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Ta có: \(x^4+16x^2+32=0\Leftrightarrow\left(x^2-8\right)^2-32=0\left(1\right)\)
Với \(x=\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)\(\Leftrightarrow x=\sqrt{3}\sqrt{2-\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)
\(\Rightarrow x^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3}\sqrt{2-\sqrt{3}}\)
Thay x vào vế phải của (1) ta được:
\(\left(x^2-8\right)^2-32=\left(8-2\sqrt{2+\sqrt{3}}-2\sqrt{3}\sqrt{2-\sqrt{3}}-8\right)^2-32\)
\(=4\left(2+\sqrt{3}\right)+4\sqrt{3}+12\left(2-\sqrt{3}\right)-32\)
\(=8+4\sqrt{3}+8\sqrt{3}+24-12\sqrt{3}-32=0\)= vế phải
Vậy \(x-\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)là 1 nghiệm của phương trình đã cho(đpcm)
a) \(\sqrt{39-12\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)
\(=\sqrt{36-12\sqrt{3}+3}+\sqrt{9-12\sqrt{3}+12}\)
\(=\sqrt{\left(6-\sqrt{3}\right)^2}+\sqrt{\left(3-\sqrt{12}\right)^2}\)
\(=6-\sqrt{3}+\sqrt{12}-3=3+\sqrt{3}\)
b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(=\frac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\frac{\sqrt{5-2\sqrt{5}+1}+\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\frac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
Bài 1:
a)Đk:\(x\ge\frac{3}{2}\)
\(pt\Leftrightarrow3-x=-\sqrt{2x-3}\)
Bình phương 2 vế ta có:
\(\left(3-x\right)^2=\left(-\sqrt{2x-3}\right)^2\)
\(\Leftrightarrow x^2-6x+9=2x-3\)
\(\Leftrightarrow x^2-8x+12=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=6\end{array}\right.\).Thay vào thấy x=2 ko thỏa mãn
Vậy x=6
b)Đk:\(x\ge1\)
\(pt\Leftrightarrow\sqrt{x-1}=\sqrt{3x-2}+\sqrt{5x-1}\)
Bình phương 2 vế của pt ta có:
\(\left(\sqrt{x-1}\right)^2=\left(\sqrt{3x-2}+\sqrt{5x-1}\right)^2\)
\(\Leftrightarrow x-1=\left(3x-2\right)+\left(5x-1\right)+2\sqrt{\left(3x-2\right)\left(5x-1\right)}\)
\(\Leftrightarrow x-1=8x-3+2\sqrt{\left(3x-2\right)\left(5x-1\right)}\)
\(\Leftrightarrow2-7x=2\sqrt{\left(3x-2\right)\left(5x-1\right)}\)
Bình phương 2 vế của pt ta có:
\(\left(2-7x\right)^2=\left[2\sqrt{\left(3x-2\right)\left(5x-1\right)}\right]^2\)
\(\Leftrightarrow49x^2-28x+4=60x^2-52x+8\)
\(\Leftrightarrow-11x^2+24x-4=0\)
\(\Leftrightarrow\left(2-x\right)\left(11x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{2}{11}\end{array}\right.\) (Loại)
Vậy pt vô nghiệm
a)\(\Leftrightarrow\)\(7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)
\(\Leftrightarrow\) \(3\sqrt{x-2}=8\)
\(\Leftrightarrow\) \(\sqrt{x-2}=24\)
\(\Leftrightarrow\)\(x-2=576\)\(\Leftrightarrow x=578\)
c)\(\Leftrightarrow GTTĐ\left(x-1\right)=\sqrt{2}-1\)\(TH1:x-1>0\)
\(\Rightarrow x-1=\sqrt{2}-1\)\(\Leftrightarrow x=\sqrt{2}\)
\(TH2:x-1< 0\)
\(\Rightarrow1-x=\sqrt{2}-1\)
\(\Leftrightarrow x=2+\sqrt{2}\)
d)\(TH1:x-10=0\Rightarrow x=10\)
\(TH2:\sqrt{x-4}=0\Rightarrow x=4\)
câu b) thì mik cần thêm time
a: Sửa đề: \(5\dfrac{1}{5}-\dfrac{1}{2}\sqrt{20}+\sqrt{5}\)
\(=5.2-\dfrac{1}{2}\cdot2\sqrt{5}+\sqrt{5}=5.2\)
b: \(=\dfrac{1}{2}\sqrt{2}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{9}{2}\sqrt{2}\)
c: \(=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+\sqrt{77}=-\sqrt{5}+9\sqrt{2}+\sqrt{77}\)
d: \(=\dfrac{1}{10}\cdot10\sqrt{2}+\dfrac{2}{5}\sqrt{2}+0.4\cdot5\sqrt{2}\)
\(=\dfrac{17}{5}\sqrt{2}\)
Câu 1:
a, \(\sqrt{50.98} = 5\sqrt{2} . 7\sqrt{2} = 70\)
b, \(\sqrt{2,5.12,1} = 30,25\)
c, \(\sqrt{17.51.27} = \sqrt{23409} = 153\)
d, \(\sqrt{32.128} = \sqrt{4096} = 64\)
e, \(\sqrt{3,2.7,2.49} = 7\sqrt{3,2.7,2} = 7\sqrt{23,04} =33,6\)
g, \(\sqrt{2,5.12,5.20} = \sqrt{625} = 25\)
mình ghi nhầm pn ơi.. bài 2 là \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{6}}\)
a: =căn 16/5*36/5*49=7*4*9/5=252/5
b: =căn 250*2,5=căn 625=25
c: =căn 3/2*2/3=1
d: =5*căn 2*7*căn 2=70