\(\widehat{BAH}=180^o-\widehat{BAC}=180^o-120^o=60^o\)

Xét tg vuông ABH có

\(\widehat{ABH}=90^o-\widehat{BAH}=90^o-60^o=30^o\)

\(\Rightarrow AH=\frac{AB}{2}\) (Trong tg vuông cạnh đối diện góc 30 bằng nửa cạnh huyền) \(\Rightarrow2AH=AB\left(dpcm\right)\)

1. Ư(3.17)= {1,3,17,51}

2.Ư(3⁵)= {1,3,3²,3³ ,3⁴,3⁵}

3.Ư(5.7²)= {1,5,7,35,45,245}

4. a=38=2.19 nên Ư(38)= {1,2,19,38}

5.a=98=2.7.7 nên Ư{98)= {1,2,7,14,49,98}

6.a=78=2.3.13  nên Ư(78)= {1,2,3,6,13,26,39,78}

7.a=138=2.3.23 nên Ư(138)= {1,2,3,6,46,69,128}

BT1: 

ĐK: \(a>0,a\ne1\).

\(A=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\left(\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\frac{a+\sqrt{a}-2-\left(a-\sqrt{a}-2\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\frac{2}{a-1}\)

ĐK: \(a\ge0,a\ne1\).

\(B=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)

\(B=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(B=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)

BT2: 

a) ĐK: \(a\ge0,a\ne1\).

\(A=\left(\frac{\sqrt{a}-2}{a-1}-\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}\right).\frac{\left(1-a\right)^2}{2}\)

\(A=\left(\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}\right).\frac{\left(1-a\right)^2}{2}\)

\(A=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\left(1-a\right)^2}{2}\)

\(A=\frac{a-\sqrt{a}-2-\left(a+\sqrt{a}-2\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\left(1-a\right)^2}{2}\)

\(A=\frac{-2\sqrt{a}}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\left(\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\right)^2}{2}\)

\(A=-\sqrt{a}\left(\sqrt{a}-1\right)=\sqrt{a}-a\)

b) \(A=\sqrt{a}-a=\sqrt{a}\left(1-\sqrt{a}\right)>0\)

(Vì \(0< a< 1\Rightarrow1-\sqrt{a}>0\))

c) \(A=\sqrt{a}-a=\frac{1}{4}-\left(\frac{1}{4}-\sqrt{a}+a\right)=\frac{1}{4}-\left(\sqrt{a}-\frac{1}{2}\right)^2\le\frac{1}{4}\)

Dấu \(=\)xảy ra khi \(\sqrt{a}-\frac{1}{2}=0\Leftrightarrow a=\frac{1}{4}\). Vậy GTLN của \(A\) là \(\frac{1}{4}\).

a)xet tg abc ab^2+ac^2=Bc^2

thay ab=6cm(gt),ac=8cm(gt) ta co

6^2+8^2=BC^2

36+64=BC^2

100=BC^2

10^2=bc^2 suy ra bc=10cm

Em làm bài này lần đầu thì máy lag mất tab, lần thứ 2 thì bảo trì._. Cứ chuẩn bị ấn gửi là toang, đây là lần 3 đấy ạ)':

Ta có:

Vì E chia (x-1) dư 15 nên E có dạng:

\(E=\left(x-1\right)\cdot P\left(x\right)+15\)

Tại x = 1 ta có: \(2.1^3-3.1^2+a.1+b=\left(1-1\right).P\left(x\right)+15\)

\(\Leftrightarrow a+b=16\)

Vì E chia (x+2) dư -18 nên E có dạng:

\(E=\left(x+2\right).Q\left(x\right)-18\)

Tại x = -2 ta có: \(2.\left(-2\right)^3-3.\left(-2\right)^2+a.\left(-2\right)+b=\left(-2+2\right).Q\left(x\right)-18\)

\(\Leftrightarrow-2a+b=10\)

Trừ vế đầu cho vế sau ta được: \(3a=6\Rightarrow a=2\Rightarrow b=14\)

E = 2x³ - 3x² + ax + b 

+) chia x - 1 dư 15

=> 2x³ - 3x² + ax + b - 15 chia hết cho x - 1

Theo định lí Bézoute ta có : E chia hết cho x - 1 <=> E(1) = 0

=> 2.1³ - 3.1² + a.1 + b - 15 = 0

=> a + b - 16 = 0

=> a + b = 16 (1)

+) chia x + 2 dư -18

=> 2x³ - 3x² + ax + b + 18 chia hết cho x + 2

Theo định lí Bézoute ta có : E chia hết cho x + 2 <=> E(-2) = 0

=> 2.(-2)³ - 3.(-2)² + a.(-2) + b + 18 = 0

=> b - 2a - 10 = 0

=> b - 2a = 10 (2)

Từ (1) và (2) => \(\hept{\begin{cases}a+b=16\\b-2a=10\end{cases}}\Rightarrow\hept{\begin{cases}a=2\\b=14\end{cases}}\)

Vậy a = 2 ; b = 14

Trong 1 phép chia có dư số dư lớn nhất = số chia -1

=> số dư = 68-1=67

Số bị chia là

68x92+67=6323

Câu 1: 

a) \(\left(x-\frac{1}{2}\right)^3=-27\) \(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow x-\frac{1}{2}=-3\)\(\Leftrightarrow x=-\frac{5}{2}\)

Vậy \(x=-\frac{5}{2}\)

b( \(2x^2+x=0\)\(\Leftrightarrow x\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

Vậy \(x=0\)hoặc \(x=-\frac{1}{2}\)

c) \(5^{x+2}=625\)\(\Leftrightarrow5^{x+2}=5^4\)

\(\Leftrightarrow x+2=4\)\(\Leftrightarrow x=2\)

Vậy \(x=2\)

d) \(\frac{7^x+7^{x+1}+7^{x+2}+7^{x+3}}{2^2.5^2.7^2}=2^2\)

\(\Leftrightarrow7^x+7^{x+1}+7^{x+2}+7^{x+3}=2^2.2^2.5^2.7^2\)

\(\Leftrightarrow7^x+7^x.7+7^x.7^2+7^x.7^3=\left(2.2.5.7\right)^2\)

\(\Leftrightarrow7^x+7^x.7+7^x.49+7^x.343=140^2\)

\(\Leftrightarrow7^x.\left(1+7+49+343\right)=19600\)

\(\Leftrightarrow7^x.400=19600\)

\(\Leftrightarrow7^x=49=7^2\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

Câu 2:

a) \(C=1+4+4^2+4^3+.......+4^{48}\)

\(\Rightarrow4C=4+4^2+4^3+4^4+........+4^{49}\)

\(\Rightarrow4C-C=4^{49}-1\)

\(\Rightarrow3C=4^{49}-1\)

\(\Rightarrow C=\frac{4^{49}-1}{3}\)

b) Ta có: \(3C+1=4^{49}-1+1=4^{49}=4^{7.7}=\left(4^7\right)^7⋮4^7\)( đpcm )

c) \(C=1+4+4^2+4^3+........+4^{48}\)

\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+........+\left(4^{46}+4^{47}+4^{48}\right)\)

\(=\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+........+4^{46}.\left(1+4+4^2\right)\)

\(=\left(1+4+4^2\right).\left(1+4^3+....+4^{46}\right)\)

\(=\left(1+4+16\right).\left(1+4^3+........+4^{46}\right)\)

\(=21.\left(1+4^3+.....+4^{46}\right)⋮21\)