=>x+y=-1
\(D=x^2\left(x+y\right)-y^2\left(x+y\right)+\left(x^2-y^2\right)+2\left(x+y\right)+3\)
\(=x^2\cdot\left(-1\right)-y^2\left(-1\right)+\left(x^2-y^2\right)+2\cdot\left(-1\right)+3\)
\(=-x^2+y^2+x^2-y^2-2+3\)
=1
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