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1.
\(\sqrt{9+4\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{9+4\sqrt{5}-\sqrt{5-2\sqrt{4.5}+4}}\)
\(=\sqrt{9+4\sqrt{5}-\sqrt{(\sqrt{5}-\sqrt{4})^2}}=\sqrt{9+4\sqrt{5}-(\sqrt{5}-\sqrt{4})}\)
\(=\sqrt{9+4\sqrt{5}-\sqrt{5}+2}=\sqrt{11+3\sqrt{5}}\)
2.
\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}=\sqrt{8-2\sqrt{7}-\sqrt{7+2\sqrt{7}+1}}\)
\(=\sqrt{8-2\sqrt{7}-\sqrt{(\sqrt{7}+1)^2}}\)
\(=\sqrt{8-2\sqrt{7}-\sqrt{7}-1}=\sqrt{7-3\sqrt{7}}\)
Vd1:
d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)
\(\Leftrightarrow x=6\)
dk \(x\ge-\frac{4}{3}\)
\(x^2-5x+4=8\sqrt{3x+4}-32\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=8\left(\sqrt{3x+4}-4\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)-8\frac{\left(\sqrt{3x+4}-4\right)\left(\sqrt{3x+4}+4\right)}{\sqrt{3x+4}+4}=0\)
\(\left(x-1\right)\left(x-4\right)-8.\frac{3\left(x-4\right)}{\sqrt{3x+4}+4}=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1-\frac{24}{\sqrt{3x+4}+4}=0\right)\)
đến đây để rồi tự làm nhé ^^
\(x^2-5x+36=8\sqrt{3x+4}\)
\(\Leftrightarrow x^2-5x+36-8\sqrt{3x+4}=0\)
\(\Leftrightarrow\left(-8\sqrt{3x+4}+32\right)+\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow-8\left(\sqrt{3x+4}-4\right)+\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow-8.\frac{3x+4-16}{\sqrt{3x+4}+4}+\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow-8.\frac{3x-12}{\sqrt{3x+4}+4}+\left(x-1\right)\left(x-4\right)=0\)
\(\left(x-4\right)\left(\frac{-24}{\sqrt{3x+4}+4}+x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\\frac{-24}{\sqrt{3x+4}+4}+x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\-\frac{24}{\sqrt{3x+4}+4}+3+x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\-3.\frac{16-3x-4}{\left(\sqrt{3x+4}+4\right)^2}+\left(x-4\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\\left(x-4\right)\left[\frac{9}{\left(\sqrt{3x+4}+4\right)^2}+1\right]=0\end{cases}}\)
Mà \(\frac{9}{\left(\sqrt{3x+4}+4\right)^2}+1>0\forall x\) nên \(x-4=0\Rightarrow x=4\)
Vật PT có nghiệm duy nhất là \(x=4\)
Anh ko ghi lại đề nha em !
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\3x^2-5x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\x_1=1;x_2=\frac{2}{3}\end{cases}}\)( vn là vô nghiệm nha )
Vậy : x = 1 hoặc x = 2/3
\(\left(x^2+1\right).\left(3x^2-5x+2\right)=0\)
\(x^2\ge0\Rightarrow x^2+1\ge1\)
\(\RightarrowĐể\left(x^2+1\right).\left(3x^2-5x+2\right)=0\)
\(\Rightarrow3x^2-5x+2=0\Rightarrow3x^2-3x-2x+2=0\)
\(\Rightarrow3x.\left(x-1\right)-2.\left(x-1\right)=0\Rightarrow\left(3x-2\right).\left(x-1\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{2}{3}\end{cases}}\)
1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=\left(\sqrt{19}\right)^2-3^2=19-9=10\)
2) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{\dfrac{8+2\sqrt{7}}{2}}-\sqrt{\dfrac{8-2\sqrt{7}}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}=\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1}{\sqrt{2}}-\dfrac{\sqrt{7}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
3) \(\sqrt{8+\sqrt{60}}+\sqrt{45}-\sqrt{12}=\sqrt{8+\sqrt{4.15}}+\sqrt{9.5}-\sqrt{4.3}\)
\(=\sqrt{8+2\sqrt{15}}+3\sqrt{5}-2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}=\left|\sqrt{5}+\sqrt{3}\right|+3\sqrt{5}-2\sqrt{3}\)
\(\sqrt{5}+\sqrt{3}+3\sqrt{5}-2\sqrt{3}=4\sqrt{5}-\sqrt{3}\)
4) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2+2.2.\sqrt{5}+2^2}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\sqrt{5}-2=-4\)
M.n giúp mk giải bài này ms:
Giải pt: \(\left(x^2-5x+1\right)\left(x^2-4\right)=6\left(x-1\right)^2\)
PT đã cho \(\Leftrightarrow\left(x^2-4-5x+5\right)\left(x^2-4\right)=6\left(x-1\right)^{2
}\)
\(\Leftrightarrow\left(x^2-4-5\left(x-1\right)\right)\left(x^2-4\right)=6\left(x-1\right)^2\)(*)
ĐẶt \(x^2-4=a.\)\(x-1=b\)
PT(*) có dạng \(\left(a-5b\right)a=6b^2\Leftrightarrow a^2-5ab-6b^2=0\Leftrightarrow\left(a+b\right)\left(a-6b\right)=0\)
\(\cdot a+b=0\Leftrightarrow x^2-4+x-1=0\Leftrightarrow x^2+x-5=0\)
\(\Rightarrow x_1=\frac{-1+\sqrt{21}}{2}.x_2=\frac{-1-\sqrt{21}}{2}\)
\(.a-6b=0\Leftrightarrow x^2-4-6\left(x-1\right)=0\Leftrightarrow x^2-6x+2=0\)
\(\Rightarrow x_3=3+\sqrt{7}.x_4=3-\sqrt{7}\)
THử lại: các nghiệm trên đều thỏa mãn pt
Vậy :....
p/s : học khuya thế ==ơ
Bạn ghi lãi đề nhá!
\(x^2-5x+36=8\sqrt{3x+4}\)
tìm x