a: góc yOz=180-60=120 độ

góc zOm=góc yOm=120/2=60 độ

b: góc xOn=góc zOm=60 độ

=>góc xOn=góc xOy

=>Ox là phân giác của góc yOn

GTNN (A)=3178+2017 khi x=0 ko co GTLN

GTLN(b)=2017 khi x=-3 va y=5 khong co GTNN

GTNN(c)=2018 khi x=-1 va y=5 khong co GTLN

neu can giai thich thi h

ko thi thoi 

em cũng muốn làm phước giúp chị lắm chứ nhưng em mới ở lớp 6 thui

giải thích ik kich ròi

\(\frac{15}{8}\)

chỉ cho mik cách giải chi tiết rõ ràng hơn đc ko bạn????

c, \(C=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

<=> \(C=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)

<=> \(C=-5\sqrt{3}:\sqrt{3}=-5\)

e. \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

\(=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(=6+2\sqrt{9-5}\)

\(=6+4=10\)

b. \(\left(\sqrt{3}+2\right)^2-\sqrt{75}\)

\(=3+4\sqrt{3}+4-5\sqrt{3}\)

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

d. \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)

\(=\left(1+\sqrt{3}\right)^2-2\)

\(=1+2\sqrt{3}+3-2\)

\(=2+2\sqrt{3}\)

f. \(\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)

\(=\sqrt{3}+2-2+\sqrt{3}\)

\(=2\sqrt{3}\)

c: Ta có: \(C=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

\(=\left(2\sqrt{3}-5\cdot3\sqrt{3}+4\cdot2\sqrt{3}\right):\sqrt{3}\)

\(=2-15+8=-5\)

d: Ta có: \(D=\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

\(=3-\sqrt{5}+3+\sqrt{5}+2\cdot\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(=6+2\cdot2=10\)

\(\Leftrightarrow7\left(2x-1\right)-15x=-3x\)

=>14x-7-15x+3x=0

=>2x=7

hay x=7/2(nhận)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{7}{3x}-\dfrac{5}{2x-1}=\dfrac{1}{1-2x}\)

\(\Rightarrow\dfrac{7}{3x}+\dfrac{5}{1-2x}=\dfrac{1}{1-2x}\)

\(\Rightarrow\dfrac{7}{3x}=\dfrac{1}{1-2x}-\dfrac{5}{1-2x}\)

\(\Rightarrow\dfrac{7}{3x}=\dfrac{-4}{1-2x}\)

\(\Rightarrow-4.3x=7\left(1-2x\right)\)

\(\Rightarrow-12x=7-14x\)

\(\Rightarrow-12x+14x=7\)

\(\Rightarrow2x=7\)

\(\Rightarrow x=\dfrac{7}{2}\left(tm\right)\)

PT⇔7/3x=5/2x-1 - 1/2x-1

     <=>7/3x=4/2x-1

     <=>7(2x-1)=4.3x

      <=>14x-7=12x

       <=>14x-12x=7

       <=>2x=7

       <=>x=7/2

Vậy tập nghiệm của pt là S={7/2}

 (26 + 53 ) +(42 -26-55-53)

=79 + ( 16 -55 - 53 )

=79 + (-39-53)

=79 -39 - 53

=40 - 53 = -13

( 26 + 53 ) + ( 42 -26 - 55 - 53 )

= 79 + ( 16 - 55 - 53 )

= 79 + ( -39 - 53 )

= 79 + ( -92 )

= -13.

\(\left(26+53\right)+\left(42-26-55-53\right)\)

\(=26+53+42-26-55-53\)

\(=\left(26-26\right)+\left(53-53\right)+\left(42-55\right)\)

\(=0+0+\left(-13\right)\)

\(=-13\)

?

a: góc xOt=góc yOt=100/2=50 độ

b: góc xOt'=180 độ-góc xOt=130 độ

Câu 3:

\(BTKL:m_{CaCO_3}=m_{CO_2}+m_{CaO}\\ m_{CO_2}=m_{CaCO_3}-m_{CaO}=1,32(tấn)\)

Câu 4:

\(a,FeCl_3+3NaOH\to Fe(OH)_3\downarrow+3NaCl\\ b,2Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O\\ c,4NH_2+3O_2\xrightarrow{t^o}2N_2+6H_2O\\ d,Na_2CO_3+CaCl_2\to CaCO_3\downarrow+2NaCl\\ e,4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ f,Fe_2(SO_4)_3+6KOH\to 2Fe(OH)_3\downarrow+3K_2SO_4\)

Câu 5:

\(BTKL:m_{O_2}+m_{Mg}=m_{MgO}\\ \Rightarrow m_{O_2}=m_{MgO}-m_{Mg}=10-6=4(g)\)