Ta có ^yBC = 180 -^B và ^zCB = 180-^C

Xét tam giác BOC có

^OBC = ^yBC/2 = (180-^B)/2

^OCB = ^zCB/2 = (180-^C)/2

^BOC = 180-(^OBC + ^OCB)=180-(180-^B)/2 - (180-^C)/2 = (^B + ^C)/2 (1)

Xét tg ABC có

^xAC = ^B+^C ( góc ngoài của 1 tam giác bằng tổng hai góc trong không kề với nó)

=> (^B+^C)/2 = ^xAC/2 (2)

Từ (1) và (2) => ^BOC = ^xAC/2 mà ^xAC là góc ngoài ở đỉnh A (dpcm)

a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)

\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)

b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)

c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)

\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)

\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)

e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)

f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)

Đề đâu bạn???? ko hỏi bt đường nào mà làm

Chúc bạn học tốt

tìm b bạn ạ

4(x+3)-2(7-3x)=-3

<=> 4x + 12 - 14 + 6x = -3

<=> 4x + 6x = -3 - 12 + 14

<=> 10x = -1

<=> x = \(-\frac{1}{10}\)

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

\(4x+12-14+6x=-3\) 

\(10x-2=-3\) 

\(10x=-3+2\) 

\(10x=1\) 

\(x=1:10\) 

\(x=\frac{1}{10}\)

                                                       Bài giải

\(A=2\left|3x-2\right|-1\ge-1\)

Dấu " = " xảy ra khi \(2\left|3x-2\right|-1=-1\text{ }\Rightarrow\text{ }2\left|3x-2\right|=0\text{ }\Rightarrow\text{ }3x-2=0\text{ }\Rightarrow\text{ }x=\frac{2}{3}\)

Vậy \(Min_A=-1\text{ khi }x=\frac{2}{3}\)

1) Góc \(\widehat{BCx}\) kề bù \(\widehat{BCA}\)  => \(\widehat{BCx}+\widehat{BCA}=180\Rightarrow\widehat{BCx}=180-40=140\)

Vì Cy là phân giác \(\widehat{BCx}\)nên \(\widehat{BCy}=\frac{1}{2}\widehat{BCx}=70\Rightarrow\widehat{BCy}=\widehat{ABC}\)ở vị trí so le trong => Cy // AB

2) Xét tam giác ABC: \(\widehat{BCA}+\widehat{ABC}+\widehat{BAC}=180\Rightarrow\widehat{BAC}=180-70-40=70\)

3) Có \(CH\perp AB\)mà \(AB//Cy\)nên \(CH\perp Cy\)

\(\frac{6^x}{2^{2000}}=3^y\Leftrightarrow\frac{2^x.3^x}{3^y}=2^{2000}\Leftrightarrow2^x.3^{x-y}=2^{2000}\)

Vì \(2^{2000}\)không chia hết cho 3 nên \(3^{x-y}=1\Leftrightarrow x-y=0\Leftrightarrow x=y\) 

Khi đó \(2^x=2^{2000}\Leftrightarrow x=2000\Rightarrow y=2000\)

\(\frac{4}{5}:\left(\frac{7}{2}+\frac{1}{4^{ }}\right)^2\)         <=>\(0,8:\left(3,5+0.25\right)^2\)          

                                             <=>\(0,8:\frac{225}{16}\)

                                              <=>\(\frac{64}{1125}\)

\(2^{x+1}.3^y=12^x\)

\(\Leftrightarrow2^{x+1}.3^y=\left(2^2.3\right)^x\)

\(\Leftrightarrow2^{x+1}.3^y=2^{2x}.3^x\)

\(\Leftrightarrow\hept{\begin{cases}2^{x+1}=2^{2x}\\3^y=3^x\end{cases}}\)

\(\Leftrightarrow x=y=1\)