2/5 - 9 - 7/2 giải thật chi tiết và từng bước ạ!
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C6: 2/5= (2 x 20)/ (5/20)= 40/100
C7: 3/4 = (3x25)/(4x25)=75/100
C8: 1/2= (1x50)/(2x50)= 50/100
C9: 5/6= (5x50/3)/(6x50/3)= (250/3)/100
Với x ≥ 0; x ≠ 9 ta có:
\(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x-3}\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
Vậy \(A=\dfrac{3}{\sqrt{x}+3}\).
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Δ=(2m-6)^2-4(m^2+3)
=4m^2-24m+36-4m^2-12=-24m+24
Để phương trình có hai nghiệm phân biệt thì -24m+24>0
=>m<1
x1^2+x2^2=36
=>(x1+x2)^2-2x1x2=36
=>(2m-6)^2-2(m^2+3)=36
=>4m^2-24m+36-2m^2-6-36=0
=>2m^2-24m-6=0
=>m^2-12m-3=0
=>\(m=6-\sqrt{39}\)
\(M=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)
\(M=\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+...+\frac{99-97}{97\cdot99}\)
\(M=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}+\frac{7}{5\cdot7}-\frac{5}{5\cdot7}+\frac{9}{7\cdot9}-\frac{7}{7\cdot9}+...+\frac{99}{97\cdot99}-\frac{97}{97\cdot99}\)
\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(M=\frac{1}{3}-\frac{1}{99}\)
\(M=\frac{33}{99}-\frac{1}{99}\)
\(M=\frac{32}{99}\)
Vậy \(M=\frac{32}{99}\)
Có 2/ 3.5 + 2/ 5.7 + 2/ 7.9 +...+ 2/ 97.99
= 1/3 -1/5 +1/5 -1/7 +1/7 -1/9 +...+ 1/ 97- 1/99
= 1/3 - 1/99
= 32/ 99
\(-\frac{1}{7}+\frac{5}{3}+\frac{5}{4}+\frac{1}{3}-\frac{3}{2}\)
\(=\left(-\frac{1}{7}+\frac{5}{3}-\frac{3}{2}\right)+\left(\frac{5}{3}+\frac{1}{3}\right)\)
\(=\frac{-6}{42}+\frac{70}{42}-\frac{63}{42}+\frac{6}{3}\)
\(=\frac{-6+70-63}{42}+2\)
\(=\frac{1}{42}+\frac{84}{42}\)
\(=\frac{85}{42}\)
Lời giải:
$2=\sqrt{4}< \sqrt{5}$
$\Rightarrow -2> -\sqrt{5}$
b. Để biểu thức trên có nghĩa thì \(\left\{\begin{matrix} 5-x\neq 0\\ \frac{10}{5-x}\geq 0\end{matrix}\right.\Leftrightarrow 5-x>0\Leftrightarrow x<5\)
\(\dfrac{2}{5}-9-\dfrac{7}{2}\)
\(=\dfrac{2}{5}-\dfrac{9}{1}-\dfrac{7}{2}\)
\(=\left(\dfrac{2}{5}-\dfrac{9}{1}\right)-\dfrac{7}{2}\)
\(=\left(\dfrac{2}{5}-\dfrac{45}{5}\right)-\dfrac{7}{2}\)
\(=-\dfrac{43}{5}-\dfrac{7}{2}\)
\(=-\dfrac{86}{10}-\dfrac{35}{10}\)
\(=-\dfrac{121}{10}\)