CMR: \(\dfrac{\sqrt{4+\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{2}}=1\)
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6 tháng 7 2021
1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)
2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{\dfrac{1}{2}\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{2}\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\dfrac{\sqrt{2}}{2}\left|\sqrt{3}-\sqrt{5}\right|+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left|\sqrt{5}-1\right|\)
\(=\dfrac{\sqrt{2}}{2}\left(\sqrt{5}-\sqrt{3}\right)+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)
3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)
\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\left|\sqrt{3}-1\right|}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)
6 tháng 7 2021
3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=2\sqrt{5}-2\left(\sqrt{5}+1\right)\)
=-2
4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
21 tháng 12 2023
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
H9
HT.Phong (9A5)
CTVHS
1 tháng 10 2023
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
16 tháng 12 2022
a: \(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=12\sqrt{2x}\)
b: \(=6-4\sqrt{3}+4\sqrt{3}-8=-2\)
c: \(=\sqrt{2}+1+2-\sqrt{2}=3\)
d: \(=\dfrac{1}{\sqrt{2}}\cdot\left(\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}+2\sqrt{3}\right)=0\)
f: \(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}=-6\sqrt{6}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
a) \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}\cdot1+1^2}+\left|\sqrt{2}-2\right|\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}-\left(\sqrt{2}-2\right)\)
\(=\left|\sqrt{2}+1\right|-\sqrt{2}+2\)
\(=\sqrt{2}+1-\sqrt{2}+2\)
\(=3\)
b) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)
\(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{\dfrac{30}{15}}+\sqrt{\dfrac{144}{6}}\)
\(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}\)
\(=-8\sqrt{6}+2\sqrt{6}\)
\(=-6\sqrt{6}\)
c) \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)
\(=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}-2\right]\left[\dfrac{4\left(1-\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+4\right]\)
\(=\left(\sqrt{5}-1-2\right)\left(\dfrac{4\left(1-\sqrt{5}\right)}{1-5}+4\right)\)
\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}-1+4\right)\)
\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)\)
\(=\left(\sqrt{5}\right)^2-3^2\)
\(=-4\)
1 tháng 9 2023
a) \(\sqrt[]{3+2\sqrt[]{2}}+\sqrt[]{\left(\sqrt[]{2}-2\right)^2}\)
\(=\sqrt[]{2+2\sqrt[]{2}.1+1}+\left|\sqrt[]{2}-2\right|\)
\(=\sqrt[]{\left(\sqrt[]{2}+1\right)^2}+\left(2-\sqrt[]{2}\right)\) \(\left(\left(\sqrt[]{2}\right)^2=2< 2^2=4\right)\)
\(=\left|\sqrt[]{2}+1\right|+2-\sqrt[]{2}\)
\(=\sqrt[]{2}+1+2-\sqrt[]{2}\)
\(=3\)
28 tháng 4 2018
a. \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
= \(\sqrt{3-2\sqrt{15}+5}-\sqrt{3+2\sqrt{15}+5}\)
= \(\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\)
= \(\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{5}\)
= \(-2\sqrt{3}\)
b. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
= \(\dfrac{\left(\sqrt{15}-\sqrt{5}\right).\left(\sqrt{3}+1\right)}{2}+\dfrac{\left(5-2\sqrt{5}\right).\left(2\sqrt{5}+4\right)}{4}\)
=\(\dfrac{\sqrt{45}+\sqrt{15}-\sqrt{15}-\sqrt{5}}{2}+\dfrac{\left(5-2\sqrt{5}\right).2\left(\sqrt{5}+2\right)}{4}\)
= \(\dfrac{3\sqrt{5}-\sqrt{5}}{2}+\dfrac{\left(5-2\sqrt{5}\right).\left(\sqrt{5}+2\right)}{2}\)
= \(\dfrac{2\sqrt{5}}{2}+\dfrac{5\sqrt{5}+10-10-4\sqrt{5}}{2}\)
= \(\sqrt{5}+\dfrac{\sqrt{5}}{2}\)
= \(\dfrac{3\sqrt{5}}{2}\)
c. \(\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}+\dfrac{1}{\sqrt{5}+\sqrt{2}}\right):\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)
= \(\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right).\left(\sqrt{5}+\sqrt{2}\right)}.\left(\sqrt{2}+1\right)^2\)
= \(\dfrac{2\sqrt{5}}{3}.\left(2+2\sqrt{2}+1\right)\)
= \(\dfrac{2\sqrt{5}}{3}.\left(3+2\sqrt{2}\right)\)
= \(\dfrac{6\sqrt{5}+4\sqrt{10}}{3}\)
d. \(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\left(\sqrt{3}+1-3\left(\sqrt{3}+2\right)+\dfrac{5\left(3+\sqrt{3}\right)}{2}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{15+5\sqrt{3}}{2}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\left(-2\sqrt{3}-5+\dfrac{15+5\sqrt{3}}{2}\right).\dfrac{1}{\sqrt{3}+5}\)
= \(\dfrac{-4\sqrt{3}-10+15+5\sqrt{3}}{2}.\dfrac{1}{\sqrt{3}+5}\)
= \(\dfrac{\sqrt{3}+5}{2}.\dfrac{1}{\sqrt{3}+5}\)
= \(\dfrac{1}{2}\)
Nếu đúng cho 1 like nhé!
Đề bài đúng: \(\dfrac{\sqrt{4-\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{2}}=1\)
Hoặc: \(\dfrac{\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}}=1\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)}{2}=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}=\dfrac{5-3}{2}=1\)