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3.
Ta có: \(VT=\)\(8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}\)
\(=8+8+\left(2\sqrt{10+2\sqrt{5}}-2\sqrt{10+2\sqrt{5}}\right)\)
\(=16\ne VP\)
⇒ Đề sai
1. Ta có: \(\sqrt{4x}\)- 3\(\sqrt{x}\)+2\(\sqrt{15x}\)=18
⇌2\(\sqrt{x}\)-3\(\sqrt{x}\) +2\(\sqrt{15x}\)=18
⇌\(-\sqrt{x}\) +2\(\sqrt{15x}\)-15 = 3
⇌-(\(\sqrt{x}\) -2\(\sqrt{15x}\)+15 )=3
⇌(\(\sqrt{x}\)-\(\sqrt{15}\))=-3 (vô lí)
Vậy không tìm được giá trị x thỏa mãn bài toán
2.Ta có: B=\(\dfrac{1}{\sqrt{11-2\sqrt{30}}}-\dfrac{3}{7-2\sqrt{10}}\)
= \(\dfrac{1}{\sqrt{6-2\sqrt{6.5}+5}}-\dfrac{3}{2-2\sqrt{2.5}+5}\)
=\(\dfrac{1}{\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}-\dfrac{3}{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
=\(\dfrac{1}{\sqrt{6}-\sqrt{5}}-\dfrac{3}{\sqrt{3}-\sqrt{2}}\)
hình như đề sai
e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\) ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))
\(=\sqrt{\left(2+\sqrt{7}+2-\sqrt{7}\right)\left(2+\sqrt{7}-2+\sqrt{7}\right)}\)
\(=\sqrt{4\cdot\sqrt{7}}\)
\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}=\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2=\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(=\left(3\sqrt{5}-5\sqrt{2}+5\sqrt{2}+3\sqrt{5}\right)\left(3\sqrt{5}-5\sqrt{2}-5\sqrt{2}-3\sqrt{5}\right)\)
\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)
\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}\)
cuối lười tính nên thôi nhá :>
\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right)\sqrt{5}-\left(3\sqrt{\frac{1}{10}}+10\right)=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\frac{3\sqrt{10}}{10}-10\)
\(=-3\sqrt{10}+10-\frac{3\sqrt{10}}{10}-10=-3\sqrt{10}-\frac{3\sqrt{10}}{10}=-3\sqrt{10}\left(1+\frac{1}{10}\right)=\frac{-33\sqrt{10}}{10}=-3,3\sqrt{10}\)
a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)
b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)
\(=-60-144\sqrt{2}+30\sqrt{2}+144\)
\(=84-114\sqrt{2}\)
a) \(\Leftrightarrow\left(\sqrt{3+\sqrt{5}}\right)^2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
\(\Leftrightarrow\sqrt{3+\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\sqrt{\left(3-\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)}=8\)
\(\Leftrightarrow\sqrt{\frac{6+2\sqrt{5}}{2}}\cdot\left(\sqrt{5}\sqrt{2}-\sqrt{2}\right)\sqrt{3^2-5}=8\).
\(\Leftrightarrow\sqrt{\frac{5+2\sqrt{5}+1}{2}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{4}=8\)
\(\Leftrightarrow\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\cdot2=8\)
\(\Leftrightarrow\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\Leftrightarrow\left(\sqrt{5}\right)^2-1=4\Leftrightarrow5-1=4\)Đúng -ĐPCM.
a. Sửa đề: \(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
biến đổi vế trái :
ta có :\(\left(3+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3^2-\left(\sqrt{5}\right)^2}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
=2(\(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\))
=2(\(\sqrt{5}+5-\sqrt{5}-1\))
=2.4=8=VP
=> đpcm
b. Đặt vế trái là A
ta có \(A^2=\sqrt{2}+1-2\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\sqrt{2}-1\)
=\(2\sqrt{2}-2\)
=2\(\left(\sqrt{2}-1\right)\)
=> A=\(\sqrt{2\left(\sqrt{2}-1\right)}\)
vậy VT=VP =>đpcm
ghi đề sai rồi