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\(\left(2\sqrt{5}-\sqrt{3}\right)\left(23+2\sqrt{15}\right)-\left(2\sqrt{5}+\sqrt{3}\right)\left(23-2\sqrt{15}\right)\)
\(=40\sqrt{5}-3\sqrt{3}-40\sqrt{5}+3\sqrt{3}\)
\(=0\)
a: Ta có: \(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\sqrt{3}-1-2-\sqrt{3}\)
=-3
b: Ta có: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
=1
c: Ta có: \(C=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
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)
a: Ta có: \(\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
b: Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{2}}\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
=9-2
=7
c: Ta có: \(\left(\sqrt{7}+\sqrt{5}\right)\cdot\sqrt{12-2\sqrt{35}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=2
Bạn nhóm hai thừa số đầu, hai thừa số cuối, sau đó áp dụng hằng đẳng thức số 3 ta dc
A=[(√2+√3)^2-5][5-(√2-√3)^2]
= Bạn khai triển các bình phương ra, thu gọn, kết quả là 24
Bạn tách các bt trong ngoặc ra về các hằng đẳng thức số1, 2 như này
8-2√15= 3-2.√3.√5+5=
(√3+√5)^2.
Nhưng căn tiếp theo phải là 21-4√5 thì mới ra được là (2√5-1)^2
\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)
\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(C=\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)
\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)
\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}\)
C=\(\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)
a.\(\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)
b.\(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
c.\(\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{2}=\left|\sqrt{2}+5\right|-\sqrt{2}=\sqrt{2}+5-\sqrt{2}=5\)
d.\(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-6\right)^2}=\left|3+\sqrt{5}\right|+\left|\sqrt{5}-6\right|=3+\sqrt{5}+6-\sqrt{5}=9\)
Sửa đề
\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\left(2+\sqrt{3}\right)\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)
a/ \(=\left(7+4\sqrt{3}+3\left(7-4\sqrt{3}\right)\right)\left(7+2\sqrt{3}\right)\)
\(=\left(28-8\sqrt{3}\right)\left(7+2\sqrt{3}\right)\)
\(=4\left(7-2\sqrt{3}\right)\left(7+2\sqrt{3}\right)\)
\(=4\left(49-12\right)=...\)
b/ \(=\left(\frac{\sqrt{15}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}-1\right)}+\frac{2\sqrt{15}}{3}\right).4\sqrt{15}\)
\(=\left(\frac{\sqrt{15}}{3}+\frac{2\sqrt{15}}{3}\right).4\sqrt{15}\)
\(=\sqrt{15}.4\sqrt{15}=4.15=...\)
c/ Bạn coi lại đề
d/ \(\sqrt{23-2\sqrt{112}}+\sqrt{23+2\sqrt{112}}\)
\(=\sqrt{\left(4-\sqrt{7}\right)^2}+\sqrt{\left(4+\sqrt{7}\right)^2}\)
\(=4-\sqrt{7}+4+\sqrt{7}=8\)