\(\left\{{}\begin{matrix}\left(x-1\right)\left(y+1\right)=xy-1\\\left(x-2\right)\left(y-2\right)=xy-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy+x-y-1=xy-1\\xy-2x-2y+4=xy-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-y=0\\-2x-2y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-2y=0\\2x+2y=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x=12\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=x=3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(x-1\right)\left(y+1\right)=xy-1\\\left(x-2\right)\left(y-2\right)=xy-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy+x-y-1=xy-1\\xy-2x-2y+4=xy-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x+y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=3\end{matrix}\right.\)

1: =(x+y-3x)(x+y+3x)

=(-2x+y)(4x+y)

2: =(3x-1-4)(3x-1+4)

=(3x+3)(3x-5)

=3(x+1)(3x-5)

3: =(2x)^2-(x^2+1)^2

=-[(x^2+1)^2-(2x)^2]

=-(x^2+1-2x)(x^2+1+2x)

=-(x-1)^2(x+1)^2

4: =(2x+1+x-1)(2x+1-x+1)

=3x(x+2)

5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

=(2x^2+2)*4x

=8x(x^2+1)

6: =(5x-5y)^2-(4x+4y)^2

=(5x-5y-4x-4y)(5x-5y+4x+4y)

=(x-9y)(9x-y)

7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)

=(x^2+2xy+y^2)(x^2-y^2)

=(x+y)^3*(x-y)

8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)

=[(x-2y)^2-4][(x+2y)^2-36]

=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)

a, x=1; y=2 => 12

x=2; y=1 => 21

b, x=1; y=5 => 15

x=5; y=1 => 51

c, x=1; y=6 => 16

x=6;y=1 => 61

x=2; y=3=> 23

x=3; y=2 => 32

d, x=1; y=8 => 18

x=2; y=4 => 24

x=4; y=2 => 42

x=8; y=1 => 81

5, 

x=3; y=4 => 34

x=4; y=3 => 43

x=2; y=6 => 26

x=6; y=2 => 62

Chọn B

1.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\dfrac{5}{4}\\x^4+y^2+xy\left(1+2x\right)=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y\right)+xy+xy\left(x^2+y\right)=-\dfrac{5}{4}\\\left(x^2+y\right)^2+xy=-\dfrac{5}{4}\end{matrix}\right.\left(1\right)\)

Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=-\dfrac{5}{4}\\a^2+b=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-a^2-\dfrac{5}{4}-a\left(a^2+\dfrac{5}{4}\right)=-\dfrac{5}{4}\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a^3-\dfrac{1}{4}a=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-a\left(a^2-a+\dfrac{1}{4}\right)=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a-\dfrac{1}{2}\right)^2=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=0\\xy=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt[3]{10}}{2}\\y=-\dfrac{5}{2\sqrt[3]{10}}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=\dfrac{1}{2}\\xy=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)

Kết luận: Phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{\sqrt[3]{10}}{2};-\dfrac{5}{2\sqrt[3]{10}}\right);\left(1;-\dfrac{3}{2}\right)\right\}\)

2.

\(\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\left(\dfrac{2}{y}\right)^3-4\left(\dfrac{2}{y}\right)\\1+\left(\dfrac{2}{y}\right)^2=5\left(x+1\right)^2+5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=u\\\dfrac{2}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-16u=v^3-4v\\v^2=5u^2+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u^3-v^3=16u-4v\\4=v^2-5u^2\end{matrix}\right.\)

\(\Rightarrow4\left(u^3-v^3\right)=\left(16u-4v\right)\left(v^2-5u^2\right)\)

\(\Leftrightarrow21u^3-5u^2v-4uv^2=0\)

\(\Leftrightarrow u\left(7u-4v\right)\left(3u+v\right)=0\Rightarrow\left[{}\begin{matrix}u=0\Rightarrow v^2=4\\u=\dfrac{4v}{7}\Rightarrow4=v^2-5\left(\dfrac{4v}{7}\right)^2\\v=-3u\Rightarrow4=\left(-3u\right)^2-5u^2\end{matrix}\right.\) 

\(\Rightarrow...\)

các bạn giúp m với =(((((

\(I=\int\limits^{-1}_{-2}\dfrac{6a}{e^x}dx-\int\limits^{-1}_{-2}\dfrac{f\left(x\right)}{e^x}dx=J-I_1\)

Xét \(I_1\) , đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=e^{-x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=-e^{-x}\end{matrix}\right.\)

\(\Rightarrow I_1=-f\left(x\right).e^{-x}|^{-1}_{-2}+\int\limits^{-1}_{-2}\dfrac{f'\left(x\right)}{e^x}dx=-f\left(-1\right).e+f\left(-2\right).e^2+I_2\)

Xét \(I_2\) , đặt \(\left\{{}\begin{matrix}u=f'\left(x\right)\\dv=e^{-x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f''\left(x\right)dx\\v=-e^{-x}\end{matrix}\right.\)

\(\Rightarrow I_2=-f'\left(x\right).e^{-x}|^{-1}_{-2}+\int\limits^{-1}_{-2}\dfrac{f''\left(x\right)}{e^x}dx=-f'\left(-1\right).e+f'\left(-2\right).e^2+I_3\)

Xét \(I_3\) , đặt \(\left\{{}\begin{matrix}u=f''\left(x\right)\\dv=e^{-x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'''\left(x\right)dx=6a.dx\\v=-e^{-x}\end{matrix}\right.\)

\(\Rightarrow I_3=-f''\left(x\right).e^{-x}|^{-1}_{-2}+\int\limits^{-1}_{-2}\dfrac{6a}{e^x}dx=-f''\left(-1\right).e+f''\left(-2\right).e^2+J\)

Do đó:

\(I=J+f\left(-1\right).e-f\left(-2\right).e^2+f'\left(-1\right).e-f'\left(-2\right).e^2+f''\left(-1\right).e-f''\left(-2\right).e^2-J\)

\(=e\left[f\left(-1\right)+f'\left(-1\right)+f''\left(-1\right)\right]-e^2\left[f\left(-2\right)+f'\left(-2\right)+f''\left(-2\right)\right]\)

\(=e.g\left(-1\right)-e^2.g\left(-2\right)=e+e^2=e\left(e+1\right)\)