\(\frac{1}{2}.\frac{2}{3}=\frac{1}{3}\)

\(\frac{66}{5}:\frac{3}{5}=\frac{66}{5}.\frac{5}{3}=22\)

\(\frac{1}{2}\cdot\frac{2}{3}=\frac{1.2}{2.3}=\frac{1.1}{1.3}=\frac{1}{3}\)

\(\frac{66}{5}:\frac{3}{5}=\frac{66}{5}\cdot\frac{5}{3}=\frac{66.5}{5.3}=\frac{22.1}{1.1}=22\)

\(\frac{1}{2}\)\(\times\)\(\frac{2}{3}\)\(=\)\(\frac{1\times2}{2\times3}\)\(=\)\(\frac{1}{3}\)

\(\frac{66}{5}\)\(\times\)\(\frac{3}{5}\)\(=\)\(\frac{66\times3}{5\times5}\)\(=\)\(\frac{198}{25}\)

\(-66\cdot\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right]+124\cdot\left[-37\right]+63\cdot\left[-1+24\right]\)

\(=-66\cdot\left[\frac{1}{6}+\frac{1}{11}\right]+\left[-4588\right]+63\cdot23\)

\(=-66\cdot\frac{17}{66}+\left[-4588\right]+1449\)

\(=-17+\left[-4588\right]+1449=-4605+1449=-3156\)

Các bạn ủng hộ cho mình nhé

C=-66.(1/2-1/3+1/11)+124.(-37)+63.(-124)

C=-66.17/66-124.(37+63)

C=-17-124.100

C=-12417

\(x:z=\frac{2}{3}:\frac{1}{2}=\frac{4}{3}\Rightarrow x=\frac{4}{3}.z\)

\(z:y=1:\frac{4}{7}=\frac{7}{4}\Rightarrow z=y.\frac{7}{4}\)

\(\Rightarrow y+z=y+y.\frac{7}{4}=66\)

\(y.\frac{11}{4}=66\Rightarrow y=24\)

\(\Rightarrow z=24.\frac{7}{4}=42\)

\(\Rightarrow x=42.\frac{4}{3}=56\)

Ta có ; \(\frac{x+1}{65}+\frac{x+2}{66}=\frac{x+3}{67}+\frac{x+4}{68}\)

\(\Leftrightarrow\left(\frac{x+1}{65}-1\right)+\left(\frac{x+2}{66}-1\right)=\left(\frac{x+3}{67}-1\right)+\left(\frac{x+4}{68}-1\right)\)

\(\Leftrightarrow\frac{x-64}{65}+\frac{x-64}{66}=\frac{x-64}{67}+\frac{x-64}{68}\)

\(\Leftrightarrow\left(x-64\right)\left(\frac{1}{65}+\frac{1}{66}-\frac{1}{67}-\frac{1}{68}\right)=0\)

Vì \(\left(\frac{1}{65}+\frac{1}{66}-\frac{1}{67}-\frac{1}{68}\right)\ne0\)nên \(x-64=0\Leftrightarrow x=64\)

Vậy nghiệm của phương trình ; \(S=\left\{64\right\}\)

\(\text{Ta có ; }\)\(\frac{x+1}{65}+\frac{x+2}{66}=\frac{x+3}{67}+\frac{x+4}{68}\)

\(\Leftrightarrow\left(\frac{x+1}{65}-1\right)+\left(\frac{x+2}{66}-1\right)=\)\(\left(\frac{x+3}{67}-1\right)+\left(\frac{x+4}{68}-1\right)\)

\(\Leftrightarrow\frac{x-64}{65}+\frac{x-64}{66}=\frac{x-64}{67}+\frac{x-64}{68}\)

\(\Leftrightarrow\left(x-64\right)\left(\frac{1}{65}+\frac{1}{66}-\frac{1}{67}-\frac{1}{68}\right)=0\)

\(\text{Vì}\)\(\left(\frac{1}{65}+\frac{1}{66}-\frac{1}{67}-\frac{1}{68}\right)\ne0\)\(\text{nên}\)\(x-64=0\Leftrightarrow x=64\)

\(\text{Vậy nghiệm của phương trình ; }S=\left\{64\right\}\)

\(A=-66.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)

\(A=-66.\left(\frac{33}{66}-\frac{22}{66}+\frac{6}{66}\right)+\left(-124\right).\left(37+63\right)\)

\(A=-66.\frac{17}{66}+\left(-124\right).100\)

\(A=-17+\left(-12400\right)\)

A = -12417

Bài giải:

a, \(11.xx-66=4.x+11\)

\(11x^2-66=4.x+11\)

\(11x^2-66-4.x-11=0\)

\(11x^2-77-4x=0\)

\(11x^2-4x-77=0\)

\(x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4.11.\left(-77\right)}}{2.11}\)

\(x=\frac{4+\sqrt{16}+3388}{22}\)

\(x=\frac{4+\sqrt{3404}}{22}\)

\(x=\frac{4+2\sqrt{851}}{22}\)

\(x=\frac{2-\sqrt{851}}{11}\)

\(\Rightarrow\)Có hai trường hợp: \(x_1=\frac{2-\sqrt{851}}{11};x_2=\frac{2+\sqrt{851}}{11}\)

Tớ bận rồi, cậu coi câu trên đã nhé ! Tớ xin lỗi, khi nào tớ sẽ làm tiếp =)) 

dấu trừ đầu tiên các bạn thay thành số 4 hộ mik nhé