a) Ta có : \(\widehat{xOy}+\widehat{O_2}=150^o\left(gt\right)\)

và \(\widehat{xOy}-\widehat{O_2}=90^o\)

\(\Rightarrow\widehat{xOy}=\left(150^o+90^o\right)\div2=120^o\)

\(\Rightarrow\widehat{O_2}=150^o-120^o=30^o\)

b) Ta có : \(\widehat{O_1}+\widehat{O_2}=\widehat{xOy}\)

                             \(\widehat{O_1}=\widehat{xOy}-\widehat{O_2}\)

                             \(\widehat{O_1}=90^o\)

Ta có : \(\widehat{O_2}+\widehat{O_3}=180^o\)( kề bù )

           \(30^o+\widehat{O_3}=180^o\)

                         \(\widehat{O_3}=180^o-30^o\)

                         \(\widehat{O_3}=150^o\)

mà \(\widehat{O_1}=90^o\left(cmt\right)\)

\(\Rightarrow\widehat{O_1}< \widehat{O_3}\)

=(-1346-0,25)-3,53-3,53:0,53053

=\(\frac{-134625}{100}-\frac{353}{100}-\frac{353}{100}\times\frac{100000}{53053}\)

\(=\frac{-134625}{100}-\frac{706}{100}\times\frac{100000}{53053}\)

\(=\frac{-135331}{100}\times\frac{100000}{53053}\)

\(=\frac{135331000}{53053}\)

Kiến thức 6,7 nó như vậy đấy.Tính vậy mệt lắm nhưng nó đúng.

1) \(B\left(x\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x^2-\frac{9}{16}\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x-\frac{3}{4}\right)\left(x+\frac{3}{4}\right)\left(x^2+3\right)=0\)

Mà \(x^2+3>0\left(\forall x\right)\)

=> 2x-5=0 hoặc x-3/4=0 hoặc x+3/4=0

=> x=5/2 hoặc x=3/4 hoặc x=-3/4

Vậy \(x\in\left\{-\frac{3}{4};\frac{3}{4};\frac{5}{2}\right\}\)

2) \(K\left(x\right)=0\)

\(\Leftrightarrow2x^2-x-10=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)

\(B\left(x\right)=\left(2x-5\right)\left(x^2-\frac{9}{16}\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow x=\frac{5}{2};x=\frac{3}{4};x=-\frac{3}{4}\)

\(K\left(x\right)=2x^2-x-10=0\Leftrightarrow\left(x+2\right)\left(2x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}\)

\(=\frac{19}{20}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\) 

\(=\frac{1}{1}-\frac{1}{20}\)             

\(=\frac{19}{20}\)            

a) \(K=\frac{2^{11}.9^2}{3^5.16^2}=\frac{2^{11}.3^4}{2^8.3^5}=\frac{2^3}{3}=\frac{8}{3}\)

b) \(N=\frac{9^3.27^2}{6^2.3^{10}}=\frac{3^6.3^6}{3^2.2^2.3^{10}}=\frac{1}{2^2}=\frac{1}{4}\)

c) \(P=\frac{27^{15}.5^3.8^4}{25^2.81^{11}.2^{11}}=\frac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\frac{2.3}{5}=\frac{6}{5}\)

a) \(\left(\frac{3}{5}-1\frac{3}{4}\right)\div2\frac{3}{10}=\left(\frac{3}{5}-\frac{7}{4}\right)\div\frac{23}{10}=\frac{-23}{20}.\frac{10}{23}=-\frac{1}{2}\)

b) \(\left(\frac{-2}{3}+\frac{3}{4}\right)^2.\frac{12}{5}-\frac{11}{5}=\left(\frac{1}{12}\right)^2.\frac{12}{5}-\frac{11}{5}=\frac{1}{60}-\frac{11}{5}=-\frac{33}{15}\)

c) \(\frac{-7}{8}\div\frac{21}{16}-\frac{5}{3}\left(\frac{1}{3}-\frac{7}{10}\right)=\frac{-7}{8}\times\frac{16}{21}-\frac{5}{3}.-\frac{11}{30}=-\frac{2}{3}-\frac{11}{18}=-\frac{23}{18}\)

\(\frac{x-1}{13}+\frac{x-2}{12}+\frac{x-3}{11}-3=0\)

\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)+\left(\frac{x-2}{12}-1\right)+\left(\frac{x-3}{11}-1\right)\)

\(\Leftrightarrow\frac{x-14}{13}+\frac{x-14}{12}+\frac{x-14}{11}=0\)

\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)=0\). Vì \(\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)>0\)

\(\Leftrightarrow x-14=0\Rightarrow x=0+14=14\). Vậy \(x=14\)

Bài làm:

Ta có: \(\frac{x-1}{13}+\frac{x-2}{12}+\frac{x-3}{11}-3=0\)

\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)+\left(\frac{x-2}{12}-1\right)+\left(\frac{x-3}{11}-1\right)=0\)

\(\Leftrightarrow\frac{x-14}{13}+\frac{x-14}{12}+\frac{x-14}{11}=0\)

\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)=0\)

\(\Leftrightarrow x-14=0\)

\(\Rightarrow x=14\)

Ta có: \(\frac{x}{3}=\frac{y}{7}\)\(\Rightarrow\)\(x=\frac{3y}{7}\)

Ta lại có: \(\frac{2}{x}=\frac{y}{-3}\)

         \(\Leftrightarrow xy=-6\)

         \(\Leftrightarrow\frac{3y}{7}.y=-6\)

         \(\Leftrightarrow y^2=-6.\frac{7}{3}\)

         \(\Leftrightarrow y^2=-14\)

Vì \(\hept{\begin{cases}y^2\ge0\forall y\\-14< 0\end{cases}}\)mà \(y^2=-14\)

      \(\Rightarrow\)\(y\in\varnothing\)\(\Rightarrow\)\(x\in\varnothing\)

Vậy x,y không có giá trị