=>  3x + 33 = 150                                                                                      => 3x+33−33=150−33                                                                            => 3x = 150- 33                                                                                         => 3x = 117                                                                                             => 3x : 3 = 117 : 3                                                                                    => x = 39

x+11=150:3

X+11=50

X=50-11

X=39

Tick mình nha 

=> 3x + 33 = 150

=> 3x + 33 - 33 = 150 - 33

=> 3x = 150 - 33

=> 3x = 117

=> 3x : 3 = 117 : 3

=> x = 39

 Đây bạn nhé chúc bạn học tốt !

=>x-3 và y-5 thuộc Ư(11)={1;11;-1;-11}

Ta có bảng kết quả:

Vậy (x;y) thuộc {(4;16);(14;6);(2;-6);(-8;4)}

(x - 3).(y - 5) = 11

=> Ta có bảng sau:

Vậy các cặp (x;y) thỏa là: (-8;4); (2;-6); (4; 16); (14;6).

\(\frac{0.75-0.6+\frac{3}{7}+\frac{3}{13}}{2.75-2.2+\frac{11}{7}+\frac{11}{13}}\)

\(=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)

\(=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(=\frac{3}{11}\)

\(-\dfrac{10}{13}+\dfrac{5}{7}-\dfrac{3}{13}+\dfrac{2}{7}-\dfrac{11}{20}\)

\(=\left(-\dfrac{10}{13}-\dfrac{3}{13}\right)+\left(\dfrac{5}{7}+\dfrac{2}{7}\right)-\dfrac{11}{20}\)

\(=-1+1-\dfrac{11}{20}\)

\(=-\dfrac{11}{20}\)

    \(\frac{7}{18}.\frac{8}{11}-\frac{7}{9}.\frac{8}{11}-11:\frac{9}{2}\)

\(=\frac{8}{11}.\left(\frac{7}{18}-\frac{7}{9}\right)-11:\frac{9}{2}\)

\(=\frac{8}{11}.\left(-\frac{7}{18}\right)-11:\frac{9}{2}\)

\(=-\frac{28}{99}-\frac{22}{9}\)

\(=-\frac{30}{11}\)

Tk mk nha!!!!!!

\(\sqrt{7-4\sqrt{3}}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)

\(\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.2+2^2}=\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|=\sqrt{7}-2\)

\(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)

\(\sqrt{9+4\sqrt{5}}=\sqrt{5}+2\)

\(\sqrt{11-4\sqrt{7}}=\sqrt{7}-2\)

\(\dfrac{5}{-6}=\dfrac{-55}{66};\dfrac{-10}{11}=\dfrac{-60}{66}\Rightarrow\dfrac{-55}{66}>\dfrac{-60}{66}\Rightarrow\dfrac{5}{-6}>\dfrac{-10}{11}\\ \dfrac{-3}{20}=\dfrac{-45}{300};\dfrac{2}{-15}=\dfrac{-40}{300}\Rightarrow\dfrac{-45}{300}< \dfrac{-40}{300}\Rightarrow\dfrac{-3}{20}< \dfrac{2}{-15}\\ -0,305>-0,36\)

\(^\circ\) \(\dfrac{5}{-6}\) và \(\dfrac{-10}{11}\)

Ta có \(:\)

\(\dfrac{5}{-6} = \dfrac{ 5 . 11 }{ -6 . 11 } = \dfrac{ 55 }{ -66} \)

\(\dfrac{-10}{11} = \dfrac{-10 . ( -6 )}{11.(-6)} = \dfrac{60}{-66}\)

Do \(55 < 60\)

\(=> \dfrac{55}{-66} > \dfrac{60}{-66}\)

Vậy \(\dfrac{55}{-66} > \dfrac{60}{-66}\)

\(\dfrac{-3}{20}\) và \(\dfrac{2}{-15}\)

Ta có :

\(\dfrac{-3}{20} = \dfrac{-3.3}{20.3}=\dfrac{-9}{60}\)

\(\dfrac{2}{-15} = \dfrac{2.4}{-15.4}= \dfrac{8}{-60}= \dfrac{-8}{60}\)

Do \(-9 <-8\)

\(=> \dfrac{-3}{20} < \dfrac{-2}{15}\)

Vậy \( \dfrac{-3}{20} < \dfrac{-2}{15}\)

:v

........cạn lời

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