Chương I : Số hữu tỉ. Số thực

a) 2, 333 - 5, 014 +3, 125 = 0,444

b) ( 8, 21 - 3, 056 + 1,2 ).3,111 = 19.767294

c) (1,5 + 3,556 ):(2,1 -3,2)= -4, 59(63)

Bạn tự vẽ hình nhé!

a) Xét \(\Delta\)ABM và \(\Delta\)CNM, ta có:

AM=MC (gt)

\(\widehat{M_1}=\widehat{M_2}\) (đối đỉnh)

BM=MN (gt)

\(\Rightarrow\) \(\Delta\)ABM = \(\Delta\)CNM (c-g-c)

\(\Rightarrow\) AB=CN (2 cạnh tương ứng)

\(\Rightarrow\)\(\widehat{CAB}\) = \(\widehat{ACN}\) (2 góc tương ứng)

b) Ta có:

Dựa vào dãy tỉ số bằng nhau mà làm !

\(\dfrac{x}{8}-\dfrac{1}{y}=\dfrac{1}{4}\)

\(\Rightarrow\dfrac{1}{y}=\dfrac{x}{8}-\dfrac{1}{4}=\dfrac{x-2}{8}\)

\(\Rightarrow y\left(x-2\right)=8\)

Ta có bảng sau:

...........................(tự lập)

Ta có: \(x-\dfrac{1}{4}< -2\Rightarrow x< -\dfrac{3}{4}\left(1\right)\)

\(-2< x\left(2\right)\)

Từ (1); (2), Suy ra:

\(-2< x< -\dfrac{3}{4}\)

Mà x là sô nguyên nên

=> x = -1

b, \(-3< x+0,5\Rightarrow-3,5< x\left(1\right)\)

\(x< -3\left(2\right)\)

Từ (1); (2) Suy ra:

\(-3,5< x< -3\)

vậy không có giá trị nguyên nào thỏa mãn

S = 2+4+6+8+...+2n = 210

= 2(1+2+3+4+...+n) = 210

= 1+2+3+4+...+n = 210:2

= 1+2+3+4+...+n = 105

Số số hạng của S là : (2n-2) : 2 +1 = n ( số hạng )

Tổng S = n(n+1) : 2

=> n(n+1) : 2 = 105

=> n(n+1) = 105.2

=> n(n+1) = 210

=> n(n+1) = 14.15

=> n = 14

1

Ta có:

\(\dfrac{a}{b}=\dfrac{a.\left(b+2017\right)}{b.\left(b+2017\right)}=\dfrac{a.b+a.2017}{b.\left(b+2017\right)}\) (1)

\(\dfrac{a+2017}{b+2017}=\dfrac{b.\left(a+2017\right)}{b.\left(b+2017\right)}=\dfrac{a.b+b.2017}{b.\left(b+2017\right)}\) (2)

Từ (1) và (2) suy ra:

+) Nếu a >b thì \(\dfrac{a.b+a.2017}{b.\left(b+2017\right)}>\dfrac{b.a+b.2017}{b.\left(b+2017\right)}\Rightarrow\dfrac{a}{b}>\dfrac{a+2017}{b+2017}\)

+) Nếu a <b thì \(\dfrac{a.b+a.2017}{b.\left(b+2017\right)}< \dfrac{a.b+b.2017}{b.\left(b+2017\right)}\) \(\Rightarrow\dfrac{a}{b}< \dfrac{a+2017}{b+2017}\)

+) Nếu a =b thì \(\dfrac{a.b+a.2017}{b.\left(b+2017\right)}=\dfrac{b.a+b.2017}{b.\left(b+2017\right)}\Rightarrow\dfrac{a}{b}=\dfrac{a+2017}{b+2017}\)

a,

b, \(\dfrac{a+5}{a+3}=\dfrac{2}{5}\Leftrightarrow\left(a+5\right)5=\left(a+3\right)2\Leftrightarrow5a+25=2a+6\)

\(\Rightarrow5a-2a=-25+6\Leftrightarrow3a=-19\Leftrightarrow a=-\dfrac{19}{3}\)

c, thay vào tính thôi

d, \(B=\dfrac{a+5}{a+3}=\dfrac{a+3+2}{a+3}=1+\dfrac{2}{a+3}\)

Vì 1 là số nguyên nên để B nguyên khi:

\(a+3\in\left\{+1;-1;+2;-2\right\}\)

\(\Rightarrow a\in\left\{-2;-4;-1;-5\right\}\)

e, Ta có B < 0, khi

\(\Rightarrow\left[{}\begin{matrix}a+5< 0\\a+3< 0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a< -5\\a< -3\end{matrix}\right.\)

Vậy B < 0; khi a < -3

Ta có : \(37^{1320}=\left(37^2\right)^{660}\)\(=1369^{660}\).

Lại có: \(11^{1979}< 11^{1980}=\left(11^3\right)^{660}=1331^{660}\).

Vì \(1369^{660}>1331^{660}\Rightarrow37^{1320}>11^{1980}>11^{1979}\)

Vậy \(11^{1979}< 37^{1320}\).

\(\dfrac{8^5\cdot\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{2^{15}\cdot5^8-2^5\cdot10^9}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{\left(2^{10}\cdot5^{18}-10^9\right)\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{\left(2^2\cdot10^8-10^9\right)\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{\left(2^2-10\right)\cdot10^8\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{\left(4-10\right)\cdot10^8\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{-6\cdot10^8\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{-3\cdot2\cdot10^8\cdot2^5}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{-3\cdot2^6\cdot10^8}{2^{16}\cdot5^7+20^8}\)

\(=\dfrac{-3\cdot64\cdot10^8}{2^{16}\cdot5^7+20^8}\)

\(\dfrac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}=\dfrac{\left(2^3\right)^5.5^8+\left(-2\right).2^4.2^9.5^9}{2^{16}.5^7+\left(2^2\right)^8.5^8}\) \(=-\dfrac{2^{15}.5^8+2^{14}.5^9}{2^{16}.5^7+2^{16}.5^8}=-\dfrac{2^{14}.5^8.\left(2+5\right)}{2^{16}.5^7.\left(1+5\right)}=-\dfrac{5.7}{4.6}=-\dfrac{35}{24}\)