10 x 5 + 5 x 10

c1 : 10 x (5 + 5)

        = 10 x 10

        = 100

c2 : (10 + 10) x 5 

       = 20 x 5

       = 100

50 + 10 + 90 + 50 + 30 + 60 + 10 + 70 + 50 + 50

c1 : (50 + 50) + (10 + 90) + (30 + 60 + 10) + (50 + 50) + 70

      =  100 + 100 + 100 + 100 + 70

      =   100 x 4 + 70

      = 400 + 70

      = 470

c2: 50 x 4 + (10 + 90) + (70 + 30) + (60 + 10)

     = 200 + 100 + 100 + 70

    = 200 + 100 x 2 + 70

    = 200 + 200 + 70

    = 200 x 2 + 70

     = 400 + 70 = 470

\(\frac{x+10}{90}+\frac{x+20}{80}+\frac{x+30}{70}+\frac{x+40}{60}+\frac{x+50}{50}=-5\)

<=> \(\frac{x+10}{90}+1+\frac{x+20}{80}+1+\frac{x+30}{70}+1+\frac{x+40}{60}+1+\frac{x+50}{50}+1=0\)

<=> \(\frac{x+100}{90}+\frac{x+100}{80}+\frac{x+100}{70}+\frac{x+100}{60}+\frac{x+100}{50}=0\)

<=> \(\left(x+100\right)\left(\frac{1}{90}+\frac{1}{80}+\frac{1}{70}+\frac{1}{60}+\frac{1}{50}\right)=0\)

<=> x + 100 = 0 

<=> x = -100

Vậy x = -100

80 + 10 = 90     30 + 40 = 70     80 + 5 = 85

90 - 80 = 10     70 - 30 = 40     85 - 5 = 80

90 - 10 = 80     70 - 40 = 30     85 - 80 = 5

80+10=90                  30+40=70                         80+5=85

90-80=10                   70-30=40                          85-5=80

90-10=80                   70-40=30                          85-80=5

80 + 10 = 90   30 + 40 = 70  80 + 5 = 85                                                                                                

90 - 80 = 10    70 + 30 = 90   85 - 5 = 80

90 - 10 = 80    70 - 40 = 30     85 - 80  = 5

toán lớp 5

2 x X + 68 = 126

2 x X = 126 - 68

2 x X = 58

      x = 58 : 2

      x = 29

a) \(2\times x+68=126\)

\(\Rightarrow2\times x=58\)

\(\Rightarrow x=29\)

b) \(3\times x-24=90\)

\(\Rightarrow3\times x=123\)

\(\Rightarrow x=41\)

c) \(84-4\times x=16\)

\(\Rightarrow4\times x=84-16=68\)

\(\Rightarrow x=17\)

d) \(\left(5\times x+10\right)+35=70\)

\(\left(5\times x+10\right)=70-35\)

\(\Rightarrow5\times x+10=35\)

\(\Rightarrow5\times x=25\)

\(\Leftrightarrow x=5\)

e) Tương tự d

f) Thiếu kết quả của đề

g)  \(45-\left(3\times x-9\right)=15\)

\(\Rightarrow(3\times x-9)=45-15\)

\(\Rightarrow3\times x-9=30\)

\(\Rightarrow3\times x=39\)

\(\Rightarrow x=13\)

-- Các câu còn lại tương tự câu g

a) 90 – ( 30 – 20) = 90 – 10

                   = 80

90 – 30 – 20 = 60 - 20

                  = 40

b) 100 – (60 + 10) = 100 – 70

                   = 30

100 - 60 + 10 = 40 + 10

                   = 50

c) 135 – (30 + 5) = 135 – 35

                   = 100

135 – 30 – 5 = 105 – 5

                   = 100

d) 70 + (40 – 10) = 70 + 30

                   = 100

70 + 40 – 10 = 110 -10

                   = 100

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b) Ta có :

\(\left(x-8\right)^2\ge0;\forall x\)

\(\Rightarrow\left(x-8\right)^2+2019\ge2019;\forall x\)

Hay\(B\ge2019;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-8\right)^2=0\)

                         \(\Leftrightarrow x=8\)

Vậy MIN B=2019 \(\Leftrightarrow x=8\)

c) Vì \(\hept{\begin{cases}-|20-x|\le0;\forall x\\-|70+y|\le0;\forall y\end{cases}}\)

\(\Rightarrow-|20-x|-|70+y|\le0;\forall x,y\)

\(\Rightarrow90-|20-x|-|70+y|\le90-0;\forall x,y\)

Hay \(C\le90;\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}|20-x|=0\\|70+y|=0\end{cases}}\)

                       \(\Leftrightarrow\hept{\begin{cases}x=20\\y=-70\end{cases}}\)

Vậy MAX C=90 \(\Leftrightarrow\hept{\begin{cases}x=20\\y=-70\end{cases}}\)

A=|2-2|+|y+5|-10

 \(=\left|y+5\right|-10\ge-10\forall x\in R\)

Dấu "=" xảy ra <=> \(\left|y+5\right|=0\Leftrightarrow y+5=0\Leftrightarrow y=-5\) 

Vậy A_min =-10 tại y=-5

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