1/x + 1/x+3 = 1/2
giải phương trình
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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
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
x-\(\dfrac{x+2}{3}\)nhỏ hơn hoặc bằng 3x+\(\dfrac{x}{2}+5\)
\(\dfrac{3}{x}=\dfrac{1}{y}=\dfrac{6}{2}\)
\(\Rightarrow\dfrac{3}{x}=\dfrac{1}{y}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{3}\end{matrix}\right.\)
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
\(\frac{1}{x}+\frac{1}{x+3}=\frac{1}{2}\)
\(\Leftrightarrow\frac{2\left(x+3\right)}{2x\left(x+3\right)}+\frac{2x}{2x\left(x+3\right)}=\frac{x\left(x+3\right)}{2x\left(x+3\right)}\)
\(\Leftrightarrow2x+6+2x=x^2+3x\)
\(\Leftrightarrow x=3\)
\(\frac{1}{x}+\frac{1}{x+3}=\frac{1}{2}\)
\(\frac{1}{x+x+3}=\frac{1}{2}\)
x+x+3=2
2x=-1
x=-1/2