1: \(\frac{1}{4+2\sqrt2}+\frac{1}{4-2\sqrt2}\)

\(=\frac{4-2\sqrt2+4+2\sqrt2}{\left(4-2\sqrt2\right)\left(4+2\sqrt2\right)}=\frac{8}{16-8}=1\)

2: \(\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}-\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\)

\(=\frac{\left(\sqrt5+\sqrt3\right)^2-\left(\sqrt5-\sqrt3\right)^2}{\left(\sqrt5+\sqrt3\right)\left(\sqrt5-\sqrt3\right)}=\frac{8+2\sqrt{15}-\left(8-2\sqrt{15}\right)}{5-3}\)

\(=\frac{4\sqrt{15}}{2}=2\sqrt{15}\)

3: \(\left(\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}+1\right):\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}=\frac{\sqrt5-\sqrt3+\sqrt5+\sqrt3}{\sqrt5+\sqrt3}\cdot\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\)

\(=\frac{2\sqrt5}{\sqrt5-\sqrt3}=\frac{2\sqrt5\left(\sqrt5+\sqrt3\right)}{5-3}=\sqrt5\left(\sqrt5+\sqrt3\right)=5+\sqrt{15}\)

4: \(\sqrt{\left(1-\sqrt3\right)^2}+\frac{1}{\sqrt{\left(1+\sqrt3\right)^2}}\)

\(=\sqrt3-1+\frac{1}{\sqrt3+1}=\frac{\left(\sqrt3-1\right)\left(\sqrt3+1\right)+1}{\sqrt3+1}=\frac{3-1+1}{\sqrt3+1}\)

\(=\frac{3\left(\sqrt3-1\right)}{2}\)

5: \(\frac{3}{\sqrt7-2}-\frac{3}{\sqrt7+2}=\frac{3\left(\sqrt7+2\right)-3\left(\sqrt7-2\right)}{\left(\sqrt7-2\right)\left(\sqrt7+2\right)}\)

\(=\frac{3\sqrt7+6-3\sqrt7+6}{7-4}=\frac{12}{3}=4\)

6: \(\frac{5\sqrt2-2\sqrt5}{\sqrt5-\sqrt2}-\frac{9}{\sqrt{10}+1}\)

\(=\frac{\sqrt{10}\left(\sqrt5-\sqrt2\right)}{\sqrt5-\sqrt2}-\frac{9\left(\sqrt{10}-1\right)}{10-1}=\sqrt{10}-\left(\sqrt{10}-1\right)=1\)

7: \(\frac{2\sqrt3}{\sqrt3+\sqrt2}+\sqrt{24}\)

\(=\frac{2\sqrt3\left(\sqrt3-\sqrt2\right)}{\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)}+2\sqrt6\)

\(=2\sqrt3\left(\sqrt3-\sqrt2\right)+2\sqrt6=6-2\sqrt6+2\sqrt6=6\)

8: \(\sqrt3\left(2\sqrt{27}-\sqrt{75}+\frac32\cdot\sqrt{12}\right)\)

\(=\sqrt3\left(2\cdot3\sqrt3-5\sqrt3+\frac32\cdot2\sqrt3\right)=\sqrt3\left(6\sqrt3-5\sqrt3+3\sqrt3\right)=\sqrt3\cdot4\sqrt3=12\)

9: \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt3-1}=\frac{\sqrt{\left(\sqrt6-\sqrt2\right)^2}}{\sqrt3-1}=\frac{\sqrt2\left(\sqrt3-1\right)}{\sqrt3-1}=\sqrt2\)

b) \(B=\sqrt{12+2\sqrt{35}}=\sqrt{12+2.\sqrt{7}.\sqrt{5}}=\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}=\left|\sqrt{7}+\sqrt{5}\right|\)

Vì \(\sqrt{7}>\sqrt{5}\) nên \(\left|\sqrt{7}+\sqrt{5}\right|=\sqrt{7}+\sqrt{5}\)

d) \(D=\sqrt{6-\sqrt{35}}+\sqrt{10}=\sqrt{2}.\sqrt{6-\sqrt{35}}+\sqrt{5}=\sqrt{12-\sqrt{35}}+\sqrt{5}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}+\sqrt{5}=\left|\sqrt{7}-\sqrt{5}\right|+\sqrt{5}=\sqrt{7}-\sqrt{5}+\sqrt{5}=\sqrt{7}\)

Bạn cần bài nào?

 Vì ∠MNO = 90° và ∠MNy' = 30° suy ra ∠ONy' = 60°. Vẽ Ox’ là tia đối của tia Ox. Khi đó, góc NOx’ kề bù với góc Nox, do đó ∠NOx' = 60°. Từ đó, x’Ox song song với yNy’ vì có một cặp góc đồng vị bằng nhau (cùng bằng 60°).

8-(-12)+10=-(-14)-x

8+12+10=14-x

x=14-8-12-10

x=-16

đáp án là -16

lớp mấy bạn ơi

Lớp 6 bạn nhé!

B= 1.99+2.98+2.97+...98.2+99.1

=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)

=1.99+2.99-1.2+3.99-2.3+...+98.99-97.98+99.99-98.99

=(1.99+2.99+3.99+...+98.99+99.99)-(1.2+2.3+3.4+...+97.98+98.99)

=99.(1+2+3+...+98+99)-(1.2+2.3+3.4+...+97.98+98.99)

=99.4950-(1.2+2.3+3.4+...+97.98+98.99)

=490050-(1.2+2.3+3.4+...+97.98+98.99)

Đặt C=1.2+2.3+3.4+...+97.98+98.99

=> 3C=1.2.3+2.3.3+3.4.3+...+97.98.3+98.99.3

=1.2.3+2.3.(4-1)+...+98.99.(100-97)

=1.2.3+2.3.4-1.2.3+...+98.99.100-97.98.99

=98.99.100

=> A=(98.99.100):3=323400

Vậy B=490050-323400=166650

=1.99+2.(99-1)+3.(99-2)+4.(99-3)+......+99.(99-98)

=99.(1+2+3+.......+99)-(2+2.3+3.4+........+98.99)

=99.(1+99).99:2-98.99.100:3

=99.50.99-98.33.100

=490050-323400=166650