a: Xét ΔAHB và ΔAHC có

AH chung

HB=HC

AB=AC

Do đó: ΔAHB=ΔAHC

b: ΔAHB=ΔAHC

=>\(\widehat{AHB}=\widehat{AHC}\)

mà \(\widehat{AHB}+\widehat{AHC}=180^0\)(hai góc kề bù)

nên \(\widehat{AHB}=\widehat{AHC}=\dfrac{180^0}{2}=90^0\)

=>AH\(\perp\)BC

c: H là trung điểm của BC

=>\(HB=HC=\dfrac{BC}{2}=3\left(cm\right)\)

ΔAHB vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(HA=\sqrt{5^2-3^2}=4\left(cm\right)\)

d: ΔAHB=ΔAHC

=>\(\widehat{HAB}=\widehat{HAC}\)

Xét ΔAEH vuông tại E và ΔAKH vuông tại K có

AH chung

\(\widehat{EAH}=\widehat{KAH}\)

Do đó: ΔAEH=ΔAKH

=>HE=HK

e: ΔAEH=ΔAKH

=>AE=AK

Xét ΔABC có \(\dfrac{AE}{AB}=\dfrac{AK}{AC}\)

nên EK//BC

Bài 3: Gọi H là giao điểm của CD với AB

\(\widehat{HCB}+\widehat{DCB}=180^0\)(hai góc kề bù)

=>\(\widehat{HCB}+143^0=180^0\)

=>\(\widehat{HCB}=180^0-143^0=37^0\)

Xét ΔHCB có \(\widehat{HCB}+\widehat{HBC}=37^0+53^0=90^0\)

nên ΔHCB vuông tại H

=>CD\(\perp\)AB tại H

Bài 2:

a: Ta có: \(\widehat{DAB}=\widehat{xAM}\)(hai góc đối đỉnh)

mà \(\widehat{xAm}=124^0\)

nên \(\widehat{DAB}=124^0\)

Ta có: \(\widehat{DAB}+\widehat{ABC}=124^0+56^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên AD//BC

=>xy//zt

b: xy//zt

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

=>\(\widehat{BCD}+90^0=180^0\)

=>\(\widehat{BCD}=90^0\)

Ak là phân giác của góc DAB

=>\(\widehat{DAC}=\dfrac{124^0}{2}=62^0\)

ΔDAC vuông tại D

=>\(\widehat{DAC}+\widehat{DCA}=90^0\)

=>\(\widehat{DCA}+62^0=90^0\)

=>\(\widehat{DCA}=28^0\)

\(1,a)\dfrac{15}{12}-\dfrac{-1}{4}\\ =\dfrac{15}{12}+\dfrac{1}{2}\\ =\dfrac{15}{12}+\dfrac{6}{12}\\ =\dfrac{21}{12}=\dfrac{7}{4}\\ b)-\dfrac{5}{12}+0,75\\ =-\dfrac{5}{12}+\dfrac{3}{4}\\ =\dfrac{-5}{12}+\dfrac{9}{12}\\ =\dfrac{4}{12}=\dfrac{1}{3}\\ c)\dfrac{15}{12}+\dfrac{5}{13}-\left(\dfrac{3}{12}+\dfrac{18}{13}\right)\\ =\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}\\ =\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\\ =\dfrac{12}{12}-\dfrac{13}{13}\\ =1-1=0\)

2: a: \(-\dfrac{16}{42}-\dfrac{5}{8}=\dfrac{-64}{168}-\dfrac{105}{168}=\dfrac{-169}{168}\)

b: \(3,5-\left(-\dfrac{2}{7}\right)=3,5+\dfrac{2}{7}=\dfrac{7}{2}+\dfrac{2}{7}=\dfrac{7^2+2^2}{14}=\dfrac{53}{14}\)

c: \(\left(-\dfrac{1}{2}+\dfrac{3}{4}\right)-\left(-\dfrac{4}{5}+\dfrac{5}{6}\right)\)

\(=\dfrac{-1}{2}+\dfrac{3}{4}+\dfrac{4}{5}-\dfrac{5}{6}\)

\(=\dfrac{-30}{60}+\dfrac{45}{60}+\dfrac{48}{60}-\dfrac{50}{60}\)

\(=\dfrac{15}{60}-\dfrac{2}{60}=\dfrac{13}{60}\)

3:

a: \(\dfrac{2}{21}-\dfrac{-1}{28}=\dfrac{2}{21}+\dfrac{1}{28}=\dfrac{8}{84}+\dfrac{3}{84}=\dfrac{11}{84}\)

b: \(-4.75-1\dfrac{7}{12}=-\dfrac{57}{12}-\dfrac{19}{12}=-\dfrac{76}{12}=-\dfrac{19}{3}\)

c: \(-\left(\dfrac{3}{5}+\dfrac{5}{4}\right)-\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\)

\(=-\dfrac{3}{5}-\dfrac{5}{4}+\dfrac{3}{4}-\dfrac{2}{5}\)

\(=-1-\dfrac{2}{4}=-\dfrac{3}{2}\)

4:

a: \(-\dfrac{2}{33}+\dfrac{5}{55}=\dfrac{-10}{165}+\dfrac{15}{165}=\dfrac{5}{165}=\dfrac{1}{33}\)

b: \(0,4+\left(-2\dfrac{4}{5}\right)=0,4-2,8=-2,4\)

c: \(-\left(\dfrac{3}{7}+\dfrac{3}{8}\right)-\left(-\dfrac{3}{8}+\dfrac{4}{7}\right)\)

\(=\dfrac{-3}{7}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{7}\)

\(=-\dfrac{3}{7}-\dfrac{4}{7}=-\dfrac{7}{7}=-1\)

\(10x^2+y^2+4z^2+6x-4y-4xz=-5\\ =>10x^2+y^2+4z^2+6x-4y-4xz+5=0\\ =>\left(9x^2+6x+1\right)+\left(x^2-4xz+4z^2\right)+\left(y^2-4y+4\right)=0\\ =>\left(3x+1\right)^2+\left(x-2z\right)^2+\left(y-2\right)^2=0\)

Mà: \(\left\{{}\begin{matrix}\left(3x+1\right)^2\ge0\forall x\\\left(x-2z\right)^2\ge0\forall x,z\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.=>\left(3x+1\right)^2+\left(x-2z\right)^2+\left(y-2\right)^2\ge0\forall x,y,z\) 

\(=>\left\{{}\begin{matrix}3x+1=0\\x-2z=0\\y-2=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\z=-\dfrac{1}{6}\\y=2\end{matrix}\right.\)

\(10x^2+y^2+4z^2+6x-4y-4xz=-5\\ \Leftrightarrow\left(x^2-4xz+4z^2\right)+\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)=0\\ \Leftrightarrow\left(x-2z\right)^2+\left(3x+1\right)^2+\left(y-2\right)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2z\right)^2\ge0\forall x,z\\\left(3x+1\right)^2\ge0\forall x\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2z\right)^2+\left(3x+1\right)^2+\left(y-2\right)^2\ge0\forall x,y,z\)

Mà: \(\left(x-2z\right)^2+\left(3x+1\right)^2+\left(y-2\right)^2=0\)

Do đó: \(\left\{{}\begin{matrix}x-2z=0\\3x+1=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=2\\z=-\dfrac{1}{6}\end{matrix}\right.\)

#$\mathtt{Toru}$

Bài 3: Gọi H là giao điểm của CD với AB

\(\widehat{HCB}+\widehat{DCB}=180^0\)(hai góc kề bù)

=>\(\widehat{HCB}+143^0=180^0\)

=>\(\widehat{HCB}=180^0-143^0=37^0\)

Xét ΔHCB có \(\widehat{HCB}+\widehat{HBC}=37^0+53^0=90^0\)

nên ΔHCB vuông tại H

=>CD\(\perp\)AB tại H

Bài 2:

a: Ta có: \(\widehat{DAB}=\widehat{xAM}\)(hai góc đối đỉnh)

mà \(\widehat{xAm}=124^0\)

nên \(\widehat{DAB}=124^0\)

Ta có: \(\widehat{DAB}+\widehat{ABC}=124^0+56^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên AD//BC

=>xy//zt

b: xy//zt

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

=>\(\widehat{BCD}+90^0=180^0\)

=>\(\widehat{BCD}=90^0\)

Ak là phân giác của góc DAB

=>\(\widehat{DAC}=\dfrac{124^0}{2}=62^0\)

ΔDAC vuông tại D

=>\(\widehat{DAC}+\widehat{DCA}=90^0\)

=>\(\widehat{DCA}+62^0=90^0\)

=>\(\widehat{DCA}=28^0\)

c;     C = \(\dfrac{28^{28}+28^{24}+...+28^4+1}{28^{30}+28^{28}+...+28^2+1}\)

        A =         1 + 28⁴ + 28⁸ + 28¹² + ...28²⁸

  28⁴A = 28⁴ + 28⁸ + 28¹² + ... + 28²⁸ + 28³²

28⁴A - A = 28⁴+ 28⁸+...+28²⁸+ 28³²- (1 + 28⁴ + 28⁸+...+28²⁸)

(28⁴ - 1)A = 28⁴ + 28⁸+ ...+ 28²⁸ + 28³² - 1 - 28⁴- ...- 28²⁸

(28⁴ - 1)A = (28³² - 1) + (28⁴ - 28⁴) + (28⁸ - 28⁸) + ... + (28²⁸ - 28²⁸)

(28⁴ - 1)A = 28³² - 1 + 0 + 0... + 0

            A = (28³² - 1): (28⁴ - 1)

  Đặt B = 28³⁰ + 28²⁸ + ... + 28² + 1

  28²B = 28³² + 28³⁰ + ... + 28⁴ + 28²

28²B - B = 28³² + 28³⁰ + ... + 28⁴ + 28² - (28³⁰ + 28²⁸ +...+1)

(28² - 1)B = 28³² + 28³⁰+...+28⁴ + 28² - 28³⁰ - 28²⁸ - ... 28²- 1

(28² - 1)B = (28³² - 1) + (28³⁰ - 28³⁰) +...+(28² - 28²)

(28² - 1)B = (28³² - 1) + 0 + 0 +...+ 0

(28² - 1)B = 28³² - 1 

             B = (28³² - 1): (28² - 1)

C = \(\dfrac{A}{B}\) = \(\dfrac{28^{32}-1}{28^4-1}\) : \(\dfrac{28^{32}-1}{28^2-1}\)

C = \(\dfrac{28^{32}-1}{28^4-1}\) \(\times\) \(\dfrac{28^2-1}{28^{32}-1}\)

C = \(\dfrac{28^2-1}{28^4-1}\)

C = \(\dfrac{1}{785}\) 

                Câu d:

 \(\dfrac{x-1}{99}\) + \(\dfrac{x-2}{98}\) + \(\dfrac{x-3}{97}\) = \(\dfrac{x-4}{96}\) + \(\dfrac{x-5}{95}\) + \(\dfrac{x-6}{94}\)

(\(\dfrac{x-1}{99}\)-1)+(\(\dfrac{x-2}{98}\)-1)+(\(\dfrac{x-3}{97}\)-1) = (\(\dfrac{x-4}{96}\)-1) + (\(\dfrac{x-5}{95}\)-1)+(\(\dfrac{x-6}{94}\)-1)

\(\dfrac{x-100}{99}\)+\(\dfrac{x-100}{98}\)+\(\dfrac{x-100}{97}\) = \(\dfrac{x-100}{96}\)+\(\dfrac{x-100}{95}\)+\(\dfrac{x-100}{94}\)

\(\dfrac{x-100}{99}\)+\(\dfrac{x-100}{98}\)+\(\dfrac{x-100}{97}\)- \(\dfrac{x-100}{96}\)-\(\dfrac{x-100}{95}\)-\(\dfrac{x-100}{94}\) = 0

(\(x-100\)).(\(\dfrac{1}{99}\)+\(\dfrac{1}{98}\)+\(\dfrac{1}{97}\) - \(\dfrac{1}{96}\)-\(\dfrac{1}{95}\)-\(\dfrac{1}{94}\)) = 0

Vì\(\dfrac{1}{98}< \dfrac{1}{98}< \dfrac{1}{97}< \dfrac{1}{96}< \dfrac{1}{95}< \dfrac{1}{94}\)

Nên (\(\dfrac{1}{99}\) + \(\dfrac{1}{98}\) + \(\dfrac{1}{97}\) )- (\(\dfrac{1}{96}\) + \(\dfrac{1}{95}\) +\(\dfrac{1}{94}\) )< 0 

⇒\(x-100\) = 0

Vậy \(x\) = 100

\(x\left(2x-3\right)-2\left(3-x^2\right)+1=0\)

=>\(2x^2-3x-6+2x^2+1=0\)

=>\(4x^2-3x-5=0\)

\(\text{Δ}=\left(-3\right)^2-4\cdot4\cdot\left(-5\right)=9+80=89>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}x=\dfrac{3-\sqrt{89}}{2\cdot4}=\dfrac{3-\sqrt{89}}{8}\\x=\dfrac{3+\sqrt{89}}{2\cdot4}=\dfrac{3+\sqrt{89}}{8}\end{matrix}\right.\)

a) \(\lim\limits_{ }\left(\sqrt{n^2-n+1}-n\right)\)

\(=\lim\limits_{ }\left[\dfrac{\left(\sqrt{n^2-n+1}-n\right)\left(\sqrt{n^2-n+1}+n\right)}{\sqrt{n^2-n+1}+n}\right]\)

\(=\lim\limits_{ }\left(\dfrac{1-n}{\sqrt{n^2-n+1}+n}\right)\)

\(=\lim\limits_{ }\left(\dfrac{\dfrac{1}{n}-1}{\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\right)\)

\(=-\dfrac{1}{2}\)

b) \(\lim\limits_{ }\left(\dfrac{-3}{4n^2-2n+1}\right)=0\)

c) \(\lim\limits_{ }\dfrac{n^2+n+5}{2n+1}=+\infty\)

d) \(\lim\limits_{ }\left(\sqrt{n^2-1}-\sqrt{3n^2+2}\right)\)

\(=\lim\limits_{ }\left(\dfrac{-2n^2-3}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)\)

\(\lim\limits_{ }\left(\dfrac{-2n-\dfrac{3}{n}}{\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}}\right)\)

\(=-\infty\)

a: \(lim\left(\sqrt{n^2-n+1}-n\right)\)

\(=\lim\limits\dfrac{n^2-n+1-n^2}{\sqrt{n^2-n+1}+n}=\lim\limits\dfrac{-n+1}{\sqrt{n^2-n+1}+n}\)

\(=\lim\limits\dfrac{-1+\dfrac{1}{n}}{\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}+1}=\dfrac{-1+0}{\sqrt{1-0+0}+1}=\dfrac{-1}{2}\)

b: \(\lim\limits\dfrac{-3}{4n^2-2n+1}\)

\(=\lim\limits\dfrac{-\dfrac{3}{n^2}}{4-\dfrac{2}{n}+\dfrac{1}{n^2}}=\dfrac{0}{4-0+0}=0\)

c: \(\lim\limits\dfrac{n^2+n+5}{2n+1}=\lim\limits\dfrac{n^2\left(1+\dfrac{1}{n}+\dfrac{5}{n^2}\right)}{n\left(2+\dfrac{1}{n}\right)}\)

\(=\lim\limits\dfrac{n\left(1+\dfrac{1}{n}+\dfrac{5}{n^2}\right)}{2+\dfrac{1}{n}}=+\infty\)

d: \(\lim\limits\left(\sqrt{n^2-1}-\sqrt{3n^2+2}\right)\)

\(=\lim\limits\left(\dfrac{n^2-1-3n^2-2}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)=\lim\limits\left(\dfrac{-2n^2-3}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)\)

\(=\lim\limits\left(\dfrac{n^2\left(-2-\dfrac{3}{n^2}\right)}{n\cdot\left(\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}\right)}\right)\)

\(=\lim\limits\left(\dfrac{n\left(-2-\dfrac{3}{n^2}\right)}{\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}}\right)=+\infty\)

Bài 4:

\(a)2,6^2+4\cdot1,3\cdot7,4+7,4^2\\ =2,6^2+2\cdot\left(2\cdot1,3\right)+7,4^2\\ =2,6^2+2\cdot2,6\cdot7,4+7,4^2\\ =\left(2,6+7,4\right)^2\\ =10^2\\ =100\\ b)2024^2-2023^2\\ =\left(2024-2023\right)\left(2024+2023\right)\\ =1\cdot4047\\ =4047\)

Bài 5:

\(a)4x^2+24x+36\\ =\left(2x\right)^2+2\cdot2x\cdot6+6^2\\ =\left(2x+6\right)^2\\ b)9x^4y^2+18x^2y+9\\ =\left(3x^2y\right)^2+2\cdot3x^2y\cdot3+3^2\\ =\left(3x^2y+3\right)^2\)

giải giúp mình bài 5 với các bạn

a: Ta có: \(\widehat{xBy}=\widehat{xAz}\)(hai góc đồng vị)

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

nên By//Az

b: AC là phân giác của góc xAz

=>\(\widehat{xAC}=\widehat{zAC}=\dfrac{\widehat{xAz}}{2}=30^0\)

=>\(\widehat{BAC}=30^0\)

Ta có: \(\widehat{CBA}+\widehat{CBx}=180^0\)(hai góc kề bù)

=>\(\widehat{CBA}+60^0=180^0\)

=>\(\widehat{CBA}=120^0\)

Xét ΔBAC có \(\widehat{BAC}+\widehat{CBA}+\widehat{ACB}=180^0\)

=>\(\widehat{ACB}+30^0+120^0=180^0\)

=>\(\widehat{ACB}=30^0\)

c: BD là phân giác của góc yBA

=>\(\widehat{ABD}=\dfrac{\widehat{yBA}}{2}=60^0\)

Xét ΔBDA có \(\widehat{DBA}+\widehat{DAB}=30^0+60^0=90^0\)

nên ΔBDA vuông tại D

=>AC\(\perp\)BD tại D