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\(VT=\sqrt{\dfrac{\sqrt{5}}{8\sqrt{5}+3\sqrt{35}}}.\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\sqrt{\dfrac{\sqrt{5}}{8\sqrt{5}+3\sqrt{5}.\sqrt{7}}}.\left(3\sqrt{2}+\sqrt{2}.\sqrt{7}\right)\)
\(=\sqrt{\dfrac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}.\left[\sqrt{2}\left(3+\sqrt{7}\right)\right]\)
\(=\sqrt{\dfrac{1}{8+3\sqrt{7}}}.\left[\sqrt{2}\left(3+\sqrt{7}\right)\right]\)
\(=\dfrac{\sqrt{2}\left(3+\sqrt{7}\right)}{\sqrt{8+3\sqrt{7}}}\)
\(=\dfrac{\sqrt{2}.\sqrt{2}\left(3+\sqrt{7}\right)}{\sqrt{2}.\sqrt{8+3\sqrt{7}}}\) (Nhân \(\sqrt{2}\) cả tử và mẫu)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{\sqrt{16+6\sqrt{7}}}\)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{\sqrt{\left(3+\sqrt{7}\right)^2}}\)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{\left|3+\sqrt{7}\right|}\)
\(=\dfrac{2\left(3+\sqrt{7}\right)}{3+\sqrt{7}}\)
\(=2=VP\left(dpcm\right)\)
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1.4:
a: CH=16^2/24=256/24=32/3
BC=24+32/3=104/3
AC=căn 32/3*104/3=16/3*căn 13
b: BC=12^2/6=24
AC=căn 24^2-12^2=12*căn 3
CH=24-6=18
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\(\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{AC}{BC}:\dfrac{AB}{BC}=\dfrac{AC}{AB}=\tan\alpha\)
\(\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{AB}{BC}:\dfrac{AC}{BC}=\dfrac{AB}{AC}=\cot\alpha\)
\(\tan\alpha\cot\alpha=\dfrac{AC}{AB}\cdot\dfrac{AB}{AC}=1\)
\(\sin^2\alpha+\cos^2\alpha=\dfrac{AC^2}{BC^2}+\dfrac{AB^2}{BC^2}=\dfrac{AB^2+AC^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\left(pytago\right)\)
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\(3\sqrt{4x+1}+4x\sqrt{3x-2}=3x^{^2}+4x+5\Leftrightarrow12x+3+12x^2-8x=3x^2+4x+5\Leftrightarrow12x+12x^2-8x-3x^2-4x=5-3\Leftrightarrow9x^2=2\Leftrightarrow x^2=\dfrac{2}{9}\Leftrightarrow x=\dfrac{\sqrt{2}}{3}\)
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ĐKXĐ: \(x^2+5x+2>=0\)
=>\(\left[{}\begin{matrix}x>=\dfrac{-5+\sqrt{17}}{2}\\x< =\dfrac{-5-\sqrt{17}}{2}\end{matrix}\right.\)
\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
=>\(x^2+5x+4-3\sqrt{x^2+5x+2}-6=0\)
=>\(x^2+5x+2-3\sqrt{x^2+5x+2}-4=0\)(1)
Đặt \(\sqrt{x^2+5x+2}=a\)(a>=0)
Phương trình (1) trở thành:
\(a^2-3a-4=0\)
=>(a-4)(a+1)=0
=>\(\left[{}\begin{matrix}a=4\left(nhận\right)\\a=-1\left(loại\right)\end{matrix}\right.\)
=>\(x^2+5x+2=4^2=16\)
=>\(x^2+5x-14=0\)
=>\(\left(x+7\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
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16:
a: ĐKXĐ: x>0
\(x+\dfrac{1}{x}+\sqrt{x}+\dfrac{1}{\sqrt{x}}=4\)
\(x+\dfrac{1}{x}>=2\cdot\sqrt{x\cdot\dfrac{1}{x}}=2\)
\(\sqrt{x}+\dfrac{1}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}=2\)
Do đó: \(x+\dfrac{1}{x}+\sqrt{x}+\dfrac{1}{\sqrt{x}}>=2+2=4\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x^2=1\\x=1\end{matrix}\right.\)
=>x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
=>5x^2-6x-11=0
=>5x^2-11x+5x-11=0
=>(5x-11)(x+1)=0
=>x=11/5 hoặc x=-1
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Câu I:
1. \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{x-3}{x-1}\left(x\ge0;x\ne1\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}-\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}+\dfrac{x-3}{x-1}\)
\(=\dfrac{x+\sqrt{x}-2x+2\sqrt{x}+x-3}{x-1}\)
\(=\dfrac{3\sqrt{x}-3}{x-1}=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{3}{\sqrt{x}+1}\)
2. \(\dfrac{1}{P}=\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{1}{\dfrac{3}{\sqrt{x}+1}}=\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{12}{\sqrt{x}+1}=3\)
\(\Leftrightarrow3\sqrt{x}+3=12\)
\(\Leftrightarrow\sqrt{x}=3\)
\(\Leftrightarrow x=9\left(Vì.x\ge0;x\ne1\right)\)
Câu II:
1. Vì đường thẳng (d) cắt trục hoành tại điểm có hoành độ bằng 2, nên đường thẳng (d) cắt trục hoành tại điểm có tọa độ (2;0)
Thay x = 2; y = 0 vào phương trình đường thẳng (d), ta được:
\(0=\left(2-m\right).2+m+1\)
\(\Leftrightarrow4-2m+m+1=0\)
\(\Leftrightarrow-m=-5\)
\(\Leftrightarrow m=5\)
Vậy nếu m = 5 thì đưởng thẳng (d) cắt trục hoành tại điểm có hoành độ bằng 2.
2. \(\left\{{}\begin{matrix}3x+2y=11\\x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=12\\x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\3-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất là (x; y) = (3; 1)