(-1)=(-3,2)

k nhé 

mình k lại!

\(\frac{-9}{5}+0,8=-1,8+0,8=-1\)

\(\text{5 - 9 + 0,8}\)

\(\text{= -4 + 0,8}\)

\(\text{= -3,2}\)

\(\frac{2}{1.4.7}+\frac{2}{4.7.10}+...+\frac{2}{58.61.64}\)

\(=\frac{1}{3}\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{58.61.64}\right)\)

\(=\frac{1}{3}\left(\frac{7-1}{1.4.7}+\frac{10-4}{4.7.10}+...+\frac{64-58}{58.61.64}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{58.61}-\frac{1}{61.64}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.4}-\frac{1}{61.64}\right)\)

\(=\frac{325}{3904}\)

giup mik gap voi :((((((((((((

a) Xét tam giác AEQ và tam giác BEC có
EQ=EC
AEQ=BEC đối đỉnh
EA=EB
=> tam giác AEQ = tam giác BEC(c.g.g).
=> AQ=BC(cạnh tuognư ứng). (1)
Xét Tam giác AFP và tam giác CFB có
AF=CF
AFP=CFB đối đỉnh
FB=FP
=> tam giác AFB = tam giác CFB(c.g.c)
=> AP = BC (2)
từ (1) và (2) suy ra AP=AQ.

\(\frac{2}{1.4.7}+\frac{2}{4.7.10}+...+\frac{2}{58.61.64}\)

\(=\frac{1}{3}.\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{58.61.64}\right)\)

\(=\frac{1}{3}.\left(\frac{7 - 1}{1.4.7}+\frac{10 - 4}{4.7.10}+...+\frac{64 - 58}{58.61.64}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{58.61}-\frac{1}{61.64}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{61.64}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{3904}\right)=\frac{1}{3}.\frac{975}{3904}=\frac{325}{3904}\)

\(\text{Giải :}\)

\(\frac{2}{1.4.7}+\frac{2}{4.7.10}+...+\frac{2}{58.61.64}=\frac{1}{3}.\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{58.61.64}\right)\)

\(=\frac{1}{3}.\left(\frac{7-1}{1.4.7}+\frac{10-4}{4.7.10}+...+\frac{64-58}{58.61.64}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{58.61}-\frac{1}{61.64}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{61.64}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{3904}\right)=\frac{1}{3}.\frac{975}{3904}=\frac{325}{3904}\)

\(\text{#Hok tốt!}\)

\(\frac{2}{3x}\) hay \(\frac{2}{3}\) .x

Đặt A = 3¹⁹⁹⁴ + 3¹⁹⁹³ - 3¹⁹⁹²

= 3¹⁹⁹²(3² + 3 - 1) 

= 3¹⁹⁹² . 11 \(⋮\)11

=> A \(⋮\)11

\(a=\left(2x-\frac{1}{30}\right)^2-2\ge0-2=-2\)(vì \(\left(2x-\frac{1}{30}\right)^2\ge0\))

Dấu \(=\)khi \(2x-\frac{1}{30}=0\Leftrightarrow x=\frac{1}{60}\)

Vậy giá trị nhỏ nhất của \(a\)là \(-2\)đạt tại \(x=\frac{1}{60}\).

Gọi số tờ tiền loại 2000đ là a, 5000đ là b, 10000đ là c => a + b + c =16

Ta có: a.2000 = b.5000 = c.10000

⇒a:12000=b:15000=c:110000⇒a:12000=b:15000=c:110000

⇒a12000=b15000=c110000⇒a12000=b15000=c110000

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

a12000=b15000=c110000=a+b+c12000+15000+110000=1611250=20000a12000=b15000=c110000=a+b+c12000+15000+110000=1611250=20000

⇒a12000=20000⇒a=20000.12000=10⇒a12000=20000⇒a=20000.12000=10

b15000=20000⇒b=20000.15000=4b15000=20000⇒b=20000.15000=4

c110000=20000⇒c=20000.110000=2c110000=20000⇒c=20000.110000=2

Vậy ...

Chúc cậu hok tốt!

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)

\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)

\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)

\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)

=> -1,44444444444........... ≤ x ≤ 0,6111111111...........

Mà x ∈ Z

=> x ∈ { -1 ; 0 }

\(x\in\varnothing\)