ĐKXĐ: \(x\ne-1;y\ne3\)

\(\left\{{}\begin{matrix}\dfrac{5x+1}{x+1}+\dfrac{y+3}{y-3}=27\\\dfrac{2x}{x+1}-\dfrac{2y}{y-3}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5\left(x+1\right)-4}{x+1}+\dfrac{y-3+6}{y-3}=27\\\dfrac{2\left(x+1\right)-2}{x+1}-\dfrac{2\left(y-3\right)+6}{y-3}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5-\dfrac{4}{x+1}+1+\dfrac{6}{y-3}=27\\2-\dfrac{2}{x+1}-2-\dfrac{6}{y-3}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{4}{x+1}+\dfrac{6}{y-3}=21\\-\dfrac{2}{x+1}-\dfrac{6}{y-3}=4\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+1}=u\\\dfrac{1}{y-3}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-4u+6v=21\\-2u-6v=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=-\dfrac{25}{6}\\v=\dfrac{13}{18}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=-\dfrac{25}{8}\\\dfrac{1}{y-3}=\dfrac{13}{18}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+1=-\dfrac{8}{25}\\y-3=\dfrac{18}{13}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{33}{25}\\y=\dfrac{57}{13}\end{matrix}\right.\)

a: Thay m=2 và n=3 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}2x+\left(3-1\right)y=5\\\left(2-1\right)x-\left(3+1\right)y=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+2y=5\\x-4y=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+2y=5\\2x-8y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10y=-9\\x-4y=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{9}{10}\\x=4y+7=4\cdot\dfrac{-9}{10}+7=\dfrac{-36+70}{10}=\dfrac{34}{10}=\dfrac{17}{5}\end{matrix}\right.\)

b: Thay x=2 và y=-3 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}2m+\left(n-1\right)\cdot\left(-3\right)=5\\2\left(m-1\right)-\left(n+1\right)\cdot\left(-3\right)=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2m-3n+3=5\\2m-2+3n-3=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2m-3n=2\\2m+3n=7+2+3=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4m=14\\2m-3n=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=\dfrac{7}{2}\\3n=2m+2=2\cdot\dfrac{7}{2}+2=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m=\dfrac{7}{2}\\n=3\end{matrix}\right.\)

thank kiu

- Với \(m=\dfrac{5}{2}\) pt trở thành pt bậc nhất nên chỉ có 1 nghiệm (loại)

- Với \(m\ne\dfrac{5}{2}\) ta có:

\(a+b+c=2m-5-2\left(m-1\right)+3=0\)

\(\Rightarrow\) Phương trình luôn có 2 nghiệm: \(\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2m-5}\end{matrix}\right.\)

Do 1 là số nguyên dương nên để pt có 2 nghiệm pb đều nguyên dương thì:

\(\left\{{}\begin{matrix}\dfrac{3}{2m-5}\ne1\\\dfrac{3}{2m-5}\in Z^+\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ne4\\2m-5=Ư\left(3\right)=\left\{1;3\right\}\end{matrix}\right.\) (do nghiệm nguyên dương và 3 dương nên ta chỉ cần xét các ước dương của 3)

\(\Rightarrow\left\{{}\begin{matrix}m\ne4\\m=\left\{3;4\right\}\end{matrix}\right.\) 

\(\Rightarrow m=3\)

Đề là "hai nghiệm dương" hay "hai nghiệm nguyên dương" vậy em?

a: \(x^3+8x=5x^2+4\)

=>\(x^3-5x^2+8x-4=0\)

=>\(x^3-x^2-4x^2+4x+4x-4=0\)

=>\(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2: \(x^3+3x^2=x+6\)

=>\(x^3+3x^2-x-6=0\)

=>\(x^3+2x^2+x^2+2x-3x-6=0\)

=>\(x^2\cdot\left(x+2\right)+x\left(x+2\right)-3\left(x+2\right)=0\)

=>\(\left(x+2\right)\left(x^2+x-3\right)=0\)

=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)

3: ĐKXĐ: x>=0

\(2x+3\sqrt{x}=1\)

=>\(2x+3\sqrt{x}-1=0\)

=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)

=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)

=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)

=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)

=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)

4: \(x^4+4x^2+1=3x^3+3x\)

=>\(x^4-3x^3+4x^2-3x+1=0\)

=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)

=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)

=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)

=>(x-1)^2=0

=>x-1=0

=>x=1

a.

\(x^3+8x=5x^2+4\)

\(\Leftrightarrow x^3-5x^2+8x-4=0\)

\(\Leftrightarrow\left(x^3-4x^2+4x\right)-\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow x\left(x-2\right)^2-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

b.

\(x^3+3x^2-x-6=0\)

\(\Leftrightarrow\left(x^3+x^2-3x\right)+\left(2x^2+2x-6\right)=0\)

\(\Leftrightarrow x\left(x^2+x-3\right)+2\left(x^2+x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2+x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)

Đề có sai không bạn? Tại theo điều kiện thì:

\(\left\{{}\begin{matrix}x_1^3\in\left[8;27\right]\\x_2^3\in\left[0;1\right]\end{matrix}\right.\Rightarrow\left(x_1^3-x_2^3\right)\in\left[8;26\right]\).

6 không thuộc khoảng trên nhé:D?

b.

Khi \(m=\dfrac{5}{2}\) pt trở thành pt bậc nhất nên chỉ có 1 nghiệm (loại)

Xét với \(m\ne\dfrac{5}{2}\):

\(\Delta'=\left(m-1\right)^2-3\left(2m-5\right)=m^2-8m+16=\left(m-4\right)^2\)

Pt đã cho luôn có 2 nghiệm \(\forall m\ne\dfrac{5}{2}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m-1\right)}{2m-5}\\x_1x_2=\dfrac{3}{2m-5}\end{matrix}\right.\)

Két hợp Viet với điều kiện đề bài:

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m-1\right)}{2m-5}\\x_1-x_2=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{8m-17}{2\left(2m-5\right)}\\x_2=\dfrac{-4m+13}{2\left(2m-5\right)}\end{matrix}\right.\)

Thế vào \(x_1x_2=\dfrac{3}{2m-5}\)

\(\Rightarrow\dfrac{\left(8m-17\right)\left(-4m+13\right)}{4\left(2m-5\right)^2}=\dfrac{3}{2m-5}\)

\(\Rightarrow32m^2-148m+161=0\)

\(\Rightarrow\left[{}\begin{matrix}m=\dfrac{7}{4}\\m=\dfrac{23}{8}\end{matrix}\right.\)

Câu b của em là 2 ý phân biệt đúng không?

\(x^2-2\left(m-4\right)x+m^2+m+3=0\)

\(\text{Δ}=\left[-2\left(m-4\right)\right]^2-4\cdot1\cdot\left(m^2+m+3\right)\)

\(=\left(2m-8\right)^2-4m^2-4m-12\)

\(=4m^2-32m+64-4m^2-4m-12\)

=-36m+52

Để phương trình có nghiệm kép thì Δ=0

=>-36m+52=0

=>-36m=-52

=>\(m=\dfrac{-52}{-36}=\dfrac{13}{9}\)

Thay m=13/9 vào phương trình, ta được:

\(x^2-2\left(\dfrac{13}{9}-4\right)x+\left(\dfrac{13}{9}\right)^2+\dfrac{13}{9}+3=0\)

=>\(x^2+\dfrac{46}{9}x+\dfrac{529}{81}=0\)

=>\(\left(x+\dfrac{23}{9}\right)^2=0\)

=>\(x+\dfrac{23}{9}=0\)

=>\(x=-\dfrac{23}{9}\)

a: Thay m=1 vào (1), ta được:

\(x^2-5x+3\cdot1+=0\)

=>\(x^2-5x+4=0\)

=>(x-1)(x-4)=0

=>\(\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

b: \(\text{Δ}=\left(-5\right)^2-4\cdot1\cdot\left(3m+1\right)\)

\(=25-12m-4=-12m+21\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>-12m+21>0

=>-12m>-21

=>\(m< \dfrac{21}{12}=\dfrac{7}{4}\)

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-5\right)}{1}=5\\x_1x_2=\dfrac{c}{a}=3m+1\end{matrix}\right.\)

\(\left|x_1^2-x_2^2\right|=15\)

=>\(\left|\left(x_1-x_2\right)\left(x_1+x_2\right)\right|=15\)

=>\(\left|5\cdot\left(x_1-x_2\right)\right|=15\)

=>\(\left|x_1-x_2\right|=3\)

=>\(\sqrt{\left(x_1-x_2\right)^2}=3\)

=>\(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=3\)

=>\(\sqrt{5^2-4\left(3m+1\right)}=3\)

=>\(25-12m-4=9\)

=>-12m+21=9

=>-12m=-12

=>m=1(nhận)

a. Em tự giải

b.

\(\Delta=25-4\left(3m+1\right)=21-12m>0\Rightarrow m< \dfrac{7}{4}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=3m+1\end{matrix}\right.\)

\(\left|x_1^2-x_2^2\right|=15\)

\(\Leftrightarrow\left(x_1^2-x_2^2\right)^2=225\)

\(\Leftrightarrow\left(x_1-x_2\right)^2\left(x_1+x_2\right)^2=225\)

\(\Leftrightarrow25\left[\left(x_1+x_2\right)^2-4x_1x_2\right]=225\)

\(\Leftrightarrow25-4\left(3m+1\right)=9\)

\(\Rightarrow m=1\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)

\(P=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{1-x}\)

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

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

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

\(=\dfrac{-\sqrt{x}+1+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

a: Thay m=3/4 vào (1), ta được:

\(x^2-2\left(\dfrac{3}{4}-1\right)x-2\cdot\dfrac{3}{4}=0\)

=>\(x^2-2\cdot\dfrac{-1}{4}x-\dfrac{3}{2}=0\)

=>\(x^2+\dfrac{1}{2}x-\dfrac{3}{2}=0\)

=>\(2x^2+x-3=0\)

=>\(\left(2x+3\right)\left(x-1\right)=0\)

=>\(\left[{}\begin{matrix}2x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)