mọi người giải giúp e với ạ :3

Bài 1:

b: Để (d) vuông góc với (d2) thì \(\left(m^2+2m\right)\cdot\dfrac{-1}{3}=-1\)

\(\Leftrightarrow m^2+2m-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-3\\m=1\end{matrix}\right.\)

\(\left(x+y-1\right)\left(x+y+1\right)=x^2+xy-x+xy+y^2-y+x+y-1\\ =x^2+\left(xy+xy\right)+\left(-x+x\right)+y^2+\left(-y+y\right)-1\\ =x^2+2xy+y^2-1\\ =>B\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{10}{7}\\b=\dfrac{3}{7}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{10}{7}\Rightarrow x=\dfrac{7}{10}\\\dfrac{1}{y}=\dfrac{3}{7}\Rightarrow y=\dfrac{7}{3}\end{matrix}\right.\)

\(b.\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{5}{\left(x+y\right)}=2\\\dfrac{3}{x}+\dfrac{1}{\left(x+y\right)}=\dfrac{17}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2a+5b=2\\3a+b=\dfrac{17}{10}\end{matrix}\right.\) (với \(\dfrac{1}{x}=a-\dfrac{1}{x+y}=b\))

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{1}{5}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{2}\Rightarrow x=2\\\dfrac{1}{x+y}=\dfrac{1}{5}\Rightarrow y=3\end{matrix}\right.\)

\(c.\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{1}{y+1}=7\\\dfrac{5}{x-1}-\dfrac{2}{y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+b=7\\5a-2b=4\end{matrix}\right.\) (với \(\dfrac{1}{x-1}=a-\dfrac{1}{y+1}=b\))

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

\(d.\left\{{}\begin{matrix}\dfrac{2}{\sqrt{x-1}}-\dfrac{1}{\sqrt{y-1}}=1\\\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{y-1}}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\a+b=2\end{matrix}\right.\) (với \(\dfrac{1}{\sqrt{x-1}}=a-\dfrac{1}{\sqrt{y-1}}=b\))

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-1}}=1\Rightarrow x=2\\\dfrac{1}{\sqrt{y-1}}=1\Rightarrow y=2\end{matrix}\right.\)

2, Gọi \(\left(d\right):y=ax+b\left(a\ne0\right)\) là đt cần tìm

PTHDGD d₁ và d₂ là \(4x-3=-x+1\Leftrightarrow x=\dfrac{4}{5}\Leftrightarrow y=\dfrac{1}{5}\Leftrightarrow A\left(\dfrac{4}{5};\dfrac{1}{5}\right)\)

Vì \(\left(d\right)//\left(d_1\right)\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b\ne1\end{matrix}\right.\Leftrightarrow\left(d\right):y=-\dfrac{1}{2}x+b\)

Vì \(A\left(\dfrac{4}{5};\dfrac{1}{5}\right)\in\left(d\right)\Leftrightarrow-\dfrac{1}{2}\cdot\dfrac{4}{5}+b=\dfrac{1}{5}\Leftrightarrow b=\dfrac{3}{5}\)

Vậy đt cần tìm là \(\left(d\right):y=-\dfrac{1}{2}x+\dfrac{3}{5}\)

a: \(=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2\)

b: \(=\left(x+y+x-y\right)^2=\left(2x\right)^2=4x^2\)

d: \(=9x^2+6x+1-9x^2+6x-1=12x\)

a: \(=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2\)

e: \(=x^3+1-x^3+1=2\)

1 a

2c

3b

4d

5c

6c