a) Ta có:

\(\left\{{}\begin{matrix}AB\perp AC\\KH\perp AC\end{matrix}\right.=>AB//KH\) 

b) Ta có: 

\(\widehat{ABK}=\widehat{BKI}\left(=60^o\right)\)

Mà hai góc này ở vị trí so le trong 

=> AB//KI 

c) AB//HK = > \(\widehat{ABK}+\widehat{HKB}=180^o\)

Mà: \(\widehat{ABK}=\widehat{BKI}\) 

\(=>\widehat{BKI}+\widehat{HKB}=180^o\)

=> \(\widehat{HKI}\) là góc bẹt hay H, K, I thẳng hàng

a) Căc cặp góc so le trong là:

\(\widehat{R_1}\) và \(\widehat{S_3}\)

\(\widehat{R_2}\) và \(\widehat{S_4}\)

Các cặp góc đồng vị là: 

\(\widehat{S_1}\) và \(\widehat{R_1}\)

\(\widehat{S_2}\) và \(\widehat{R_2}\)

\(\widehat{S_3}\) và \(\widehat{R_3}\)

\(\widehat{S_4}\) và \(\widehat{R_4}\)

Các cặp góc trong cùng phía là:

\(\widehat{S_4}\) và \(\widehat{R_1}\)

\(\widehat{S_3}\) và \(\widehat{R_2}\)

b) Ta có: 

\(\widehat{R_2}=120^o=>\widehat{S_2}=120^o\) (đồng vị)

\(\widehat{R_2}=120^o=>\widehat{S_3}=180^o-\widehat{R_2}=180^o-120^o=60^o\) (trong cùng phía) 

\(\widehat{S_4}=120^o=>\widehat{R_4}=120^o\) (đồng vị) 

\(\widehat{S_4}=120^o=>\widehat{R_1}=180^o-\widehat{S_4}=180^o-120^o=60^o\) (trong cùng phía)  

\(\widehat{R_1}=60^o=>\widehat{S_1}=60^o\) (đồng vị)  

\(\widehat{S_3}=60^o=>\widehat{R_3}=60^o\) (đồng vị) 

a) Ta có: 

\(\widehat{MAB}=\widehat{ABC}\left(=55^o\right)\)

Mà hai góc này ở vị trí so le trong 

=> AM//BC 

b) Ta có:
\(\widehat{NAC}=\widehat{ACB}\left(=40^o\right)\)

Mà hai góc này ở vị trí so le trong

=> AN//BC 

c) Xét tam giác ABC có:

\(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^o\\ =>\widehat{BAC}=180^o-\widehat{ABC}-\widehat{ACB}\\ =>\widehat{BAC}=180^o-55^o-40^o=85^o\) 

\(\widehat{MAB}+\widehat{BAC}+\widehat{NAC}=55^o+85^o+40^o=180^o\) 

=> \(\widehat{MAN}\) là góc bẹt => M, A, N thẳng hàng

a) Ta có:

\(\widehat{ADE}=\widehat{ABC}\left(=45^o\right)\)

Mà hai góc này ở vị trí đồng vị

=> DE//BC

b) Ta có: 

\(\widehat{FEC}=\widehat{ECB}\left(gt\right)\)

Mà hai góc này ở vị trí so le trong 

=> EF//BC

c) Ta có: DE//BC

=> \(\widehat{DEC}+\widehat{ECB}=180^o\) (trong cùng phía) 

Mà: \(\widehat{FEC}=\widehat{ECB}\left(gt\right)\)

\(=>\widehat{FEC}+\widehat{ECB}=180^o\)

\(=>\widehat{DEF}\) là góc bẹt

=> D, E, F thẳng hàng

\(e)2^{5x-4}=64\\ \Rightarrow2^{5x-4}=2^6\\ \Rightarrow5x-4=6\\ \Rightarrow5x=6+4=10\\ \Rightarrow x=\dfrac{10}{5}\\ \Rightarrow x=2\\ f)2^{3x+2}=4^{x+6}\\ \Rightarrow2^{3x+2}=\left(2^2\right)^{x+2}\\ \Rightarrow2^{3x+2}=2^{2x+4}\\ \Rightarrow3x+2=2x+4\\ \Rightarrow3x-2x=4-2\\ \Rightarrow x=2\\ g)4^x=5\cdot4^3-4\cdot4^3\\ \Rightarrow4^x=4^3\cdot\left(5-4\right)\\ \Rightarrow4^x=4^3\\ \Rightarrow x=3\\ h)4^{5x-3}=16^{2x-1}\\ \Rightarrow4^{5x-3}=\left(4^2\right)^{2x-2}\\ \Rightarrow4^{5x-3}=4^{4x-4}\\ \Rightarrow5x-3=4x-4\\ \Rightarrow5x-4x=-4+3\\ \Rightarrow x=-1\\ i)5^{7x-2}=5^{3x+10}\\ \Rightarrow7x-2=3x+10\\ \Rightarrow7x-3x=10+2\\ \Rightarrow4x=12\\ \Rightarrow x=12:4\\ \Rightarrow x=3\)

e: \(2^{5x-4}=64\)

=>\(2^{5x-4}=2^6\)

=>5x-4=6

=>5x=10

=>x=10/5=2

f: \(2^{3x+2}=4^{x+6}\)

=>\(2^{3x+2}=2^{2x+12}\)

=>3x+2=2x+12

=>3x-2x=12-2

=>x=10

g: \(4^x=5\cdot4^3-4\cdot4^3\)

=>\(4^x=4^3\)

=>x=3

h: \(4^{5x-3}=16^{2x-1}\)

=>\(4^{5x-3}=\left(4^2\right)^{2x-1}=4^{4x-2}\)

=>5x-3=4x-2

=>5x-4x=-2+3

=>x=1

i: \(5^{7x-2}=5^{3x+10}\)

=>7x-2=3x+10

=>4x=12

=>x=4

l: \(\dfrac{16}{2^x}=2\)

=>\(2^x=\dfrac{16}{2}=8=2^3\)

=>x=3

m: \(\dfrac{\left(-3\right)^x}{81}=-27\)

=>\(\left(-3\right)^x=\left(-3\right)^3\cdot\left(-3\right)^4=\left(-3\right)^7\)

=>x=7

a) Ta có: 

\(VT=\left(a+b\right)^2-4ab=\left(a^2+2ab+b^2\right)-4ab\\ =a^2+2ab+b^2-4ab=a^2-2ab+b^2\\ =\left(a-b\right)^2=VP\)

=> Đpcm 

b) Ta có:

\(VT=\left(a-b\right)^3=\left[-\left(b-a\right)\right]^3=\left[\left(-1\right)\cdot\left(b-a\right)\right]^3\\ =\left(-1\right)^3\left(b-a\right)^3=\left(-1\right)\cdot\left(b-a\right)^3=-\left(b-a\right)^3=VP\)

=> Đpcm  

c) Ta có: 

\(\left(n+2\right)^2-n^2=\left(n^2+4n+4\right)-n^2\\ =n^2+4n+4-n^2=4n+4=4\left(n+1\right)⋮4\forall n\in N\) 

=> Đpcm 

a: \(\left(a+b\right)^2-4ab\)

\(=a^2+2ab+b^2-4ab\)

\(=a^2-2ab+b^2=\left(a-b\right)^2\)

b: \(\left(a-b\right)^3=\left[-\left(b-a\right)\right]^3=-\left(b-a\right)^3\)

c: \(\left(n+2\right)^2-n^2=\left(n+2+n\right)\left(n+2-n\right)\)

\(=2\left(2n+2\right)=4\left(n+1\right)⋮4\)

a: ta có: \(\widehat{MNS}=\widehat{HNQ}\)(hai góc đối đỉnh)

mà \(\widehat{HNQ}=60^0\)

nên \(\widehat{MNS}=60^0\)

b: Ta có: \(\widehat{QNH}=\widehat{PMN}\left(=60^0\right)\)

mà hai góc này là hai góc ở vị trí đồng vị

nên PI//QS

=>MP//NQ

c: ta có: MP//NQ

KP\(\perp\)MP

Do đó: KP\(\perp\)QN

d: ta có: MI//SN

=>\(\widehat{MIS}+\widehat{S}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{S}+100^0=180^0\)

=>\(\widehat{S}=80^0\)

Bài 3:

\(a)\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\\ \Leftrightarrow x^2+3x+2x+6-\left(x^2+5x-2x-10\right)=0\\ \Leftrightarrow x^2+5x+6-x^2-3x+10=0\\ \Leftrightarrow2x+16=0\\ \Leftrightarrow2x=-16\\ \Leftrightarrow x=-\dfrac{16}{2}=-8\\ b)\left(x-3\right)\left(x-2\right)-\left(x+1\right)\left(x-5\right)=0\\ \Leftrightarrow\left(x^2-2x-3x+6\right)-\left(x^2-5x+x-5\right)=0\\ \Leftrightarrow x^2-5x+6-x^2+4x+5=0\\ \Leftrightarrow-x+11=0\\ \Leftrightarrow x=11\\ c)x\left(2x-5\right)-2x\left(x-6\right)=42\\ \Leftrightarrow2x^2-5x-2x^2+12x=42\\ \Leftrightarrow7x=42\\ \Leftrightarrow x=\dfrac{42}{7}\\ \Leftrightarrow x=6\\ d)\left(x-1\right)\left(2x+3\right)-2x\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(2x+3-2x\right)=0\\ \Leftrightarrow3\left(x-1\right)=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)

Bài 2:

b: 

c:

d:

Bài 4:

a: \(A\left(x\right)=x^7-3x^2-x^5+x^4-x^2+2x-7\)

\(=x^7-x^5+x^4+\left(-3x^2-x^2\right)+2x-7\)

\(=x^7-x^5+x^4-4x^2+2x-7\)

\(B\left(x\right)=x-2x^2+x^4-x^5-x^7-4x^2-1\)

\(=-x^7-x^5+x^4+\left(-2x^2-4x^2\right)+x-1\)

\(=-x^7-x^5+x^4-6x^2+x-1\)

b: A(x)+B(x)

\(=x^7-x^5+x^4-4x^2+2x-7-x^7-x^5+x^4-6x^2+x-1\)

\(=-2x^5-10x^2+3x-8\)

A(x)-B(x)

\(=x^7-x^5+x^4-4x^2+2x-7+x^7+x^5-x^4+6x^2-x+1\)

\(=2x^7+2x^2+x-6\)

c: C(x)=A(x)+B(x

=>\(C\left(x\right)=-2x^5-10x^2+3x-8\)

Thay x=-1 vào C(x), ta được:

\(C\left(-1\right)=-2\cdot\left(-1\right)^5-10\cdot\left(-1\right)^2+3\cdot\left(-1\right)-8\)

=2-10-3-8

=-1-10-8=-19

a, Ta có DE vuông AB 

AH vuông AB 

=> DE // AH 

b, Ta có DE // AH => ^BDE = ^ACB ( 2 góc đồng vị ) 

=> ^BDE = ^DCH = 40⁰

c, Ta có DH vuông AC 

AB vuông AC 

=> DH // AB 

Ta có DH // AB; ED//AH ; ^EAH = ^AED = ^AHD = 90⁰

Vậy tứ giác AEDH là hình vuông 

=> DE vuông DH