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\(\dfrac{1}{\sqrt{\dfrac{5}{7}}+\sqrt{\dfrac{5}{13}}+1}+\dfrac{1}{\sqrt{\dfrac{7}{13}}+\sqrt{\dfrac{7}{5}}+1}+\dfrac{1}{\sqrt{1\dfrac{6}{7}}+\sqrt{2\dfrac{3}{5}}+1}\\ =\dfrac{1}{\dfrac{\sqrt{5}}{\sqrt{7}}+\dfrac{\sqrt{5}}{\sqrt{13}}+\dfrac{\sqrt{5}}{\sqrt{5}}}+\dfrac{1}{\dfrac{\sqrt{7}}{\sqrt{13}}+\dfrac{\sqrt{7}}{\sqrt{5}}+\dfrac{\sqrt{7}}{\sqrt{7}}}+\dfrac{1}{\dfrac{\sqrt{13}}{\sqrt{7}}+\dfrac{\sqrt{13}}{\sqrt{5}}+\dfrac{\sqrt{13}}{\sqrt{13}}}\\ =\left(\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{7}}+\dfrac{1}{\sqrt{13}}\right)\cdot\dfrac{1}{\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{7}}+\dfrac{1}{\sqrt{13}}}\\ =1\)
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bạn giải thích giúp mình bước 1 mấy bước sau mình sẽ tham khảo thêm cảm ơn nhiều 🙏
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Bài 2:
a: Ta có: \(\sqrt{64x+64}-\sqrt{25x+25}+\sqrt{4x+4}=20\)
\(\Leftrightarrow8\sqrt{x+1}-5\sqrt{x+1}+2\sqrt{x+1}=20\)
\(\Leftrightarrow5\sqrt{x+1}=20\)
\(\Leftrightarrow x+1=16\)
hay x=15
b: Ta có: \(\sqrt{3x}+5\sqrt{27x}-16=\sqrt{432x}\)
\(\Leftrightarrow\sqrt{3x}+15\sqrt{3x}-12\sqrt{3x}=16\)
\(\Leftrightarrow4\sqrt{3x}=16\)
\(\Leftrightarrow3x=16\)
hay \(x=\dfrac{16}{3}\)
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a.
Đặt \(x+2y+1=a\)
\(\Rightarrow P=a^2+\left(a+4\right)^2=2a^2+8a+16=2\left(a+2\right)^2+8\ge8\)
\(P_{min}=8\) khi \(a=-2\) hay \(x+2y+3=0\)
b.
\(\sqrt{x}-1=a\ge0\Rightarrow\sqrt{x}=a+1\Rightarrow x=a^2+2a+1\)
\(Q=\dfrac{\left(a^2+2a+1\right)+\left(a+1\right)+1}{a}=\dfrac{a^2+3a+3}{a}=a+\dfrac{3}{a}+3\ge2\sqrt{\dfrac{3a}{a}}+3=3+2\sqrt{3}\)
\(Q_{min}=3+2\sqrt{3}\) khi \(a=\sqrt{3}\) hay \(x=4+2\sqrt{3}\)
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a) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-\sqrt{x}}\right):\dfrac{1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{x+2}{\sqrt{x}}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-\sqrt{x}}\right):\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)
\(=\dfrac{x+2}{\sqrt{x}}\)
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b,\(\sqrt{\left(3x-2\right)^2}=4\)
\(\Leftrightarrow3x-2=4\)
\(\Leftrightarrow3x=6\Leftrightarrow x=2\)
vậy......
c,\(\dfrac{2\sqrt{x}-19}{4-\sqrt{x}}=\dfrac{1}{5}\) ĐKXĐ: x <16
\(\Rightarrow2\sqrt{x}-19=\dfrac{1}{5}\left(4-\sqrt{x}\right)\)
\(\Leftrightarrow2\sqrt{x}-19=\dfrac{4}{5}-\dfrac{1}{5}\sqrt{x}\)
\(\Leftrightarrow\dfrac{11}{5}\sqrt{x}=\dfrac{99}{5}\)
\(\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\left(KTMĐK\right)\)
vậy........
a/ ĐKXĐ: \(x\ge2\)
\(2\sqrt{4x-8}-\sqrt{9x-18}+\sqrt{36x-72}=14\)
\(\Leftrightarrow4\sqrt{x-2}-3\sqrt{x-2}+6\sqrt{x-2}=14\)
\(\Leftrightarrow7\sqrt{x-2}=14\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\) ( tmđk)
Vậy phương trình đã cho có nghiệm x=6