x=0

x=-3

(x+1)(x+2) = 2

x² + 2x + x + 2 = 2

            x² + 3x = 2 - 2

           x(x + 3) = 2

⇒ x = 0 hoặc x + 3 = 0

⇒ x = 0 hoặc x = -3

Vậy tập nghiệm S = {-3}

m=1

`hpt`:$\begin{cases}x+y=1\\x+4y=2\\\end{cases}$

`<=>` $\begin{cases}3y=1\\x=1-y\\\end{cases}$

`<=>` $\begin{cases}y=\dfrac13\\x=\dfrac23\\\end{cases}$

⇔m2x−mx−2x+m−2=0

⇔m2x−4x−mx+2x+m−2=0

⇔x(m−2)(m+2)−x(m−2)+(m−2)=0

⇔(mx+2x−x+1)(m−2)=0

⇔((m+1)x+1)(m−2)=0

⇒[x=−1m+1  m=2thì TM mọi x thuôộc R

     m=2

X = - 1000/507

\(\dfrac{x}{40}-\dfrac{x+4}{52}=2\)

\(\Rightarrow52x-40x-160=4160\Leftrightarrow12x=4320\Leftrightarrow x=360\)

Bài 1:

\(a,3\left(x-11\right)-2\left(x+11\right)=2011\)

\(\Leftrightarrow3x-33-2x-22=2011\)

\(\Leftrightarrow x-55=2011\)

\(\Leftrightarrow x=2066\)

Vậy pt có nghiệm x = 2066

\(b,\left(x-1\right)\left(3x-7\right)=\left(x-1\right)\left(x+30\right)\)

\(\Leftrightarrow\left(x-1\right)\left(3x-7\right)-\left(x-1\right)\left(x+30\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-7-x-30\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-37\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-37=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{37}{2}\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{1;\dfrac{37}{2}\right\}\)

\(c,\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) (1)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)

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

\(\Rightarrow x^2+2x-x+2-2=0\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

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

Vậy pt có tập nghiệm \(S=\left\{0;-1\right\}\)

\(d,\left|2x-3\right|=x+1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{4;\dfrac{2}{3}\right\}\)

Bài 2:

\(a,2\left(x-1\right)< x+1\)

\(\Leftrightarrow2x-2< x+1\)

\(\Leftrightarrow2x-x< 1+2\)

\(\Leftrightarrow x< 3\)

Vậy bpt có nghiệm x < 3

b, Đề bài ko rõ

\(\dfrac{3}{x}=\dfrac{1}{y}=\dfrac{6}{2}\)

\(\Rightarrow\dfrac{3}{x}=\dfrac{1}{y}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{3}\end{matrix}\right.\)

x = 1

y = 1/3

x = 1 

y = 1/3. 

Sửa đề; \(A=\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\)

a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}+1-2}{x-1}=\dfrac{2\sqrt{x}-2}{x-1}=\dfrac{2}{\sqrt{x}+1}\)

b: Khi x=3+2căn 2 thì \(A=\dfrac{2}{\sqrt{2}+1+1}=\dfrac{2}{\sqrt{2}+2}=2-\sqrt{2}\)

1 a,\(4-5\sqrt{x}=-1\)=>\(5\sqrt{x}=5\)\(\Rightarrow\sqrt{x}=1\)\(\Rightarrow x=1\)

b,\(\Leftrightarrow\)\(\sqrt{x-1}=0\)hoâc \(\sqrt{x+3}=0\)

<=> x=1   hoâc x= -3

2,

a,=> \(x=\frac{2}{3}\)

b=>,\(x^2=\frac{9}{25}\)\(\Rightarrow x=\frac{3}{5}\)

c,=>\(4x^2=1\)\(\Rightarrow x^2=\frac{1}{4}\)\(\Rightarrow x=\frac{1}{2}\)

d,=>x+1=\(\sqrt{2}\)

   =>x    =\(\sqrt{2}-1\)

nhân dúng cho mk nha