Ta có : \(\hept{\begin{cases}\left|x-2007\right|\ge0\forall x\\\left|y-2008\right|\ge0\forall y\end{cases}}\Rightarrow\left|x-2007\right|+\left|y+2008\right|\ge0\forall x;y\)

Kết hợp với đề bài => đẳng thức chỉ thỏa mãn khi |x - 2007| + |y - 2008| = 0

=> \(\hept{\begin{cases}x-2007=0\\y-2008=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2007\\y=2008\end{cases}}\)

Vậy x = 2007 ; y = 2008

Vì | x - 2007 | lớn hơn hoặc bằng 0 với mọi x

| y - 2008 | lớn hơn hoặc bằng 0 với mọi y

=>  | x - 2007 | + | y - 2008 | lớn hơn hoặc bằng 0 với mọi x ; y

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-2007\right|=0\\\left|y-2008\right|=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2007\\y=2008\end{cases}}\)

a) \(\frac{-6}{21}.\frac{3}{2}=-\frac{3}{7}\)          b) \(\left(-3\right).\left(\frac{-7}{12}\right)=\frac{21}{12}=\frac{7}{4}\)

c) \(\left(\frac{11}{12}:\frac{33}{16}\right).\frac{3}{5}=\frac{11}{12}.\frac{16}{33}.\frac{3}{5}=\frac{4}{15}\)

d) \(\sqrt{\left(-7\right)^2}+\sqrt{\frac{2}{16}}=7+\sqrt{\frac{1}{8}}\)

c) \(\frac{1}{2}.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0=\frac{1}{2}.10-\frac{1}{4}+1=5\frac{3}{4}\)

U(45)= 5 .3 . 3                                          

U (204) = 2. 2 . 3. 17

U(126) = 2. 3. 3. 7

UCLN ( 45,204,126) = 2.3 = 6

BCNN ( 45,204,126 ) =3.3.2.2.5.17.7 = 21420

Vậy UCLN = 6 , BCNN = 21420

a. 3/7+(-5/2)+(-3/5)

= 3/7-5/2-3/5

=30/70-175/70-42/70

=-145/70-42/70

=-187/70 

b. -8/18-15/27

= -4/9-15/27

= -12/27-15/27

= -27/27

= -1 

c. 4/5-(-2/7)-7/10

=4/5+2/7-7/10

=56/70+20/70-49/70

=76/70-49/70

=27/70

d. 3,5-(-2/7)

=7/2-(-2/7) 

=7/2+2/7

=49/14+4/14

=53/114

a. \(\left(\frac{2}{3}\right)^3-\left(\frac{3}{4}\right)^2.\left(-1\right)^5=\frac{8}{27}-\frac{9}{16}.\left(-1\right)=\frac{8}{27}+\frac{9}{16}=\frac{371}{432}\)

b. \(12:\left(\frac{3}{4}-\frac{5}{6}\right)^2=12:\left(-\frac{1}{12}\right)^2=12:\frac{1}{144}=12.144=1728\)

c. \(\frac{7}{22}:\frac{3}{11}+\frac{7}{22}:\frac{4}{11}=\frac{7}{22}.\frac{11}{3}+\frac{7}{22}.\frac{11}{4}=\frac{7}{22}\left(\frac{11}{3}+\frac{11}{4}\right)\)

\(=\frac{7}{22}.\frac{77}{12}=\frac{49}{24}\)

d. \(\frac{12}{35}\left(\frac{7}{4}+\frac{13}{4}\right)-\frac{1}{3}=\frac{12}{35}.5-\frac{1}{3}=\frac{12}{7}-\frac{1}{3}=\frac{29}{21}\)

HAI CÂU KHÁC NHAU NHA

1 + (-3) + 5 + (-7) + ... + 17 + (-19)

= [1 + (-3)] + [5 + (-7)] + ... + [17 + (-19)]   (có 10 cặp)

= (-2) + (-2) + ... + (-2)

= -2 . 10 = -20

-2 + 4 + (-6) + 8 + ... + (-18) + 20

= (-2 + 4) + (-6 + 8) + ... + (-18 + 20)   (có 10 cặp)

= -2 + (-2) + ... + (-2)

= -2 . 10 = -20

a. \(\left(2-\frac{3}{4}\right)^2:\frac{11}{16}=\frac{5}{4}^2.\frac{11}{16}=\frac{25}{16}.\frac{16}{11}=\frac{25}{11}\)

b. \(2^3.\frac{7}{20}+\frac{7}{10}=8.\frac{7}{20}+\frac{7}{10}=\frac{14}{5}+\frac{7}{10}=\frac{7}{2}\)

c. \(\sqrt{3^2+4^2}-\sqrt{1^3+2^3+3^3}=\sqrt{9+16}-\sqrt{1+8+27}\)

\(=\sqrt{25}-\sqrt{36}=5-6=-1\)

d. \(21^3:\left(-7\right)^3=\left(21:\left(-7\right)\right)^3=-3^3=-27\)

a) \(\left(2-\frac{3}{4}\right)^2\div\frac{11}{16}=\left(\frac{5}{4}\right)^2.\frac{16}{11}=\frac{25}{16}.\frac{16}{11}=\frac{25}{11}\)

b) \(2^3.\frac{7}{20}+\frac{7}{10}=8.\frac{7}{20}+\frac{7}{10}=\frac{14}{5}+\frac{7}{10}=\frac{7}{2}\)

c) \(\sqrt{3^2+4^2}-\sqrt{1^3+2^3+3^3}=\sqrt{9+16}-\sqrt{1+8+27}\)

\(=\sqrt{25}-\sqrt{36}=5-6=-1\)

d) \(\frac{21^3}{\left(-7\right)^3}=\frac{9261}{-343}=-27\)

Bài 1 : em chưa học 

Bài 2 : \(A=a+\left(42-70+18\right)-\left(42+18+a\right)\)

\(=a-10-60-a=-70\)

Phải trả lời đúng 2 câu mới được nha !