\(Tìm\)\(x,y\in N\)\(thỏa\)\(mãn\)
\(a.\)\(\left(3x-2\right)\cdot\left(2y-3\right)=1\)
\(b.\)\(\left(x+1\right)\cdot\left(2y-1\right)=12\)
Những câu hỏi liên quan
Rút gọn biểu thức sau:
A=\(\left(2x+y\right)^2-\left(y-2x\right)^2\)
B=\(\left(3x+2\right)^2+2\cdot\left(2+3x\right)\cdot\left(1-2y\right)+\left(2y-1\right)^2\)
a: Ta có: \(A=\left(2x+y\right)^2-\left(2x-y\right)^2\)
\(=\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)\)
\(=4x\cdot2y=8xy\)
b: Ta có: \(B=\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(2y-1\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
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\(\hept{\begin{cases}\left(x-3\right)\cdot\left(2y+5\right)=\left(2x+7\right)\cdot\left(y-1\right)\\\left(4x+1\right)\cdot\left(3y-6\right)=\left(6x-1\right)\cdot\left(2y+3\right)\end{cases}}\)
rút gọn biểu thức sau bằng cách nhanh nhấtA left(a^2+b^2-c^2right)^2-left(a^2-b^2+c^2right)^2-4a^2b^2B left(3x^3+3x+1right)cdotleft(3x^3-3x+1right)-left(3x^3+1right)^2C left(2-6xright)^2+left(2-5xright)^2+2cdotleft(6x-2right)cdotleft(2-5xright)D 5cdotleft(3x-1right)^2+4cdotleft(5x+1right)^2-12cdotleft(5x-2right)left(5x+2right)E left(3x-1right)^2+left(2x+4right)cdotleft(1-3xright)+left(x+2right)^2G left(x-1right)^3+4cdotleft(x+1right)cdotleft(1-xright)+3cdotleft(x-1right)cdotleft(x^2+x+1rig...
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rút gọn biểu thức sau bằng cách nhanh nhất
A = \(\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
B = \(\left(3x^3+3x+1\right)\cdot\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
C = \(\left(2-6x\right)^2+\left(2-5x\right)^2+2\cdot\left(6x-2\right)\cdot\left(2-5x\right)\)
D = \(5\cdot\left(3x-1\right)^2+4\cdot\left(5x+1\right)^2-12\cdot\left(5x-2\right)\left(5x+2\right)\)
E = \(\left(3x-1\right)^2+\left(2x+4\right)\cdot\left(1-3x\right)+\left(x+2\right)^2\)
G = \(\left(x-1\right)^3+4\cdot\left(x+1\right)\cdot\left(1-x\right)+3\cdot\left(x-1\right)\cdot\left(x^2+x+1\right)\)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
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Thực hiện phép tính :
a, \(\left(x^2+\dfrac{2}{5}y\right)\cdot\left(x^2-\dfrac{2}{5}y\right)\)
b,\(\left(3x-2y\right)\cdot\left(3x+2y\right)\cdot\left(9x^2+4y^2\right)\)
a, \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=x^4-\dfrac{4}{25}y^2\)
b, \(\left(3x-2y\right)\left(3x+2y\right)\left(9x^2+4y^2\right)\)
\(=\left(9x^2-4y^2\right)\left(9x^2+4y^2\right)\)
\(=81x^4-16y^4\)
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Bài 1 : Tìm x;y biết :
\(x^2=y^2+2y+12\)
\(4x^2=y^2-2y=16\)
Bài 2 : Tìm n \(\in Z\)để biểu thức sau là phân số
a) \(\frac{3}{\left(n+1\right)\cdot\left(n-3\right)}\)
b) \(-\frac{4}{\left(n^2+1\right)\cdot\left(n+4\right)}\)
\(\hept{\begin{cases}\left(2x-3\right)\cdot\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\cdot\left(3y-3\right)=3y\left(x+1\right)-12\end{cases}}\) giải hệ phương trình
\(\hept{\begin{cases}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{cases}}\)
\(\hept{\begin{cases}4xy+8x-6y-12=4xy-12x+54\\3xy-3x+3y-3=3xy+3y-12\end{cases}}\)
\(\hept{\begin{cases}4xy-4xy+8x+12x-6y-12-54=0\\3xy-3xy-3x+3y-3y-3+12=0\end{cases}}\)
\(\hept{\begin{cases}20x-6y-66=0\\-3x+9=0\end{cases}}\)
\(\hept{\begin{cases}2\left(10x-3y\right)=66\\-3\left(x-3\right)=0\end{cases}}\)
\(\hept{\begin{cases}10x-3y=33\\x-3=0\end{cases}}\)
\(\hept{\begin{cases}10x-3y=33\\x=3\end{cases}}\)
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rút gọn biểu thức:
\(3x^2\cdot\left(2y-1\right)-2x^2\cdot\left(5y-3\right)-2x\cdot\left(x-1\right)\)
Giải hệ pt và pt sau:
a.left{{}begin{matrix}left(2x-3right)cdotleft(2y+4right)4xcdotleft(y-3right)+54left(x+1right)cdotleft(3y-3right)3yleft(x+1right)-12end{matrix}right.
b.left{{}begin{matrix}x+y-10x^2+xy+30end{matrix}right.
c.left{{}begin{matrix}2x-3y5x^2-y^240end{matrix}right.
d.left{{}begin{matrix}3x+2y36left(x-2right)left(y-3right)18end{matrix}right.
e.left{{}begin{matrix}2x+y5m-1x-2y2end{matrix}right. . Tìm m để hệ có nghiệm (x;y) t/m x^2-2y^21
f. frac{t^2}{t-1}+tfrac{2t^2+5t}{t+1}...
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Giải hệ pt và pt sau:
a.\(\left\{{}\begin{matrix}\left(2x-3\right)\cdot\left(2y+4\right)=4x\cdot\left(y-3\right)+54\\\left(x+1\right)\cdot\left(3y-3\right)=3y\left(x+1\right)-12\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}x+y-1=0\\x^2+xy+3=0\end{matrix}\right.\)
c.\(\left\{{}\begin{matrix}2x-3y=5\\x^2-y^2=40\end{matrix}\right.\)
d.\(\left\{{}\begin{matrix}3x+2y=36\\\left(x-2\right)\left(y-3\right)=18\end{matrix}\right.\)
e.\(\left\{{}\begin{matrix}2x+y=5m-1\\x-2y=2\end{matrix}\right.\) . Tìm m để hệ có nghiệm (x;y) t/m x\(^2\)-2y\(^2\)=1
f. \(\frac{t^2}{t-1}+t=\frac{2t^2+5t}{t+1}\)
g.\(\frac{x^2+2x-3}{x^2-9}+\frac{2x^2-2}{x^2-3x+2}=8\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}4xy+8x-6y-12=4xy-12x+54\\3xy-3x+3y-3=3xy+3y-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20x-6y=66\\-3x=-9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}y=1-x\\x^2+xy+3=0\end{matrix}\right.\)
\(\Leftrightarrow x^2+x\left(1-x\right)+3=0\)
\(\Leftrightarrow x+3=0\Rightarrow x=-3\Rightarrow y=4\)
c.
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{2x-5}{3}\\x^2-y^2=40\end{matrix}\right.\)
\(\Rightarrow x^2-\left(\frac{2x-5}{3}\right)^2-40=0\)
\(\Leftrightarrow9x^2-\left(4x^2-20x+25\right)-360=0\)
\(\Leftrightarrow5x^2+20x-385=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\Rightarrow y=3\\x=-11\Rightarrow y=-9\end{matrix}\right.\)
d.
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{36-3x}{2}\\\left(x-2\right)\left(y-3\right)=18\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)\left(\frac{36-3x}{2}-3\right)=18\)
\(\Leftrightarrow\left(x-2\right)\left(10-x\right)=12\)
\(\Leftrightarrow-x^2+12x-32=0\Rightarrow\left[{}\begin{matrix}x=4\Rightarrow y=12\\x=8\Rightarrow y=6\end{matrix}\right.\)
e.
\(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=10m-2\\x-2y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=10m\\x-2y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2m\\y=m-1\end{matrix}\right.\)
\(x^2-2y^2=1\)
\(\Leftrightarrow4m^2-2\left(m-1\right)^2=1\)
\(\Leftrightarrow4m^2-\left(2m^2-4m+2\right)-1=0\)
\(\Leftrightarrow2m^2+4m-3=0\Rightarrow m=\frac{-2\pm\sqrt{10}}{2}\)
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\(\left(x-1\right)\cdot\left(2x-2\sqrt{x^2-9}\right)+y\cdot\left(3y-2\sqrt{2y^2-4}\right)=12\)12