b, x-1/x+2=4/5 (x khác -2)
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a, \(\frac{x+2}{5}=\frac{1}{x-2}\Rightarrow\left(x+2\right)\left(x-2\right)=5\Rightarrow x^2-2x+2x-4=5\Rightarrow x^2=9\Rightarrow x=\pm3\)
b, \(\frac{3}{x-4}=\frac{x+4}{3}\Rightarrow\left(x+4\right)\left(x-4\right)=9\Rightarrow x^2-4x+4x-16=9\Rightarrow x^2=25\Rightarrow x=\pm5\)
c, \(\frac{x+2}{2}=\frac{1}{1-x}\Rightarrow\left(x+2\right)\left(1-x\right)=2\Rightarrow x-x^2+2-2x=2\Rightarrow-x^2-x=0\Rightarrow-x\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
a: Ta có: \(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\)
\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{x-1+3x}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{4x-1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\left(2x-2\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)
Câu a đã làm: F=(2√x/2√x-1 - 1/√x) ( √x+1/√x-1 + 3x/x-2√x+1) với x >0, x khác 1, x khác 1/4 a) rút gọn F - Hoc24
\(b,F=2\Leftrightarrow\dfrac{\left(2\sqrt{x}+1\right)\left(2x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}=2\\ \Leftrightarrow2\sqrt{x}\left(x-2\sqrt{x}+1\right)=2x\sqrt{x}-4x+2\sqrt{x}+2x-2\sqrt{x}+1\\ \Leftrightarrow2x\sqrt{x}-4x+2\sqrt{x}=2x\sqrt{x}-2x+1\\ \Leftrightarrow2x-2\sqrt{x}+1=0\\ \Leftrightarrow2\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{2}=0\\ \Leftrightarrow2\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{2}=0\\ \Leftrightarrow x\in\varnothing\)
Bài 1: a/b=b/c=c/a chứ không phải c/d
áp dụng tính chất dãy tỉ số bằng nhau, ta có:
a/b=b/c=c/a=(a+b+c)/(b+c+a)=1
a/b=1 => a=b
b/c=1 => b=c
Vậy a=b=c
\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)
\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)
\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)
a)2x+5-(x-7)=18
2x+5-x+7=18
2x-x+(5+7)=18
x+12=18
x=18-12
x=6
b)2(x+1)+4^2=2^4
2(x+1)+(2^2)^2=2^4
2(x+1)+2^4=2^4
2(x+1)=2^4-2^4
2(x+1)=0
x+1=0:2=0
x=0-1
x=-1
c) mk đang nghĩ tí mk gửi qua tin nhắn cho
a) \(2x+5-\left(x-7\right)=18\)
\(2x+5-x+7=18\)
\(2x-x=18-5-7\)
\(x=6\)
b) \(2\left(x+1\right)+4^2=2^4\)
\(2x+2+16=16\)
\(2x=16-16-2\)
\(2x=-2\)
\(x=-1\)
c) \(\frac{x-3}{x-5}=\frac{5}{7}\)
\(\left(x-3\right)\cdot7=\left(x+5\right)\cdot5\)
\(7x-21=5x+25\)
\(7x-5x=25+21\)
\(2x=46\)
\(x=\frac{46}{2}=23\)
\(\frac{x-1}{x+2}=\frac{4}{5}\)\(ĐKXĐ\left(x\ne-2\right)\)
\(\Leftrightarrow\left(x-1\right)5=\left(x+2\right)4\)
\(\Leftrightarrow5x-5=4x+8\)
\(\Leftrightarrow5x-4x=8+5\)
\(\Leftrightarrow x=12\)
x-1/x+2=4/5
<=> 5(x-1)=4(x+2)
<=>5x-5=4x+8
<=>5x-5-4x-8=0
<=>x-13=0
<=>x=13
Vậy x=13