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P=\(\frac{\sqrt{10+2\sqrt{25-9x^2}}}{x}\)
P=\(\frac{\sqrt{10+2\sqrt{\left(5+3x\right)\left(5-3x\right)}}}{x}\)
P=\(\frac{\sqrt{10+10-a^2}}{x}\)(Vì a2=\(\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2\)=10-2\(\sqrt{\left(5+3x\right)\left(5-3x\right)}\))
\(\sqrt{5+3x}-\sqrt{5-3x}=a\)
\(\Leftrightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\)
\(\Leftrightarrow5+3x+5-3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}=a^2\)
\(\Leftrightarrow10-2\sqrt{\left(5+3x\right)\left(5-3x\right)}=a^2\)
\(\Leftrightarrow2\sqrt{\left(5+3x\right)\left(5-3x\right)}=10-a^2\)
Thế vào P ta được:
\(P=\frac{\sqrt{10+2\sqrt{25-9x^2}}}{x}=\frac{\sqrt{10+2\sqrt{\left(5-3x\right)\left(5+3x\right)}}}{x}\)
\(=\frac{\sqrt{10+10-a^2}}{x}\)
\(=\frac{\sqrt{20-a^2}}{x}\)
P/s: nếu em có sai sót, xin bỏ qua
\(\sqrt{5+3x}-\sqrt{5-3x}=a\left(x\le\dfrac{5}{3}\right)\)
\(\Rightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\)
\(\Rightarrow5+3x+5-3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}=a^2\)
\(\Rightarrow10-2\sqrt{25-9x^2}=a^2\)
\(\Rightarrow-2\sqrt{25-9x^2}=a^2-10\)
\(\Rightarrow2\sqrt{25-9x^2}=10-a^2\)
\(\Rightarrow10+2\sqrt{25-9x^2}=20-a^2\)
\(\Rightarrow P=\dfrac{\sqrt{10+2\sqrt{25-9x^2}}}{x}=\dfrac{\sqrt{20-a^2}}{x}\)
\(\sqrt{5+3x}-\sqrt{5-3x}=a\\ \Rightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\\ \Rightarrow5+3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}+5-3x=a^2\\ \Rightarrow2\sqrt{25-9x^2}=10-a^2\\ \Rightarrow4\left(25-9x^2\right)=\left(10-a^2\right)^2\\ \Rightarrow100-36x^2=100-20a^2+a^4\\ \Rightarrow36x^2=20a^2-a^4\\ \Rightarrow x^2=\dfrac{20a^2-a^4}{36}\\ \Rightarrow x=\dfrac{\sqrt{a^2\left(20-a^2\right)}}{6}\)
\(\Rightarrow P=\dfrac{\sqrt{10+2\sqrt{25-9x^2}}}{x}\\ =\dfrac{\sqrt{10+10-a^2}}{\dfrac{\sqrt{a^2\left(20-a^2\right)}}{6}}=6\sqrt{\dfrac{20-a^2}{a^2\left(20-a^2\right)}}=\dfrac{6}{\left|a\right|}\)
\(A=\sqrt{80}+\sqrt{45}+\sqrt{5}=\sqrt{16.5}+\sqrt{9.5}+\sqrt{5}\)
\(=4\sqrt{5}+3\sqrt{5}+\sqrt{5}=8\sqrt{5}\)
\(B=\frac{5}{\sqrt{10}}+3,5\sqrt{40}=\sqrt{\frac{25}{10}}+3,5\sqrt{16.2,5}\)
\(=\sqrt{2,5}+3,5.4\sqrt{2,5}=15\sqrt{2,5}\)
\(C=\frac{1}{\sqrt{3}-2}+\frac{\sqrt{300}}{10}-\sqrt{12}\)
\(=\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{\sqrt{100.3}}{10}-\sqrt{4.3}\)
\(=-\sqrt{3}-2+\sqrt{3}-2\sqrt{3}=-2\sqrt{3}-2\)
\(D=4\sqrt{x}+2\sqrt{x^2}-\sqrt{16x}=4\sqrt{x}+2x-4\sqrt{x}=2x\) ( do \(x\ge0\))
\(F=\frac{a-2\sqrt{a}}{\sqrt{a}-2}=\frac{\sqrt{a}.\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=\sqrt{a}\)
mk chỉnh đề
\(E=\sqrt{25x+25}-\sqrt{9x+9}+\sqrt{4x+4}\)
\(=\sqrt{25\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}\)
\(=5\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=4\sqrt{x+1}\)
\(G=\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{2}{\sqrt{5}-\sqrt{7}}=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}-\frac{2\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)
\(=\sqrt{3}-\sqrt{5}-\sqrt{5}-\sqrt{7}=\sqrt{3}-\sqrt{7}\)
Câu a :
Ta có : \(\sqrt{5+3x}-\sqrt{5-3x}=a\)
\(\Leftrightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\)
\(\Leftrightarrow5+3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}+5-3x=a^2\)
\(\Leftrightarrow10-2\sqrt{25-9x^2}=a^2\)
\(\Leftrightarrow2\sqrt{25-9x^2}=10-a^2\)
\(\Leftrightarrow\sqrt{25-9x^2}=\dfrac{10-a^2}{2}\)
\(\Leftrightarrow25-9x^2=\dfrac{\left(a^2-10\right)^2}{2}\)
\(\Leftrightarrow9x^2=25-\dfrac{\left(a^2-10\right)^2}{2}\)
\(\Leftrightarrow3x=\sqrt{\dfrac{50-\left(a^2-10\right)^2}{2}}\)
\(\Leftrightarrow x=\dfrac{\sqrt{50-\left(a^2-10\right)^2}}{3\sqrt{2}}\)
\(P=\dfrac{3\sqrt{2}.\sqrt{10+2\sqrt{\dfrac{10-a^2}{2}}}}{\sqrt{50-\left(a^2-10\right)^2}}\)
Bạn tự rút gọn nữa nhé :))
Câu b : \(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-24}{z}\)
\(=\dfrac{x-3}{x}+\dfrac{y-3}{y}+\dfrac{z-12}{z}\)
\(=3-3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{4}{z}\right)\le3-3\left[\dfrac{\left(1+1+2\right)^2}{12}\right]=-1\)